Exercise 0.1: Norms, dot products and angles: ten direct calculations

For each line, compute the requested quantity from the components, in an orthonormal coordinate system of space. Give exact values when they are integers or fractions, irrational norms and components to three decimals (the distance in c) to two decimals), and angles to the tenth of a degree.

Four formulas are enough: $\|\vec{u}\|=\sqrt{a^{2}+b^{2}+c^{2}}$ , $\vec{u}\cdot\vec{v}=aa'+bb'+cc'$ , $\cos\theta=\frac{\vec{u}\cdot\vec{v}}{\|\vec{u}\|\,\|\vec{v}\|}$ , and the projection of $\vec{u}$ onto $\vec{v}$ , scalar $\frac{\vec{u}\cdot\vec{v}}{\|\vec{v}\|}$ or vector $\frac{\vec{u}\cdot\vec{v}}{\|\vec{v}\|^{2}}\,\vec{v}$ . Always compute the dot product and the norms first, divide next, and take the inverse cosine only at the very end, with the calculator in degrees.

a) The components of the vector $\vec{AB}$ , for $A(2, -3, 4)$ and $B(-1, 5, 6)$ .
b) The norm of $\vec{u}=(4, -7, 4)$ .
c) The distance between the points $P(-1, 3, 2)$ and $Q(2, -1, 7)$ , as an exact value, then to two decimals.
d) The dot product of $\vec{u}=(3, -2, 5)$ and $\vec{v}=(-4, 1, 2)$ .
e) The value of $k$ for which $\vec{u}=(k, 3, -2)$ and $\vec{v}=(5, k, 6)$ are orthogonal.
f) The norm of $\vec{w}=(1, -2, 3)$ , then the third component of the unit vector in the same direction as $\vec{w}$ .
g) The angle between $\vec{u}=(1, 2, -2)$ and $\vec{v}=(3, 0, 4)$ , to the tenth of a degree.
h) The direction cosine $\cos\beta$ of the vector $\vec{v}=(1, -4, 8)$ , then the direction angle $\beta$ it makes with the $y$ -axis.
i) The scalar projection of $\vec{u}=(5, 1, -2)$ onto $\vec{v}=(2, -2, 1)$ , that is, the signed length of its shadow on the direction of $\vec{v}$ .
j) The first and third components of the vector projection of $\vec{u}=(3, -1, 4)$ onto $\vec{v}=(1, 1, 2)$ .

Type your answers, the page tells you right or wrong 0/15

a)

first component

second component

third component

b)

\|\vec{u}\|

c)

PQ

d)

\vec{u}\cdot\vec{v}

e)

k

f)

\|\vec{w}\|

third component

g)

\theta

in degrees

h)

\cos\beta

\beta

in degrees

i)

scalar projection

j)

first component

third component

Show the solution

Answers

a) $\vec{AB}=(-3, 8, 2)$
b) $\|\vec{u}\|=\sqrt{81}=9$
c) $PQ=\sqrt{50}=5\sqrt{2}\approx 7.07$
d) $\vec{u}\cdot\vec{v}=-4$
e) $8k-12=0$ ; $k=\frac{3}{2}$
f) $\|\vec{w}\|=\sqrt{14}\approx 3.742$ ; $\frac{3}{\sqrt{14}}\approx 0.802$
g) $\cos\theta=-\frac{1}{3}$ ; $\theta\approx 109.5^{\circ}$
h) $\cos\beta=-\frac{4}{9}\approx-0.444$ ; $\beta\approx 116.4^{\circ}$
i) $\frac{6}{3}=2$
j) $\frac{10}{6}(1, 1, 2)=\left(\frac{5}{3}, \frac{5}{3}, \frac{10}{3}\right)$

The move that decides everything: know what the formula returns BEFORE applying it. A norm and a dot product return a number, a vector projection returns a vector, an angle goes through a cosine. And in the two projections, the only difference is in the denominator: $\|\vec{v}\|$ for a length, $\|\vec{v}\|^{2}$ for a vector.

a)

$\vec{AB}=B-A=(-1-2, 5-(-3), 6-4)=(-3, 8, 2)$ . Pitfall: computing $A-B=(3, -8, -2)$ , which is the vector $\vec{BA}$ ; it is the head minus the tail.

b)

$\|\vec{u}\|=\sqrt{4^{2}+(-7)^{2}+4^{2}}=\sqrt{16+49+16}=\sqrt{81}=9$ . The square of a negative component is positive.

c)

$\vec{PQ}=(3, -4, 5)$ , so $PQ=\sqrt{9+16+25}=\sqrt{50}=5\sqrt{2}\approx 7.07$ .

d)

$\vec{u}\cdot\vec{v}=3\times(-4)+(-2)\times 1+5\times 2=-12-2+10=-4$ . A negative dot product signals an obtuse angle.

e)

$\vec{u}\cdot\vec{v}=5k+3k-12=8k-12$ , which is zero for $k=\frac{3}{2}$ . Pitfall: forgetting the $k$ in the second component of $\vec{v}$ , which gives $5k=12$ and $k=\frac{12}{5}$ .

f)

$\|\vec{w}\|=\sqrt{1+4+9}=\sqrt{14}\approx 3.742$ , and the unit vector is $\frac{1}{\sqrt{14}}(1, -2, 3)$ : its third component is $\frac{3}{\sqrt{14}}\approx 0.802$ . EVERY component is divided by the norm.

g)

$\vec{u}\cdot\vec{v}=3+0-8=-5$ , $\|\vec{u}\|=3$ and $\|\vec{v}\|=5$ , so $\cos\theta=-\frac{5}{15}=-\frac{1}{3}$ and $\theta=\arccos\left(-\frac{1}{3}\right)\approx 109.5^{\circ}$ . Pitfall: dropping the sign, which gives $70.5^{\circ}$ , the supplementary angle.

h)

$\|\vec{v}\|=\sqrt{1+16+64}=9$ and $\cos\beta=\frac{-4}{9}\approx-0.444$ , the second component over the norm, since $\beta$ is measured from the $y$ -axis. So $\beta=\arccos\left(-\frac{4}{9}\right)\approx 116.4^{\circ}$ . Pitfall: taking the third component, the one of the $z$ -axis.

i)

$\vec{u}\cdot\vec{v}=10-2-2=6$ and $\|\vec{v}\|=\sqrt{4+4+1}=3$ , so the scalar projection is $\frac{6}{3}=2$ . Pitfall: dividing by $\|\vec{u}\|=\sqrt{30}$ , which projects $\vec{v}$ onto $\vec{u}$ .

j)

$\vec{u}\cdot\vec{v}=3-1+8=10$ and $\|\vec{v}\|^{2}=1+1+4=6$ , so the projection is $\frac{10}{6}(1, 1, 2)=\left(\frac{5}{3}, \frac{5}{3}, \frac{10}{3}\right)$ : first component $\frac{5}{3}\approx 1.667$ , third $\frac{10}{3}\approx 3.333$ . Pitfall: dividing by $\|\vec{v}\|=\sqrt{6}$ instead of its square.

Check: a norm can be recomputed backwards (the unit vector of f) has norm $1$ ), a zero dot product is checked by substituting $k$ into both vectors, and the projection in j) must leave a remainder $\vec{u}-\vec{p}=\left(\frac{4}{3}, -\frac{8}{3}, \frac{2}{3}\right)$ orthogonal to $\vec{v}$ , which takes one line: $\frac{4}{3}-\frac{8}{3}+\frac{4}{3}=0$ .

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Exercise 0.2: Cross products and scalar triple products: eight direct calculations

For each line, compute the cross product, the area, the scalar triple product, the volume or the parameter requested, in an orthonormal coordinate system of space. Give exact values when they are integers, the area in c) to two decimals and the volume in g) to three decimals.

The cross product is computed with the symbolic determinant $\begin{vmatrix} \vec{i} & \vec{j} & \vec{k} \\ a & b & c \\ a' & b' & c' \end{vmatrix}$ , expanded along the first row, CHANGING the sign of the $\vec{j}$ term. Its norm is the area of the parallelogram built on the two vectors, half of it the area of the triangle. The scalar triple product $\vec{u}\cdot(\vec{v}\times\vec{w})$ is the determinant of the three vectors: its absolute value gives the volume of the parallelepiped, one sixth of it that of the tetrahedron, and it is zero exactly when the three vectors are coplanar.

a) The cross product $\vec{u}\times\vec{v}$ , for $\vec{u}=(1, 3, -2)$ and $\vec{v}=(2, -1, 4)$ .
b) The cross product $\vec{k}\times\vec{u}$ , for $\vec{k}=(0, 0, 1)$ and $\vec{u}=(4, -1, 5)$ .
c) The area of the parallelogram built on $\vec{u}=(2, 0, 1)$ and $\vec{v}=(1, 3, -1)$ , to two decimals.
d) The area of the triangle with vertices $A(0, 1, 2)$ , $B(2, 3, 2)$ and $C(1, 1, 4)$ .
e) The scalar triple product $\vec{u}\cdot(\vec{v}\times\vec{w})$ , for $\vec{u}=(1, 2, 0)$ , $\vec{v}=(0, 1, 3)$ and $\vec{w}=(2, 0, 1)$ .
f) The volume of the parallelepiped built on $\vec{u}=(2, 1, -1)$ , $\vec{v}=(1, -3, 2)$ and $\vec{w}=(-1, 2, 1)$ .
g) The volume of the tetrahedron with vertices $A(1, 0, 1)$ , $B(3, 1, 1)$ , $C(1, 2, 2)$ and $D(2, 1, 4)$ , to three decimals.
h) The value of $m$ for which $\vec{u}=(1, -1, 2)$ , $\vec{v}=(2, 1, m)$ and $\vec{w}=(0, 3, 1)$ are coplanar.

Type your answers, the page tells you right or wrong 0/12

a)

first component

second component

third component

b)

first component

second component

third component

c)

area

d)

area

e)

triple product

f)

volume

g)

volume

h)

m

Show the solution

Answers

a) $\vec{u}\times\vec{v}=(10, -8, -7)$
b) $\vec{k}\times\vec{u}=(1, 4, 0)$
c) $\|(-3, 3, 6)\|=3\sqrt{6}\approx 7.35$
d) $\frac{1}{2}\|(4, -4, -2)\|=3$
e) $\vec{u}\cdot(1, 6, -2)=13$
f) triple product $-16$ ; volume $16$
g) $\frac{11}{6}\approx 1.833$
h) $15-3m=0$ ; $m=5$

The move that decides everything: the MINUS sign in front of the $\vec{j}$ component of the symbolic determinant. Every other error in this list is caught by an orthogonality check; this one spreads silently into the area, the volume and the parameter, unless you check.

a)

$\vec{u}\times\vec{v}=\left(3\times 4-(-2)(-1), -\left(1\times 4-(-2)\times 2\right), 1\times(-1)-3\times 2\right)=(12-2, -8, -7)=(10, -8, -7)$ . Check: $\vec{u}\cdot(10, -8, -7)=10-24+14=0$ . Pitfall: forgetting the minus on the $\vec{j}$ component, which gives $8$ instead of $-8$ .

b)

$\vec{k}\times\vec{u}=\left(0\times 5-1\times(-1), -(0\times 5-1\times 4), 0\times(-1)-0\times 4\right)=(1, 4, 0)$ . Pitfall: the order. $\vec{u}\times\vec{k}=(-1, -4, 0)$ , the opposite: the cross product is anticommutative.

c)

$\vec{u}\times\vec{v}=\left(0\times(-1)-1\times 3, -\left(2\times(-1)-1\times 1\right), 2\times 3-0\times 1\right)=(-3, 3, 6)$ , so the area is $\sqrt{9+9+36}=\sqrt{54}=3\sqrt{6}\approx 7.35$ .

d)

$\vec{AB}=(2, 2, 0)$ , $\vec{AC}=(1, 0, 2)$ , and $\vec{AB}\times\vec{AC}=(4-0, -(4-0), 0-2)=(4, -4, -2)$ , with norm $\sqrt{16+16+4}=6$ . The area of the triangle is $\frac{6}{2}=3$ . Pitfall: stopping at $6$ , the area of the parallelogram.

e)

$\vec{v}\times\vec{w}=(1\times 1-3\times 0, -(0\times 1-3\times 2), 0\times 0-1\times 2)=(1, 6, -2)$ , then $\vec{u}\cdot(1, 6, -2)=1+12+0=13$ . You can also expand the determinant of the three vectors along the first row: $1\times(1-0)-2\times(0-6)+0=13$ .

f)

$\vec{v}\times\vec{w}=\left(-3-4, -(1+2), 2-3\right)=(-7, -3, -1)$ , then $\vec{u}\cdot(\vec{v}\times\vec{w})=-14-3+1=-16$ . The volume is $|-16|=16$ . Pitfall: answering $-16$ ; the sign of the scalar triple product gives the orientation of the triple, not an amount of volume.

g)

Three edges from $A$ : $\vec{AB}=(2, 1, 0)$ , $\vec{AC}=(0, 2, 1)$ , $\vec{AD}=(1, 1, 3)$ . $\vec{AC}\times\vec{AD}=(6-1, -(0-1), 0-2)=(5, 1, -2)$ , then $\vec{AB}\cdot(5, 1, -2)=10+1+0=11$ . The volume is $\frac{11}{6}\approx 1.833$ . Pitfall: taking $\frac{11}{3}$ or $\frac{11}{2}$ ; the tetrahedron is one SIXTH of the parallelepiped.

h)

$\vec{v}\times\vec{w}=(1-3m, -(2-0), 6-0)=(1-3m, -2, 6)$ , then $\vec{u}\cdot(\vec{v}\times\vec{w})=1-3m+2+12=15-3m$ . The vectors are coplanar when this product is zero: $m=5$ .

Check: a cross product is orthogonal to both factors, so two zero dot products validate it, as in a). A scalar triple product can be recomputed by a cyclic permutation, $\vec{v}\cdot(\vec{w}\times\vec{u})$ must give the same value, and in h) check that $\vec{v}=(2, 1, 5)$ does make the determinant zero.

Exercise 1: Components, norm and operations in space

Consider the points $A(1, 2, -1)$ , $B(3, 5, 5)$ and $C(2, 6, 7)$ in space.

a) Find the components of the vectors $\vec{AB}$ and $\vec{AC}$ .
b) Compute $\|\vec{AB}\|$ and $\|\vec{AC}\|$ .
c) Find the components of the vector $2\vec{AB}-3\vec{AC}$ .
d) Find the unit vector with the same direction and the same sense as $\vec{AB}$ .

Show the solution

Answers

a) $\vec{AB}=(2, 3, 6)$ , $\vec{AC}=(1, 4, 8)$
b) 7 and 9
c) $(1, -6, -12)$
d) $\left(\frac27, \frac37, \frac67\right)$

a)

$\vec{AB}=(3-1, 5-2, 5-(-1))=(2, 3, 6)$ and $\vec{AC}=(2-1, 6-2, 7-(-1))=(1, 4, 8)$ .

b)

$\|\vec{AB}\|=\sqrt{2^{2}+3^{2}+6^{2}}=\sqrt{4+9+36}=\sqrt{49}=7$ and $\|\vec{AC}\|=\sqrt{1^{2}+4^{2}+8^{2}}=\sqrt{1+16+64}=\sqrt{81}=9$ .

c)

$2\vec{AB}=(4, 6, 12)$ and $3\vec{AC}=(3, 12, 24)$ , so $2\vec{AB}-3\vec{AC}=(4-3, 6-12, 12-24)=(1, -6, -12)$ .

d)

The unit vector is $\dfrac{\vec{AB}}{\|\vec{AB}\|}=\dfrac{1}{7}(2, 3, 6)=\left(\dfrac{2}{7}, \dfrac{3}{7}, \dfrac{6}{7}\right)$ . Check: $\sqrt{\dfrac{4}{49}+\dfrac{9}{49}+\dfrac{36}{49}}=\sqrt{\dfrac{49}{49}}=1$ .

Exercise 2: Dot product, angle and orthogonal projection

Consider the vectors $\vec{u}=(2, -1, 3)$ and $\vec{v}=(1, 4, -2)$ .

a) Compute $\vec{u}\cdot\vec{v}$ . Without computing the angle, say whether the angle between $\vec{u}$ and $\vec{v}$ is acute, right or obtuse, and justify.
b) Compute the measure of the angle $\theta$ between $\vec{u}$ and $\vec{v}$ , to the nearest tenth of a degree.
c) Find the value of $k$ for which $\vec{u}$ is orthogonal to $\vec{w}=(k, 2, 1)$ .
d) Find the vector projection of $\vec{u}$ onto $\vec{v}$ .

Show the solution

Answers

a) $-8$ : obtuse angle
b) $\theta\approx 117.8^\circ$
c) $k=-\frac12$
d) $-\frac{8}{21}(1, 4, -2)$

a)

$\vec{u}\cdot\vec{v}=(2)(1)+(-1)(4)+(3)(-2)=2-4-6=-8$ . The dot product is negative, so $\cos\theta<0$ and the angle is obtuse. No angle computation is needed: the sign is enough.

b)

$\|\vec{u}\|=\sqrt{4+1+9}=\sqrt{14}$ and $\|\vec{v}\|=\sqrt{1+16+4}=\sqrt{21}$ . So $\cos\theta=\dfrac{-8}{\sqrt{14}\sqrt{21}}=\dfrac{-8}{\sqrt{294}}\approx -0.4666$ , hence $\theta\approx 117.8^{\circ}$ . It is indeed an obtuse angle, consistent with a).

c)

$\vec{u}\cdot\vec{w}=2k+(-1)(2)+(3)(1)=2k+1$ . Orthogonality requires $2k+1=0$ , so $k=-\dfrac{1}{2}$ .

d)

$\vec{u_{v}}=\dfrac{\vec{u}\cdot\vec{v}}{\|\vec{v}\|^{2}}\,\vec{v}=\dfrac{-8}{21}(1, 4, -2)=\left(-\dfrac{8}{21}, -\dfrac{32}{21}, \dfrac{16}{21}\right)$ . The projection points in the direction opposite to $\vec{v}$ , which is expected since the angle is obtuse.

Exercise 3: Cross product, orthogonality and areas

Take again $\vec{u}=(2, -1, 3)$ and $\vec{v}=(1, 4, -2)$ .

a) Compute $\vec{u}\times\vec{v}$ .
b) Check by computation that $\vec{u}\times\vec{v}$ is orthogonal to $\vec{u}$ and to $\vec{v}$ .
c) Compute the area of the parallelogram built on $\vec{u}$ and $\vec{v}$ , then the area of the corresponding triangle.
d) Compute $\vec{v}\times\vec{u}$ and compare it with the result of a). What can you conclude about the commutativity of the cross product?

Show the solution

Answers

a) $(-10, 7, 9)$
b) Both dot products are zero
c) $\sqrt{230}\approx 15.17$ ; triangle $7.58$
d) $\vec v\times\vec u=-\vec u\times\vec v$ : anticommutative

a)

$\vec{u}\times\vec{v}=((-1)(-2)-(3)(4), (3)(1)-(2)(-2), (2)(4)-(-1)(1))=(2-12, 3+4, 8+1)=(-10, 7, 9)$ .

b)

$(\vec{u}\times\vec{v})\cdot\vec{u}=(-10)(2)+(7)(-1)+(9)(3)=-20-7+27=0$ and $(\vec{u}\times\vec{v})\cdot\vec{v}=(-10)(1)+(7)(4)+(9)(-2)=-10+28-18=0$ . Both dot products are zero, so $\vec{u}\times\vec{v}$ is indeed orthogonal to both vectors.

c)

$\|\vec{u}\times\vec{v}\|=\sqrt{(-10)^{2}+7^{2}+9^{2}}=\sqrt{100+49+81}=\sqrt{230}\approx 15.17$ . The area of the parallelogram is $\sqrt{230}\approx 15.17$ square units, and that of the triangle $\dfrac{\sqrt{230}}{2}\approx 7.58$ square units.

d)

$\vec{v}\times\vec{u}=(10, -7, -9)=-(\vec{u}\times\vec{v})$ . So the cross product is NOT commutative: swapping the order of the factors reverses the direction of the resulting vector. It is anticommutative. This is the costliest mistake of the chapter, because the norm (and therefore the area) stays the same: the error does not show on an area, but it ruins any oriented normal vector.

Exercise 4: Scalar triple product, volume and coplanarity

Take again $\vec{u}=(2, -1, 3)$ and $\vec{v}=(1, 4, -2)$ , and let $\vec{w}=(3, 2, 1)$ .

Recall that $\vec{u}\times\vec{v}=(-10, 7, 9)$ .

a) Compute the scalar triple product $(\vec{u}\times\vec{v})\cdot\vec{w}$ .
b) Are the vectors $\vec{u}$ , $\vec{v}$ and $\vec{w}$ coplanar? Justify.
c) Compute the volume of the parallelepiped built on $\vec{u}$ , $\vec{v}$ and $\vec{w}$ , then the volume of the corresponding tetrahedron.
d) Replace $\vec{w}$ by $\vec{w'}=(3, 2, m)$ . Find the value of $m$ for which $\vec{u}$ , $\vec{v}$ and $\vec{w'}$ become coplanar.

Show the solution

Answers

a) $-7$
b) Not coplanar
c) 7; tetrahedron $\frac76\approx 1.17$
d) $m=\frac{16}{9}$

a)

$(\vec{u}\times\vec{v})\cdot\vec{w}=(-10)(3)+(7)(2)+(9)(1)=-30+14+9=-7$ .

b)

The scalar triple product is $-7\neq 0$ , so the three vectors are NOT coplanar. Watch the sign: coplanarity is characterized by the product being zero, not by its sign. A negative scalar triple product only indicates the orientation of the triple.

c)

Volume of the parallelepiped: $V=|(\vec{u}\times\vec{v})\cdot\vec{w}|=|-7|=7$ cubic units. Volume of the tetrahedron: $V_{t}=\dfrac{7}{6}\approx 1.17$ cubic units.

d)

$(\vec{u}\times\vec{v})\cdot\vec{w'}=(-10)(3)+(7)(2)+(9)(m)=-30+14+9m=9m-16$ . Coplanarity requires $9m-16=0$ , so $m=\dfrac{16}{9}\approx 1.78$ .

Exercise 5: Direction angles and direction cosines

The direction angles of a non-zero vector $\vec{v}=(a, b, c)$ are the angles $\alpha$ , $\beta$ , $\gamma$ it makes with the $x$ -, $y$ - and $z$ -axes. Their cosines, called direction cosines, describe the orientation of the vector in space completely.

a) Express $\cos\alpha$ , $\cos\beta$ and $\cos\gamma$ in terms of the components of $\vec{v}$ , then compute them for $\vec{v}=(2, 3, 6)$ . Give the three direction angles to the tenth of a degree.
b) Prove that $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=1$ for every non-zero vector, and explain the link with the unit vector.
c) Find a vector that makes the SAME angle with all three axes, and compute this common angle.
d) Can a vector make a $45^{\circ}$ angle with each of the three axes? And if it makes $60^{\circ}$ with the $x$ -axis and $60^{\circ}$ with the $y$ -axis, what angle must it make with the $z$ -axis?

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Answers

a) $\frac27$ , $\frac37$ , $\frac67$ ; $73.4^\circ$ , $64.6^\circ$ , $31.0^\circ$
b) Components of the unit vector
c) $(1, 1, 1)$ : $54.7^\circ$
d) No; $\gamma=45^\circ$

a)

The cosine of the angle between $\vec{v}$ and the $x$ -axis is the cosine of the angle between $\vec{v}$ and the vector $\vec{i}=(1, 0, 0)$ : $\cos\alpha=\dfrac{\vec{v}\cdot\vec{i}}{\|\vec{v}\|\,\|\vec{i}\|}=\dfrac{a}{\|\vec{v}\|}$ . Likewise $\cos\beta=\dfrac{b}{\|\vec{v}\|}$ and $\cos\gamma=\dfrac{c}{\|\vec{v}\|}$ . For $\vec{v}=(2, 3, 6)$ , $\|\vec{v}\|=\sqrt{4+9+36}=\sqrt{49}=7$ , so $\cos\alpha=\dfrac{2}{7}$ , $\cos\beta=\dfrac{3}{7}$ , $\cos\gamma=\dfrac{6}{7}$ . The direction angles are $\alpha=\arccos\dfrac{2}{7}\approx 73.4^{\circ}$ , $\beta=\arccos\dfrac{3}{7}\approx 64.6^{\circ}$ and $\gamma=\arccos\dfrac{6}{7}\approx 31.0^{\circ}$ . The vector is closest to the $z$ -axis, which is consistent with its largest component.

b)

Adding the squares: $\cos^{2}\alpha+\cos^{2}\beta+\cos^{2}\gamma=\dfrac{a^{2}}{\|\vec{v}\|^{2}}+\dfrac{b^{2}}{\|\vec{v}\|^{2}}+\dfrac{c^{2}}{\|\vec{v}\|^{2}}=\dfrac{a^{2}+b^{2}+c^{2}}{\|\vec{v}\|^{2}}$ . By definition of the norm, $a^{2}+b^{2}+c^{2}=\|\vec{v}\|^{2}$ , so the sum is $1$ . The link with the unit vector is direct and illuminating: the unit vector of $\vec{v}$ is $\dfrac{\vec{v}}{\|\vec{v}\|}=\left(\dfrac{a}{\|\vec{v}\|}, \dfrac{b}{\|\vec{v}\|}, \dfrac{c}{\|\vec{v}\|}\right)=(\cos\alpha, \cos\beta, \cos\gamma)$ . The direction cosines ARE the components of the unit vector, and the identity is nothing but the fact that a unit vector has norm $1$ . Check on a): $\dfrac{4}{49}+\dfrac{9}{49}+\dfrac{36}{49}=\dfrac{49}{49}=1$ .

c)

Making the same angle with all three axes means $\cos\alpha=\cos\beta=\cos\gamma$ , so, by a), $\dfrac{a}{\|\vec{v}\|}=\dfrac{b}{\|\vec{v}\|}=\dfrac{c}{\|\vec{v}\|}$ , that is, $a=b=c$ . The vector $\vec{v}=(1, 1, 1)$ works, as do all its multiples: it is the diagonal of the cube. Its common angle satisfies $\cos^{2}\theta+\cos^{2}\theta+\cos^{2}\theta=1$ , that is $3\cos^{2}\theta=1$ , so $\cos\theta=\dfrac{1}{\sqrt{3}}$ and $\theta=\arccos\dfrac{1}{\sqrt{3}}\approx 54.7^{\circ}$ . This is the well-known angle between the diagonal of a cube and its edges.

d)

No: if all three angles were $45^{\circ}$ , the sum of the squared cosines would be $3\cos^{2}45^{\circ}=3\times\dfrac{1}{2}=\dfrac{3}{2}$ , which exceeds $1$ and contradicts the identity of b). So no vector can be at $45^{\circ}$ from all three axes at once: the identity imposes a constraint, and the three angles are not independent. On the other hand, $60^{\circ}$ with the $x$ -axis and $60^{\circ}$ with the $y$ -axis is possible: $\cos^{2}\gamma=1-\cos^{2}60^{\circ}-\cos^{2}60^{\circ}=1-\dfrac{1}{4}-\dfrac{1}{4}=\dfrac{1}{2}$ , so $\cos\gamma=\dfrac{1}{\sqrt{2}}$ and $\gamma=45^{\circ}$ . The third angle is completely determined (up to the sign of $\cos\gamma$ , which distinguishes the two possible orientations): the constraint of b) fixes it.

Exercise 6: Proofs: Lagrange's identity and the coplanarity criterion

This exercise asks for almost no numerical computation: you work with the vectors themselves. Let $\vec{u}$ and $\vec{v}$ be two non-zero vectors in space, and $\theta$ the angle between them.

a) Starting from $\vec{u}\cdot\vec{v}=\|\vec{u}\|\,\|\vec{v}\|\cos\theta$ and $\|\vec{u}\times\vec{v}\|=\|\vec{u}\|\,\|\vec{v}\|\sin\theta$ , show that $\|\vec{u}\times\vec{v}\|^{2}+(\vec{u}\cdot\vec{v})^{2}=\|\vec{u}\|^{2}\,\|\vec{v}\|^{2}$ . This equality is Lagrange's identity.
b) Check this identity on the vectors $\vec{u}=(2, -1, 3)$ and $\vec{v}=(1, 4, -2)$ of Part A.
c) Deduce from Lagrange's identity that if $\vec{u}$ and $\vec{v}$ are orthogonal, then $\|\vec{u}\times\vec{v}\|=\|\vec{u}\|\,\|\vec{v}\|$ . Interpret this result geometrically in terms of area.
d) Show that three vectors $\vec{u}$ , $\vec{v}$ and $\vec{w}$ in space are coplanar if and only if $\vec{u}\cdot(\vec{v}\times\vec{w})=0$ . Reason on the direction of $\vec{v}\times\vec{w}$ .

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Answers

a) $\sin^2\theta+\cos^2\theta=1$
b) $230+64=294=14\times 21$
c) Area of the rectangle $\|\vec u\|\,\|\vec v\|$
d) $\vec u\perp\vec v\times\vec w$ : coplanar

a)

Square both relations: $(\vec{u}\cdot\vec{v})^{2}=\|\vec{u}\|^{2}\|\vec{v}\|^{2}\cos^{2}\theta$ and $\|\vec{u}\times\vec{v}\|^{2}=\|\vec{u}\|^{2}\|\vec{v}\|^{2}\sin^{2}\theta$ . Adding: $\|\vec{u}\times\vec{v}\|^{2}+(\vec{u}\cdot\vec{v})^{2}=\|\vec{u}\|^{2}\|\vec{v}\|^{2}(\sin^{2}\theta+\cos^{2}\theta)=\|\vec{u}\|^{2}\|\vec{v}\|^{2}$ , since $\sin^{2}\theta+\cos^{2}\theta=1$ . The identity is proved.

b)

We have $\vec{u}\times\vec{v}=(-10, 7, 9)$ so $\|\vec{u}\times\vec{v}\|^{2}=230$ , and $\vec{u}\cdot\vec{v}=-8$ so $(\vec{u}\cdot\vec{v})^{2}=64$ . The sum is $230+64=294$ . On the other side, $\|\vec{u}\|^{2}=14$ and $\|\vec{v}\|^{2}=21$ , so $\|\vec{u}\|^{2}\|\vec{v}\|^{2}=14\times 21=294$ . Both sides are equal: the identity checks out.

c)

If $\vec{u}$ and $\vec{v}$ are orthogonal, then $\vec{u}\cdot\vec{v}=0$ , and the identity becomes $\|\vec{u}\times\vec{v}\|^{2}=\|\vec{u}\|^{2}\|\vec{v}\|^{2}$ , hence $\|\vec{u}\times\vec{v}\|=\|\vec{u}\|\,\|\vec{v}\|$ (norms being non-negative). Geometrically: the parallelogram built on two orthogonal vectors is a rectangle, and its area is indeed the product of its two sides. For fixed norms $\|\vec{u}\|$ and $\|\vec{v}\|$ , the area is therefore largest when the vectors are perpendicular.

d)

The vector $\vec{v}\times\vec{w}$ is orthogonal to the plane spanned by $\vec{v}$ and $\vec{w}$ . Saying that $\vec{u}$ , $\vec{v}$ and $\vec{w}$ are coplanar amounts to saying that $\vec{u}$ lies in this plane, that is, that $\vec{u}$ is orthogonal to the normal $\vec{v}\times\vec{w}$ , which is exactly $\vec{u}\cdot(\vec{v}\times\vec{w})=0$ . Conversely, if this dot product is zero, $\vec{u}$ is orthogonal to $\vec{v}\times\vec{w}$ and therefore lies in the plane of $\vec{v}$ and $\vec{w}$ : the three vectors are coplanar. An equivalent reading: the volume of the parallelepiped is zero, so the solid is flattened.

Exercise 7: Problem: the tetrahedron, two methods for one height

In an orthonormal coordinate system of space, consider the tetrahedron with vertices $P(1, 1, 1)$ , $Q(3, 2, 1)$ , $R(2, 4, 3)$ and $S(4, 3, 5)$ .

The goal is to compute the height of the tetrahedron from $S$ , that is, the distance from $S$ to the plane $PQR$ , along two independent paths that must give the same number.

a) Find the components of $\vec{PQ}$ , $\vec{PR}$ and $\vec{PS}$ .
b) Compute $\vec{PQ}\times\vec{PR}$ , then the area of the triangle $PQR$ .
c) Compute the volume of the tetrahedron $PQRS$ .
d) First method: deduce from b) and c) the height $h$ from $S$ , using $V=\dfrac{1}{3}\times\mathcal{A}_{\text{base}}\times h$ .
e) Second method: find a Cartesian equation of the plane $PQR$ , then compute the distance from $S$ to this plane. Check that you get the height of d) again.
f) Does the point $T(0, 5, 1)$ lie in the plane $PQR$ ? Compute the volume of the tetrahedron $PQRT$ , and explain why the value obtained by testing $T$ in the equation of the plane made it predictable.

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Answers

a) $(2, 1, 0)$ , $(1, 3, 2)$ , $(3, 2, 4)$
b) $(2, -4, 5)$ ; area $\frac{3\sqrt5}{2}\approx 3.35$
c) $V=3$
d) $h=\frac{6\sqrt5}{5}\approx 2.68$
e) $2x-4y+5z-3=0$ ; distance $2.68$
f) No; volume 3

a)

$\vec{PQ}=(3-1, 2-1, 1-1)=(2, 1, 0)$ , $\vec{PR}=(2-1, 4-1, 3-1)=(1, 3, 2)$ and $\vec{PS}=(4-1, 3-1, 5-1)=(3, 2, 4)$ .

b)

$\vec{PQ}\times\vec{PR}=((1)(2)-(0)(3), (0)(1)-(2)(2), (2)(3)-(1)(1))=(2, -4, 5)$ . Its norm is $\sqrt{4+16+25}=\sqrt{45}=3\sqrt{5}$ . The area of the triangle $PQR$ is half the area of the parallelogram: $\mathcal{A}_{PQR}=\dfrac{3\sqrt{5}}{2}\approx 3.35$ square units.

c)

Scalar triple product: $(\vec{PQ}\times\vec{PR})\cdot\vec{PS}=(2)(3)+(-4)(2)+(5)(4)=6-8+20=18$ . The volume of the tetrahedron is $V=\dfrac{|18|}{6}=3$ cubic units.

d)

From $V=\dfrac{1}{3}\mathcal{A}_{PQR}\,h$ we get $h=\dfrac{3V}{\mathcal{A}_{PQR}}=\dfrac{3\times 3}{\frac{3\sqrt{5}}{2}}=\dfrac{9\times 2}{3\sqrt{5}}=\dfrac{18}{3\sqrt{5}}=\dfrac{6}{\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\approx 2.68$ .

e)

The vector $\vec{n}=\vec{PQ}\times\vec{PR}=(2, -4, 5)$ is normal to the plane $PQR$ , which passes through $P(1, 1, 1)$ . Its equation is $2(x-1)-4(y-1)+5(z-1)=0$ , that is $2x-4y+5z-3=0$ . Quick check on $Q$ : $6-8+5-3=0$ , and on $R$ : $4-16+15-3=0$ . Distance from $S(4, 3, 5)$ : $d=\dfrac{|2(4)-4(3)+5(5)-3|}{\sqrt{2^{2}+(-4)^{2}+5^{2}}}=\dfrac{|8-12+25-3|}{\sqrt{45}}=\dfrac{18}{3\sqrt{5}}=\dfrac{6\sqrt{5}}{5}\approx 2.68$ . This is exactly the height of d), which validates both paths.

f)

Test $T(0, 5, 1)$ in the equation of the plane: $2(0)-4(5)+5(1)-3=0-20+5-3=-18\neq 0$ . So $T$ does NOT lie in the plane $PQR$ . The volume of the tetrahedron $PQRT$ is therefore not zero: $\vec{PT}=(-1, 4, 0)$ and $(\vec{PQ}\times\vec{PR})\cdot\vec{PT}=(2)(-1)+(-4)(4)+(5)(0)=-2-16=-18$ , hence $V=\dfrac{|-18|}{6}=3$ cubic units. The prediction without any volume computation was possible: the value $-18$ obtained by testing $T$ in the equation of the plane IS exactly the scalar triple product $\vec{n}\cdot\vec{PT}$ , because $2x-4y+5z-3$ equals $\vec{n}\cdot\vec{PM}$ for every point $M$ : the equation of the plane is built on the same normal vector. A point off the plane gives a non-zero volume; only a point in the plane would have given a zero volume.

Exercise 8: Orthogonal decomposition and the distance from a point to a line

Every vector splits in a unique way into a part PARALLEL to a given direction and a part ORTHOGONAL to it. This decomposition, combined with the cross product, gives two ways of measuring the distance from a point to a line. Consider the line $\Delta$ through $A(1, 0, 2)$ with direction vector $\vec{d}=(2, 1, -2)$ , and the point $P(4, 4, 3)$ .

The figure shows the plane that contains the line and the point.

a) Compute $\vec{AP}$ , then decompose it as $\vec{AP}=\vec{p}+\vec{q}$ , where $\vec{p}$ is the projection of $\vec{AP}$ onto $\vec{d}$ and $\vec{q}$ the orthogonal component.
b) Check that $\vec{q}$ is indeed orthogonal to $\vec{d}$ , then check the decomposition with the Pythagorean theorem $\|\vec{AP}\|^{2}=\|\vec{p}\|^{2}+\|\vec{q}\|^{2}$ .
c) First method: deduce the distance from the point $P$ to the line $\Delta$ .
d) Second method: compute the same distance with the formula $d=\dfrac{\|\vec{AP}\times\vec{d}\|}{\|\vec{d}\|}$ . Check that it agrees with c), and explain why the two methods must give the same result.

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Answers

a) $\vec p=\frac89\vec d$ , $\vec q=\left(\frac{11}{9}, \frac{28}{9}, \frac{25}{9}\right)$
b) $\frac{64}{9}+\frac{170}{9}=26$
c) $\frac{\sqrt{170}}{3}\approx 4.35$
d) Same distance with the cross product

a)

$\vec{AP}=(4-1, 4-0, 3-2)=(3, 4, 1)$ . The projection of $\vec{AP}$ onto $\vec{d}$ is $\vec{p}=\dfrac{\vec{AP}\cdot\vec{d}}{\|\vec{d}\|^{2}}\,\vec{d}$ . Compute $\vec{AP}\cdot\vec{d}=(3)(2)+(4)(1)+(1)(-2)=6+4-2=8$ and $\|\vec{d}\|^{2}=4+1+4=9$ , so $\vec{p}=\dfrac{8}{9}(2, 1, -2)=\left(\dfrac{16}{9}, \dfrac{8}{9}, -\dfrac{16}{9}\right)$ . The orthogonal component is $\vec{q}=\vec{AP}-\vec{p}=\left(3-\dfrac{16}{9}, 4-\dfrac{8}{9}, 1+\dfrac{16}{9}\right)=\left(\dfrac{11}{9}, \dfrac{28}{9}, \dfrac{25}{9}\right)$ .

b)

Orthogonality: $\vec{q}\cdot\vec{d}=\dfrac{1}{9}\big[(11)(2)+(28)(1)+(25)(-2)\big]=\dfrac{1}{9}(22+28-50)=\dfrac{0}{9}=0$ . The component $\vec{q}$ is indeed orthogonal to $\vec{d}$ , which confirms the decomposition. Pythagorean check: $\|\vec{AP}\|^{2}=9+16+1=26$ . Also $\|\vec{p}\|^{2}=\dfrac{\big(\vec{AP}\cdot\vec{d}\big)^{2}}{\|\vec{d}\|^{2}}=\dfrac{64}{9}$ and $\|\vec{q}\|^{2}=\dfrac{11^{2}+28^{2}+25^{2}}{81}=\dfrac{121+784+625}{81}=\dfrac{1530}{81}=\dfrac{170}{9}$ . The sum is $\dfrac{64}{9}+\dfrac{170}{9}=\dfrac{234}{9}=26$ , which matches: $\vec{p}$ and $\vec{q}$ are indeed the two legs of a right triangle whose hypotenuse is $\vec{AP}$ .

c)

The distance from the point $P$ to the line $\Delta$ is the length of the perpendicular segment dropped from $P$ to $\Delta$ , that is, exactly $\|\vec{q}\|$ , the component of $\vec{AP}$ orthogonal to the direction of the line. We get $d=\|\vec{q}\|=\sqrt{\dfrac{170}{9}}=\dfrac{\sqrt{170}}{3}\approx 4.35$ . The parallel part $\vec{p}$ only slides along the line and contributes nothing to the distance.

d)

Second method: $\vec{AP}\times\vec{d}=\big((4)(-2)-(1)(1), (1)(2)-(3)(-2), (3)(1)-(4)(2)\big)=(-8-1, 2+6, 3-8)=(-9, 8, -5)$ . Its norm is $\|\vec{AP}\times\vec{d}\|=\sqrt{81+64+25}=\sqrt{170}$ , and $\|\vec{d}\|=3$ , so $d=\dfrac{\sqrt{170}}{3}\approx 4.35$ . This is exactly the value of c). The agreement is no accident: the norm of the cross product is $\|\vec{AP}\times\vec{d}\|=\|\vec{AP}\|\,\|\vec{d}\|\sin\theta$ , where $\theta$ is the angle between $\vec{AP}$ and $\vec{d}$ . Dividing by $\|\vec{d}\|$ gives $\|\vec{AP}\|\sin\theta$ , which is precisely the length of the side opposite $\theta$ in the right triangle, that is, the perpendicular distance $\|\vec{q}\|$ . Both methods compute the same height, one by subtracting the projection, the other through the area of the parallelogram divided by its base $\|\vec{d}\|$ .

Exercise 9: The vector triple product and non-associativity

The cross product of a vector with a cross product, called the vector triple product, obeys a remarkable identity that reduces it to dot products. It also reveals that the cross product, unlike the multiplication of numbers, is NOT associative. Work with $\vec{u}=(1, 2, 3)$ , $\vec{v}=(0, 1, -1)$ and $\vec{w}=(2, 1, 0)$ .

a) Compute $\vec{v}\times\vec{w}$ , then $\vec{u}\times(\vec{v}\times\vec{w})$ .
b) Check on these vectors the vector triple product identity $\vec{u}\times(\vec{v}\times\vec{w})=(\vec{u}\cdot\vec{w})\,\vec{v}-(\vec{u}\cdot\vec{v})\,\vec{w}$ .
c) Compute $(\vec{u}\times\vec{v})\times\vec{w}$ and compare it with $\vec{u}\times(\vec{v}\times\vec{w})$ . What can you conclude about the associativity of the cross product?
d) Using the identity of b), prove that $\vec{u}\times(\vec{v}\times\vec{w})$ always lies in the plane spanned by $\vec{v}$ and $\vec{w}$ . Is this result compatible with the fact that this vector is orthogonal to $\vec{u}$ ?

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Answers

a) $(2, 5, -4)$
b) $4\vec v+\vec w=(2, 5, -4)$
c) $(-1, 2, -7)$ : not associative
d) Combination of $\vec v$ and $\vec w$

a)

$\vec{v}\times\vec{w}=\big((1)(0)-(-1)(1), (-1)(2)-(0)(0), (0)(1)-(1)(2)\big)=(0+1, -2-0, 0-2)=(1, -2, -2)$ . Then $\vec{u}\times(\vec{v}\times\vec{w})=(1, 2, 3)\times(1, -2, -2)=\big((2)(-2)-(3)(-2), (3)(1)-(1)(-2), (1)(-2)-(2)(1)\big)=(-4+6, 3+2, -2-2)=(2, 5, -4)$ .

b)

Compute the two dot products: $\vec{u}\cdot\vec{w}=(1)(2)+(2)(1)+(3)(0)=4$ and $\vec{u}\cdot\vec{v}=(1)(0)+(2)(1)+(3)(-1)=2-3=-1$ . The identity gives $(\vec{u}\cdot\vec{w})\,\vec{v}-(\vec{u}\cdot\vec{v})\,\vec{w}=4(0, 1, -1)-(-1)(2, 1, 0)=(0, 4, -4)+(2, 1, 0)=(2, 5, -4)$ . This is exactly the result of a). The identity, remembered as the “BAC minus CAB” rule because it reads $\vec{a}\times(\vec{b}\times\vec{c})=\vec{b}(\vec{a}\cdot\vec{c})-\vec{c}(\vec{a}\cdot\vec{b})$ , turns a computation of two cross products into two dot products, which are much faster.

c)

First, $\vec{u}\times\vec{v}=\big((2)(-1)-(3)(1), (3)(0)-(1)(-1), (1)(1)-(2)(0)\big)=(-2-3, 0+1, 1-0)=(-5, 1, 1)$ . Then $(\vec{u}\times\vec{v})\times\vec{w}=(-5, 1, 1)\times(2, 1, 0)=\big((1)(0)-(1)(1), (1)(2)-(-5)(0), (-5)(1)-(1)(2)\big)=(0-1, 2-0, -5-2)=(-1, 2, -7)$ . Compare: $(\vec{u}\times\vec{v})\times\vec{w}=(-1, 2, -7)$ while $\vec{u}\times(\vec{v}\times\vec{w})=(2, 5, -4)$ . The two results DIFFER, so the cross product is NOT associative: where the parentheses go changes the result. This is a fundamental difference from the multiplication of numbers, and it forbids writing $\vec{u}\times\vec{v}\times\vec{w}$ without parentheses, a notation that has no meaning.

d)

The identity of b) reads $\vec{u}\times(\vec{v}\times\vec{w})=(\vec{u}\cdot\vec{w})\,\vec{v}-(\vec{u}\cdot\vec{v})\,\vec{w}$ , that is, a LINEAR COMBINATION of $\vec{v}$ and $\vec{w}$ : the vector has the form $\lambda\vec{v}+\mu\vec{w}$ with $\lambda=\vec{u}\cdot\vec{w}$ and $\mu=-(\vec{u}\cdot\vec{v})$ . Every linear combination of $\vec{v}$ and $\vec{w}$ lies in the plane they span: the vector triple product therefore lives in the plane of $\vec{v}$ and $\vec{w}$ . This result is perfectly compatible with orthogonality to $\vec{u}$ . Indeed, $\vec{u}\times(\vec{v}\times\vec{w})$ is orthogonal to $\vec{u}$ , as every cross product is to both its factors. There is no contradiction: the vector can lie in the plane of $\vec{v}$ and $\vec{w}$ and be perpendicular to $\vec{u}$ at the same time, as long as $\vec{u}$ is not itself perpendicular to that plane. It can also be read another way: unlike $\vec{v}\times\vec{w}$ , which is perpendicular to the plane of $\vec{v}$ and $\vec{w}$ , crossing it once more with $\vec{u}$ brings the result back INTO that plane, folding it down.

Exercise 10: Problem: two skew lines, classification and distance

In space, two lines can be parallel, intersecting, or neither: they are then called SKEW (not coplanar). The three products of the chapter are enough to decide and to measure. Consider the line $\Delta_{1}$ through $P(1, 0, 1)$ with direction vector $\vec{d_{1}}=(1, 2, -1)$ , and the line $\Delta_{2}$ through $Q(2, 1, 0)$ with direction vector $\vec{d_{2}}=(2, -1, 1)$ .

a) Compute $\vec{d_{1}}\times\vec{d_{2}}$ . What can you already conclude about whether the two lines are parallel?
b) Compute the scalar triple product $\vec{PQ}\cdot(\vec{d_{1}}\times\vec{d_{2}})$ and conclude: are the lines intersecting or skew?
c) Compute the distance between the two lines with the formula $d=\dfrac{|\vec{PQ}\cdot(\vec{d_{1}}\times\vec{d_{2}})|}{\|\vec{d_{1}}\times\vec{d_{2}}\|}$ , and interpret this quotient.
d) Move $\Delta_{2}$ so that it now passes through the point $Q'(2, 1, m)$ , its direction $\vec{d_{2}}$ remaining unchanged. Find the value of $m$ for which $\Delta_{1}$ and $\Delta_{2}$ become intersecting.

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Answers

a) $(1, -3, -5)$ : not parallel
b) Scalar triple product 3: skew
c) $\frac{3}{\sqrt{35}}\approx 0.51$
d) $m=\frac35$

a)

$\vec{d_{1}}\times\vec{d_{2}}=\big((2)(1)-(-1)(-1), (-1)(2)-(1)(1), (1)(-1)-(2)(2)\big)=(2-1, -2-1, -1-4)=(1, -3, -5)$ . This vector is not zero, so $\vec{d_{1}}$ and $\vec{d_{2}}$ are not collinear: the two lines are NOT parallel. Two possibilities remain, intersecting or skew, and question b) decides between them.

b)

$\vec{PQ}=(2-1, 1-0, 0-1)=(1, 1, -1)$ . The scalar triple product is $\vec{PQ}\cdot(\vec{d_{1}}\times\vec{d_{2}})=(1)(1)+(1)(-3)+(-1)(-5)=1-3+5=3$ . This scalar triple product is NON-ZERO, so the vectors $\vec{PQ}$ , $\vec{d_{1}}$ and $\vec{d_{2}}$ are not coplanar. Now two lines are coplanar if and only if this scalar triple product is zero: here the lines do not lie in one plane, so they are SKEW. Being neither parallel nor intersecting, they never meet even though they do not have the same direction.

c)

The norm of the cross product is $\|\vec{d_{1}}\times\vec{d_{2}}\|=\sqrt{1+9+25}=\sqrt{35}$ , so $d=\dfrac{|3|}{\sqrt{35}}=\dfrac{3}{\sqrt{35}}\approx 0.51$ . Interpretation: the scalar triple product $|\vec{PQ}\cdot(\vec{d_{1}}\times\vec{d_{2}})|$ is the volume of the parallelepiped built on $\vec{PQ}$ , $\vec{d_{1}}$ and $\vec{d_{2}}$ ; the norm $\|\vec{d_{1}}\times\vec{d_{2}}\|$ is the area of the base built on $\vec{d_{1}}$ and $\vec{d_{2}}$ . The quotient is therefore the HEIGHT of the parallelepiped, measured perpendicular to the base, that is, the shortest gap between the two lines, along their common perpendicular. A volume divided by an area does give a length.

d)

Replacing $Q$ by $Q'(2, 1, m)$ , the vector becomes $\vec{PQ'}=(1, 1, m-1)$ , while $\vec{d_{1}}\times\vec{d_{2}}=(1, -3, -5)$ does not change, since the directions are unchanged. The two lines become coplanar when their scalar triple product vanishes: $\vec{PQ'}\cdot(\vec{d_{1}}\times\vec{d_{2}})=(1)(1)+(1)(-3)+(m-1)(-5)=1-3-5m+5=3-5m$ . Solve $3-5m=0$ : $m=\dfrac{3}{5}$ . For this value the lines are coplanar; and since they are still not parallel (their directions remain non-collinear), two coplanar non-parallel lines must INTERSECT. They therefore meet at a single point when $m=\dfrac{3}{5}$ .

Exercise 11: The cube in coordinates: angles, area and distance

Place a cube with edge $1$ in an orthonormal coordinate system, with vertices $O(0, 0, 0)$ , $A(1, 0, 0)$ , $B(0, 1, 0)$ , $C(0, 0, 1)$ and $G(1, 1, 1)$ , the vertex opposite $O$ .

a) Compute the angle between the two space diagonals $\vec{OG}$ and $\vec{AA'}$ , where $A'(0, 1, 1)$ is the vertex opposite $A$ . Round to the tenth of a degree.
b) Compute the angle between the face diagonal $(1, 1, 0)$ and the space diagonal $\vec{OG}$ , to the tenth of a degree.
c) Compute the area of the triangle $ABC$ to three decimals. What kind of triangle is it?
d) Compute the volume of the tetrahedron $OABC$ .
e) Deduce from c) and d) the distance from $O$ to the plane $ABC$ , to three decimals. What fraction of the diagonal $OG$ is this distance?

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Answers

a) $\cos\theta=\frac13$ : $70.5^\circ$
b) $35.3^\circ$
c) $\frac{\sqrt3}{2}\approx 0.866$ , equilateral
d) $\frac16$
e) $\frac{1}{\sqrt3}\approx 0.577$ : one third

a)

$\vec{OG}=(1, 1, 1)$ and $\vec{AA'}=(-1, 1, 1)$ , both with norm $\sqrt{3}$ . $\cos\theta=\frac{-1+1+1}{3}=\frac{1}{3}$ , so $\theta\approx 70.5^{\circ}$ . Two space diagonals of a cube are not perpendicular.

b)

$\cos\theta=\frac{1+1+0}{\sqrt{2}\sqrt{3}}=\frac{2}{\sqrt{6}}\approx 0.8165$ , so $\theta\approx 35.3^{\circ}$ .

c)

$\vec{AB}=(-1, 1, 0)$ and $\vec{AC}=(-1, 0, 1)$ , $\vec{AB}\times\vec{AC}=(1\cdot 1-0\cdot 0, 0\cdot(-1)-(-1)\cdot 1, (-1)\cdot 0-1\cdot(-1))=(1, 1, 1)$ . Area $\frac{\sqrt{3}}{2}\approx 0.866$ . The three sides are face diagonals, of length $\sqrt{2}$ : the triangle is EQUILATERAL, and $\frac{\sqrt{3}}{4}(\sqrt{2})^{2}=\frac{\sqrt{3}}{2}$ confirms it.

d)

$(\vec{OA}\times\vec{OB})\cdot\vec{OC}=(0, 0, 1)\cdot(0, 0, 1)=1$ , so $V=\frac{1}{6}\approx 0.167$ .

e)

$V=\frac{1}{3}\,\mathcal{A}\,h$ gives $h=\frac{3V}{\mathcal{A}}=\frac{1/2}{\sqrt{3}/2}=\frac{1}{\sqrt{3}}\approx 0.577$ . The diagonal has length $\sqrt{3}$ , so $\frac{h}{\sqrt{3}}=\frac{1}{3}$ : the plane $ABC$ , with equation $x+y+z=1$ , cuts the space diagonal at one third of its length, and the normal $(1, 1, 1)$ found in c) points precisely along this diagonal.

The thread: in the cube, every geometry question comes down to a dot product, a cross product or a scalar triple product on vectors with coordinates $0$ and $1$ .

Exercise 12: A parallelogram in space

Consider the points $A(1, 0, 2)$ , $B(3, 1, 4)$ , $C(4, 3, 3)$ and $D(2, 2, 1)$ .

a) Show that $ABCD$ is a parallelogram.
b) Is this parallelogram a rectangle?
c) Compute its area, to two decimals.
d) Compute the angle $\angle BAD$ , to the tenth of a degree.
e) Compute the lengths of the two diagonals, to two decimals, and check the parallelogram law $AC^{2}+BD^{2}=2(AB^{2}+AD^{2})$ .

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Answers

a) $\vec{AB}=\vec{DC}=(2, 1, 2)$
b) No: dot product 2
c) $\sqrt{50}\approx 7.07$
d) $74.2^\circ$
e) $\sqrt{19}$ and $\sqrt{11}$ : $30=30$

a)

$\vec{AB}=(2, 1, 2)$ and $\vec{DC}=(4-2, 3-2, 3-1)=(2, 1, 2)$ : $\vec{AB}=\vec{DC}$ , so $ABCD$ is a parallelogram. The four points must also not be collinear: $\vec{AD}=(1, 2, -1)$ is not a multiple of $\vec{AB}$ .

b)

$\vec{AB}\cdot\vec{AD}=2+2-2=2\neq 0$ : the angle at $A$ is not a right angle, so it is not a rectangle.

c)

$\vec{AB}\times\vec{AD}=(1\cdot(-1)-2\cdot 2, 2\cdot 1-2\cdot(-1), 2\cdot 2-1\cdot 1)=(-5, 4, 3)$ , with norm $\sqrt{25+16+9}=\sqrt{50}\approx 7.07$ . This is the area of the parallelogram, with no division by two.

d)

$\|\vec{AB}\|=3$ and $\|\vec{AD}\|=\sqrt{6}$ , so $\cos\angle BAD=\frac{2}{3\sqrt{6}}\approx 0.2722$ and $\angle BAD\approx 74.2^{\circ}$ . Check with the area: $3\sqrt{6}\sin 74.2^{\circ}\approx 7.07$ .

e)

$\vec{AC}=(3, 3, 1)$ , $AC=\sqrt{19}\approx 4.36$ ; $\vec{BD}=(-1, 1, -3)$ , $BD=\sqrt{11}\approx 3.32$ . $AC^{2}+BD^{2}=19+11=30$ and $2(AB^{2}+AD^{2})=2(9+6)=30$ : the law checks out. The diagonals are unequal, which confirms that it is not a rectangle.

The thread: two equal vectors make the parallelogram, the dot product decides the right angle, the cross product gives the area.

Exercise 13: Decomposing a vector with scalar triple products

Write $[\vec{p},\vec{q},\vec{r}]=\vec{p}\cdot(\vec{q}\times\vec{r})$ for the scalar triple product. Consider $\vec{a}=(1, 0, 1)$ , $\vec{b}=(0, 1, 1)$ , $\vec{c}=(1, 1, 0)$ and $\vec{v}=(3, 5, 4)$ .

a) Compute $[\vec{a},\vec{b},\vec{c}]$ . What can you conclude about the set $\{\vec{a}, \vec{b}, \vec{c}\}$ ?
b) Write $\vec{v}=\alpha\vec{a}+\beta\vec{b}+\gamma\vec{c}$ . By taking the dot product with $\vec{b}\times\vec{c}$ , show that $\alpha=\dfrac{[\vec{v},\vec{b},\vec{c}]}{[\vec{a},\vec{b},\vec{c}]}$ .
c) Compute $\alpha$ .
d) By the same reasoning, compute $\beta=\dfrac{[\vec{a},\vec{v},\vec{c}]}{[\vec{a},\vec{b},\vec{c}]}$ and $\gamma=\dfrac{[\vec{a},\vec{b},\vec{v}]}{[\vec{a},\vec{b},\vec{c}]}$ .
e) Check the decomposition, and recognize the rule for solving systems that these formulas express.

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Answers

a) $-2$ : basis
b) The other two terms vanish
c) $\alpha=1$
d) $\beta=3$ , $\gamma=2$
e) Cramer's rule

a)

$\vec{b}\times\vec{c}=(1\cdot 0-1\cdot 1, 1\cdot 1-0\cdot 0, 0\cdot 1-1\cdot 1)=(-1, 1, -1)$ , then $[\vec{a},\vec{b},\vec{c}]=-1+0-1=-2\neq 0$ . The three vectors are not coplanar: they form a basis of $\mathbb{R}^{3}$ , and the decomposition of $\vec{v}$ exists and is unique.

b)

$\vec{b}\times\vec{c}$ is orthogonal to $\vec{b}$ and to $\vec{c}$ , so $\vec{b}\cdot(\vec{b}\times\vec{c})=\vec{c}\cdot(\vec{b}\times\vec{c})=0$ . Taking the dot product of the equality with $\vec{b}\times\vec{c}$ leaves $\vec{v}\cdot(\vec{b}\times\vec{c})=\alpha\,\vec{a}\cdot(\vec{b}\times\vec{c})$ , hence the formula, the denominator being non-zero.

c)

$[\vec{v},\vec{b},\vec{c}]=(3, 5, 4)\cdot(-1, 1, -1)=-3+5-4=-2$ , so $\alpha=\frac{-2}{-2}=1$ .

d)

$\vec{v}\times\vec{c}=(5\cdot 0-4\cdot 1, 4\cdot 1-3\cdot 0, 3\cdot 1-5\cdot 1)=(-4, 4, -2)$ and $[\vec{a},\vec{v},\vec{c}]=-4+0-2=-6$ , so $\beta=3$ . $\vec{b}\times\vec{v}=(1\cdot 4-1\cdot 5, 1\cdot 3-0\cdot 4, 0\cdot 5-1\cdot 3)=(-1, 3, -3)$ and $[\vec{a},\vec{b},\vec{v}]=-1+0-3=-4$ , so $\gamma=2$ .

e)

$\vec{a}+3\vec{b}+2\vec{c}=(1+0+2, 0+3+2, 1+3+0)=(3, 5, 4)=\vec{v}$ . These formulas are CRAMER'S RULE: the scalar triple product is the determinant of the three vectors, and each unknown is the quotient of the determinant in which $\vec{v}$ replaces the corresponding vector by the determinant of the basis.

The thread: the scalar triple product isolates one coordinate by cancelling the other two at once; that is the geometric idea behind Cramer's rule.

Exercise 14: Problem: methane and the tetrahedral angle

In the methane molecule $CH_{4}$ , place the carbon atom at the origin and the four hydrogen atoms at $H_{1}(1, 1, 1)$ , $H_{2}(1, -1, -1)$ , $H_{3}(-1, 1, -1)$ and $H_{4}(-1, -1, 1)$ , in a unit of length to be determined.

a) Compute the distance between the carbon and each hydrogen, to three decimals. What do you notice?
b) Compute the angle $\angle H_{1}CH_{2}$ , to the tenth of a degree.
c) Compute the distance $H_{1}H_{2}$ to two decimals, and show that the six distances between hydrogens are equal. What is the shape of the solid $H_{1}H_{2}H_{3}H_{4}$ ?
d) The carbon-hydrogen bond is actually $1.09$ angstroms long. What length, in angstroms, is one unit of the coordinate system? Deduce the actual distance between two hydrogens, to two decimals.
e) Compute $\vec{CH_{1}}+\vec{CH_{2}}+\vec{CH_{3}}+\vec{CH_{4}}$ . What does this tell you about the position of the carbon, and about the polarity of the molecule?

Show the solution

Answers

a) $\sqrt3\approx 1.732$ for all four
b) $109.5^\circ$
c) $\sqrt8\approx 2.83$ : regular tetrahedron
d) $0.629$ Å; $1.78$ Å
e) Zero sum: nonpolar

a)

$CH_{i}=\sqrt{1+1+1}=\sqrt{3}\approx 1.732$ for each of the four: the hydrogens are all at the same distance from the carbon.

b)

$\vec{CH_{1}}\cdot\vec{CH_{2}}=1-1-1=-1$ , so $\cos\theta=\frac{-1}{3}$ and $\theta\approx 109.5^{\circ}$ . This is the TETRAHEDRAL angle, the same for all six pairs, since every dot product between two distinct hydrogens equals $-1$ .

c)

$\vec{H_{1}H_{2}}=(0, -2, -2)$ , with norm $\sqrt{8}\approx 2.83$ . Two distinct hydrogens always have two opposite coordinates and one equal coordinate: each difference contains two $\pm 2$ and one $0$ , and the six distances all equal $\sqrt{8}$ . The solid is a REGULAR TETRAHEDRON, with the carbon at its centre.

d)

$\sqrt{3}$ units correspond to $1.09$ angstroms, so one unit is $\frac{1.09}{\sqrt{3}}\approx 0.629$ angstroms. $H_{1}H_{2}=\sqrt{8}\times 0.629\approx 1.78$ angstroms, in agreement with the value measured for methane.

e)

The sum is $(1+1-1-1, 1-1+1-1, 1-1-1+1)=(0, 0, 0)$ : the carbon is the centroid of the four hydrogens. The four bonds being identical and of equal length, their dipole moments, carried by these four vectors, cancel out: the molecule is NONPOLAR.

The thread: the symmetry of a molecule can be read off its coordinates; dot products and norms give the bond angles and the bond lengths.

Exercise 15: Problem: the power received by a solar panel

A rectangular solar panel has the vectors $\vec{a}=(2, 0, 0)$ and $\vec{b}=(0, 1, 1)$ as sides, in metres, the $z$ -axis being vertical. Sunlight arrives along the unit direction $\vec{s}=\frac{1}{3}(1, -2, 2)$ , pointing towards the Sun, with an irradiance of $1000\ \mathrm{W}$ per square metre perpendicular to the rays.

Take as given that the power received is $1000\times(\vec{a}\times\vec{b})\cdot\vec{s}$ watts: the area of the panel times the cosine of the angle between its normal and the direction of the Sun.

The orange arrow shows the direction of the Sun.

a) Check that the panel is indeed rectangular, then compute its area to two decimals.
b) Compute the normal vector $\vec{a}\times\vec{b}$ and check that $\vec{s}$ is a unit vector.
c) Compute the angle of incidence, between the normal and the direction of the Sun, to the tenth of a degree.
d) Compute the power received by the panel, to the nearest watt.
e) What power would the same panel receive lying flat, with normal $(0, 0, 1)$ ? And facing the Sun exactly?

Show the solution

Answers

a) Rectangle, $2\sqrt2\approx 2.83$ m²
b) $(0, -2, 2)$
c) $19.5^\circ$
d) 2667 W
e) 1886 W lying flat; 2828 W facing the Sun

a)

$\vec{a}\cdot\vec{b}=0$ : the sides are perpendicular, so the parallelogram is a rectangle. Area $\|\vec{a}\|\,\|\vec{b}\|=2\sqrt{2}\approx 2.83\ \mathrm{m^{2}}$ .

b)

$\vec{a}\times\vec{b}=(0\cdot 1-0\cdot 1, 0\cdot 0-2\cdot 1, 2\cdot 1-0\cdot 0)=(0, -2, 2)$ , with norm $\sqrt{8}$ , equal to the area. $\|\vec{s}\|=\frac{1}{3}\sqrt{1+4+4}=1$ .

c)

$(\vec{a}\times\vec{b})\cdot\vec{s}=\frac{1}{3}(0+4+4)=\frac{8}{3}$ , so $\cos i=\frac{8/3}{\sqrt{8}}\approx 0.9428$ and $i\approx 19.5^{\circ}$ .

d)

$P=1000\times\frac{8}{3}\approx 2667\ \mathrm{W}$ . The scalar triple product directly gives the APPARENT area of the panel seen from the Sun, $\frac{8}{3}\approx 2.67\ \mathrm{m^{2}}$ .

e)

Lying flat, the normal of the same length is $(0, 0, 2\sqrt{2})$ and $(0, 0, 2\sqrt{2})\cdot\vec{s}=\frac{4\sqrt{2}}{3}\approx 1.886$ : $P\approx 1886\ \mathrm{W}$ . Facing the Sun, $\cos i=1$ and $P=1000\times 2\sqrt{2}\approx 2828\ \mathrm{W}$ . The chosen tilt already collects $94\%$ of the maximum, against $67\%$ lying flat.

The thread: the cross product of the sides gives a normal whose norm is the area; its dot product with a direction gives the area seen from that direction.

Corrected exercises: vectors in space and the three products (201-SN4-RE)

Course recap

Calculations: norms, dot products and angles

Exercise 0.1: Norms, dot products and angles: ten direct calculations

Calculations: cross products and scalar triple products

Exercise 0.2: Cross products and scalar triple products: eight direct calculations

Part A: The basics in space (/50)

Exercise 1: Components, norm and operations in space

Exercise 2: Dot product, angle and orthogonal projection

Exercise 3: Cross product, orthogonality and areas

Exercise 4: Scalar triple product, volume and coplanarity

Exercise 5: Direction angles and direction cosines

Part B: Exam level (/50)

Exercise 6: Proofs: Lagrange's identity and the coplanarity criterion

Exercise 7: Problem: the tetrahedron, two methods for one height

Exercise 8: Orthogonal decomposition and the distance from a point to a line

Exercise 9: The vector triple product and non-associativity

Exercise 10: Problem: two skew lines, classification and distance

Part C: the classics (/50)

Exercise 11: The cube in coordinates: angles, area and distance

Exercise 12: A parallelogram in space

Exercise 13: Decomposing a vector with scalar triple products

Exercise 14: Problem: methane and the tetrahedral angle

Exercise 15: Problem: the power received by a solar panel

See also

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