Exercise 0.1: Computing equations, intersections and distances in space: ten direct calculations

For each item, compute the requested quantity. Give fractions exactly, distances to two decimal places and angles to the nearest tenth of a degree. A plane is written $ax+by+cz+d=0$ .

Each line comes down to a formula once its ingredients are named: a plane is a point and a normal vector $\vec{n}=(a, b, c)$ ; a line is a point and a direction vector $\vec{u}$ . To find an intersection, substitute the parametric equations into the equation of the plane; for parallelism, write $\vec{u}\cdot\vec{n}=0$ ; use $\frac{|ax_{0}+by_{0}+cz_{0}+d|}{\|\vec{n}\|}$ for the distance from a point to a plane and $\frac{\|\vec{AM}\times\vec{u}\|}{\|\vec{u}\|}$ for the distance from a point to a line. Angles come from a dot product: a cosine between two normals, a sine between a direction and a normal.

a) The plane passes through $A(2, -1, 3)$ and has normal vector $\vec{n}=(4, 1, -2)$ : write its equation $4x+y-2z+d=0$ , then give $d$ .
b) The plane passes through $A(1, 1, 0)$ , $B(2, 0, 1)$ and $C(0, 3, 2)$ : write its equation in the form $4x+by+cz+d=0$ , then give $c$ and $d$ .
c) The line $(x, y, z)=(0, 4, 1)+t(3, -1, 2)$ meets the plane $2x+y-z=9$ : give the value of $t$ at the point of intersection.
d) The line $x=2-t$ , $y=1+3t$ , $z=-1+2t$ meets the plane $2x+y+z=5$ : give the $x$ - and $z$ -coordinates of the point of intersection.
e) Find $k$ so that the line with direction vector $(2, k, -1)$ is parallel to the plane $3x-2y+4z=7$ .
f) Compute the distance from the point $P(2, -3, 1)$ to the plane $6x-2y+3z=4$ .
g) Compute the distance between the parallel planes $2x-3y+6z=5$ and $-4x+6y-12z=18$ .
h) Compute the acute angle between the planes $x-2y+2z=1$ and $3x+4z=2$ , in degrees.
i) Compute the distance from the point $M(3, 0, -1)$ to the line through $A(1, 2, 1)$ with direction vector $\vec{u}=(2, 1, -2)$ .
j) Compute the angle between the line with direction vector $(2, 1, 2)$ and the plane $2x+3y+6z=1$ , in degrees.

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

a)

d

b)

c

d

c)

t

d)

x

z

e)

k

f)

distance

g)

distance

h)

angle in degrees

i)

distance

j)

angle in degrees

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Answers

a) $4x+y-2z-1=0$ ; $d=-1$
b) $4x+3y-z-7=0$ ; $c=-1$ , $d=-7$
c) $3t+3=9$ ; $t=2$
d) $t=\frac{1}{3}$ ; $\left(\frac{5}{3}, 2, -\frac{1}{3}\right)$
e) $6-2k-4=0$ ; $k=1$
f) $\frac{17}{7}\approx 2.43$
g) $\frac{|-5-9|}{7}=2$
h) $\arccos\frac{11}{15}\approx 42.8$ degrees
i) $\frac{6\sqrt{2}}{3}=2\sqrt{2}\approx 2.83$
j) $\arcsin\frac{19}{21}\approx 64.8$ degrees

The move that decides everything: read the normal vector off the coefficients of $x$ , $y$ and $z$ , and the direction vector off the coefficient of the parameter, BEFORE choosing a formula. Once those two vectors are written down, every item is a dot product, a cross product or a substitution.

a)

The point lies in the plane: $4(2)+(-1)-2(3)+d=0$ , that is $8-1-6+d=0$ , so $d=-1$ . The plane is $4x+y-2z-1=0$ .

Pitfall: $-2\times 3=-6$ , the sign of the coefficient of $z$ .

b)

$\vec{AB}=(1, -1, 1)$ and $\vec{AC}=(-1, 2, 2)$ , so $\vec{AB}\times\vec{AC}=\left((-1)(2)-(1)(2), (1)(-1)-(1)(2), (1)(2)-(-1)(-1)\right)=(-4, -3, 1)$ . Take the opposite vector $\vec{n}=(4, 3, -1)$ ; then $A$ gives $4+3+d=0$ : the plane is $4x+3y-z-7=0$ , so $c=-1$ and $d=-7$ . Check: $B$ gives $8+0-1-7=0$ and $C$ gives $0+9-2-7=0$ .

Pitfall: the middle component of the cross product.

c)

Substitute $x=3t$ , $y=4-t$ , $z=1+2t$ : $2(3t)+(4-t)-(1+2t)=9$ , that is $3t+3=9$ , so $t=2$ . The point is $(6, 2, 5)$ , and $12+2-5=9$ .

Pitfall: $-(1+2t)$ , the minus sign applies to both terms.

d)

$2(2-t)+(1+3t)+(-1+2t)=5$ , that is $4+3t=5$ , so $t=\frac{1}{3}$ . The point is $\left(\frac{5}{3}, 2, -\frac{1}{3}\right)$ : $x=\frac{5}{3}$ and $z=-\frac{1}{3}$ .

Pitfall: giving $t$ instead of the coordinates.

e)

Parallel to the plane means orthogonal to $\vec{n}=(3, -2, 4)$ : $3(2)-2k+4(-1)=0$ , that is $2-2k=0$ , so $k=1$ .

Pitfall: making the direction a multiple of $\vec{n}$ , which would give a line perpendicular to the plane.

f)

$\|\vec{n}\|=\sqrt{36+4+9}=7$ , and the plane is written $6x-2y+3z-4=0$ . So $\frac{|6(2)-2(-3)+3(1)-4|}{7}=\frac{|12+6+3-4|}{7}=\frac{17}{7}\approx 2.43$ .

Pitfall: dropping the $-4$ when moving the right-hand side to the left.

g)

Divide the second equation by $-2$ : $2x-3y+6z=-9$ , that is $2x-3y+6z+9=0$ , with the SAME normal $(2, -3, 6)$ as $2x-3y+6z-5=0$ . So $\frac{|-5-9|}{\sqrt{4+9+36}}=\frac{14}{7}=2$ .

Pitfall: comparing $5$ and $18$ before both equations have the same normal.

h)

Normals $(1, -2, 2)$ and $(3, 0, 4)$ , with norms $3$ and $5$ . $\cos\theta=\frac{|3+0+8|}{3\times 5}=\frac{11}{15}$ , so $\theta=\arccos\frac{11}{15}\approx 42.8$ degrees. The absolute value guarantees the acute angle.

i)

$\vec{AM}=(2, -2, -2)$ and $\vec{AM}\times\vec{u}=\left((-2)(-2)-(-2)(1), (-2)(2)-(2)(-2), (2)(1)-(-2)(2)\right)=(6, 0, 6)$ , with norm $6\sqrt{2}$ . With $\|\vec{u}\|=3$ , the distance is $\frac{6\sqrt{2}}{3}=2\sqrt{2}\approx 2.83$ .

Pitfall: dividing by $\|\vec{AM}\|$ instead of $\|\vec{u}\|$ .

j)

The angle between a line and a plane is the complement of the angle between the direction and the normal, hence a SINE: $\sin\varphi=\frac{|4+3+12|}{3\times 7}=\frac{19}{21}$ , so $\varphi=\arcsin\frac{19}{21}\approx 64.8$ degrees.

Pitfall: taking the inverse cosine, which gives the angle with the normal, $25.2$ degrees.

Check: substitute an intersection point into the equation of the plane, check the equation of a plane on each of the given points, and remember that the acute angle between two planes stays between $0$ and $90$ degrees. For a point-to-plane distance, the projection $P-\frac{17}{49}\vec{n}$ must land in the plane.

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Exercise 1: The equations of a line in space

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

a) Find a direction vector $\vec{u}$ of the line $d$ through $A$ and $B$ .
b) Write the vector equation, the parametric equations and the symmetric equations of $d$ .
c) Does the point $P(5, 8, 3)$ lie on $d$ ? What about the point $Q(0, 0, 0)$ ?

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Answers

a) $\vec u=(2, 3, 2)$
b) $\frac{x-1}{2}=\frac{y-2}{3}=\frac{z+1}{2}$
c) $P\in d$ ( $t=2$ ); $Q\notin d$

a)

$\vec{u}=\vec{AB}=(3-1, 5-2, 1-(-1))=(2, 3, 2)$ .

b)

Vector equation: $(x, y, z)=(1, 2, -1)+t(2, 3, 2)$ . Parametric equations: $x=1+2t,\ y=2+3t,\ z=-1+2t$ . Symmetric equations: $\dfrac{x-1}{2}=\dfrac{y-2}{3}=\dfrac{z+1}{2}$ .

c)

For $P(5, 8, 3)$ : the first equation gives $5=1+2t$ , so $t=2$ ; check $y=2+3(2)=8$ and $z=-1+2(2)=3$ . All three coordinates agree for the same $t$ , so $P\in d$ . For $Q(0, 0, 0)$ : $0=1+2t$ gives $t=-\tfrac{1}{2}$ , but then $y=2+3(-\tfrac{1}{2})=\tfrac{1}{2}\neq 0$ . The coordinates do not agree, so $Q\notin d$ .

Exercise 2: The scalar equation of a plane

Consider the points $P(1, 0, 2)$ , $Q(3, 1, 1)$ and $R(2, 2, 3)$ , assumed not collinear.

a) Find two direction vectors of the plane $\pi$ through $P$ , $Q$ and $R$ .
b) Find a normal vector $\vec{n}$ to $\pi$ , then a scalar equation of $\pi$ .
c) Check that $Q$ and $R$ satisfy your equation. Does the point $S(0, 0, 0)$ lie in $\pi$ ?

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Answers

a) $\vec{PQ}=(2, 1, -1)$ , $\vec{PR}=(1, 2, 1)$
b) $x-y+z-3=0$
c) $Q$ and $R$ check; $S\notin\pi$

a)

$\vec{PQ}=(2, 1, -1)$ and $\vec{PR}=(1, 2, 1)$ are two direction vectors of the plane.

b)

A normal vector is $\vec{n}=\vec{PQ}\times\vec{PR}=((1)(1)-(-1)(2), (-1)(1)-(2)(1), (2)(2)-(1)(1))=(3, -3, 3)$ . It can be simplified to $\vec{n}=(1, -1, 1)$ . The plane passes through $P(1, 0, 2)$ : $1(x-1)-1(y-0)+1(z-2)=0$ , that is $x-y+z-3=0$ .

c)

For $Q(3, 1, 1)$ : $3-1+1-3=0$ ; for $R(2, 2, 3)$ : $2-2+3-3=0$ . Both points satisfy the equation. For $S(0, 0, 0)$ : $0-0+0-3=-3\neq 0$ , so $S\notin\pi$ .

Exercise 3: Relative positions and the intersection of a line and a plane

Take again the line $d$ with direction vector $\vec{u}=(2, 3, 2)$ through $A(1, 2, -1)$ , and the plane $\pi : x-y+z-3=0$ with normal vector $\vec{n}=(1, -1, 1)$ .

a) Show that $d$ is not parallel to $\pi$ , then find their point of intersection.
b) Consider the line $d_{2}$ with equation $(x, y, z)=(0, 1, 0)+s(1, 2, 1)$ . Show that $d_{2}$ is parallel to $\pi$ , then say whether it lies in $\pi$ or is strictly parallel to it.

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Answers

a) $t=5$ : point $(11, 17, 9)$
b) $\vec v\cdot\vec n=0$ : strictly parallel

a)

$\vec{u}\cdot\vec{n}=(2)(1)+(3)(-1)+(2)(1)=2-3+2=1\neq 0$ , so $d$ crosses the plane at a single point. Substitute the parametric equations of $d$ into the equation of $\pi$ : $(1+2t)-(2+3t)+(-1+2t)-3=0$ , that is $-5+t=0$ , so $t=5$ . The point of intersection is $(1+10, 2+15, -1+10)=(11, 17, 9)$ . Check: $11-17+9-3=0$ .

b)

The direction vector of $d_{2}$ is $\vec{v}=(1, 2, 1)$ and $\vec{v}\cdot\vec{n}=(1)(1)+(2)(-1)+(1)(1)=1-2+1=0$ : $d_{2}$ is therefore parallel to $\pi$ . It remains to test one point of $d_{2}$ , for example $(0, 1, 0)$ : $0-1+0-3=-4\neq 0$ , so this point is not in $\pi$ . The line $d_{2}$ is strictly parallel to $\pi$ : they have no point in common.

The figure sums up the test: if the direction vector of the line is ORTHOGONAL to $\vec{n}$ , the line is parallel to the plane, with either no point in common or all of its points in the plane.

Exercise 4: Distances in space

Keep the plane $\pi : x-y+z-3=0$ and the line $d$ through $A(1, 2, -1)$ with direction vector $\vec{u}=(2, 3, 2)$ . Consider the point $S(4, 1, 5)$ .

a) Compute the distance from the point $S$ to the plane $\pi$ .
b) Compute the distance from the point $S$ to the line $d$ .
c) Consider the plane $\pi_{2} : x-y+z+1=0$ . Show that $\pi$ and $\pi_{2}$ are parallel, then compute the distance between them.

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Answers

a) $\frac{5\sqrt3}{3}\approx 2.89$
b) $\sqrt{\frac{557}{17}}\approx 5.72$
c) $\frac{4\sqrt3}{3}\approx 2.31$

a)

$\text{dist}(S,\pi)=\dfrac{|4-1+5-3|}{\sqrt{1^{2}+(-1)^{2}+1^{2}}}=\dfrac{|5|}{\sqrt{3}}=\dfrac{5}{\sqrt{3}}=\dfrac{5\sqrt{3}}{3}\approx 2.89$ .

b)

$\vec{AS}=(4-1, 1-2, 5-(-1))=(3, -1, 6)$ . Compute $\vec{AS}\times\vec{u}=((-1)(2)-(6)(3), (6)(2)-(3)(2), (3)(3)-(-1)(2))=(-20, 6, 11)$ , with norm $\sqrt{(-20)^{2}+6^{2}+11^{2}}=\sqrt{400+36+121}=\sqrt{557}$ . Since $\|\vec{u}\|=\sqrt{4+9+4}=\sqrt{17}$ , we get $\text{dist}(S,d)=\dfrac{\sqrt{557}}{\sqrt{17}}=\sqrt{\dfrac{557}{17}}\approx 5.72$ .

c)

The two planes have the same normal vector $\vec{n}=(1, -1, 1)$ , so they are parallel. Writing $\pi : x-y+z=3$ and $\pi_{2} : x-y+z=-1$ , the distance between them is $\dfrac{|3-(-1)|}{\sqrt{3}}=\dfrac{4}{\sqrt{3}}=\dfrac{4\sqrt{3}}{3}\approx 2.31$ .

Exercise 5: The intersection of two planes and the angle between them

Consider the planes $\pi_{1} : x+y+z-1=0$ with normal vector $\vec{n_{1}}=(1, 1, 1)$ and $\pi_{2} : 2x-y+z-2=0$ with normal vector $\vec{n_{2}}=(2, -1, 1)$ .

a) Show that $\pi_{1}$ and $\pi_{2}$ are not parallel.
b) Find a direction vector of their line of intersection $\Delta$ .
c) Find a point of $\Delta$ , then write a vector equation of $\Delta$ .
d) Compute the acute angle between the two planes, to the nearest hundredth of a degree.

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Answers

a) Normals not parallel
b) $(2, 1, -3)$
c) $(1, 0, 0)+t(2, 1, -3)$
d) $61.87^\circ$

a)

Two planes are parallel if and only if their normal vectors are scalar multiples of each other. But $\vec{n_{1}}=(1, 1, 1)$ and $\vec{n_{2}}=(2, -1, 1)$ are not, since $\dfrac{2}{1}\neq\dfrac{-1}{1}$ . The planes are therefore not parallel: they meet along a line.

b)

The line of intersection lies in both planes, so its direction vector is perpendicular to both normals at once. Take it equal to their cross product: $\vec{d}=\vec{n_{1}}\times\vec{n_{2}}=((1)(1)-(1)(-1), (1)(2)-(1)(1), (1)(-1)-(1)(2))=(2, 1, -3)$ .

c)

Look for a point common to both planes by fixing one coordinate, for example $z=0$ . The system becomes $\begin{cases} x+y=1 \\ 2x-y=2 \end{cases}$ . Adding: $3x=3$ , so $x=1$ and $y=0$ . The point $I(1, 0, 0)$ does lie in both planes: $1+0+0-1=0$ and $2-0+0-2=0$ . A vector equation of $\Delta$ is therefore $(x, y, z)=(1, 0, 0)+t(2, 1, -3)$ .

d)

The angle between two planes is the angle between their normals, brought back to the acute angle by the absolute value: $\cos\theta=\dfrac{|\vec{n_{1}}\cdot\vec{n_{2}}|}{\|\vec{n_{1}}\|\,\|\vec{n_{2}}\|}$ . Here $\vec{n_{1}}\cdot\vec{n_{2}}=(1)(2)+(1)(-1)+(1)(1)=2$ , $\|\vec{n_{1}}\|=\sqrt{3}$ and $\|\vec{n_{2}}\|=\sqrt{6}$ , so $\cos\theta=\dfrac{2}{\sqrt{18}}=\dfrac{2}{3\sqrt{2}}=\dfrac{\sqrt{2}}{3}\approx 0.4714$ , hence $\theta\approx 61.87^{\circ}$ . The absolute value is what guarantees an acute angle: without it, a negative dot product would give the supplementary obtuse angle.

The figure shows the two planes edge-on: they meet along a line, and the angle between the planes is the angle between their normals.

Exercise 6: Two skew lines and the distance between them

Consider the line $d_{1}$ through $A(1, 0, 0)$ with direction vector $\vec{u}=(1, 2, 1)$ , and the line $d_{2}$ through $C(0, 1, 2)$ with direction vector $\vec{v}=(2, -1, 1)$ .

a) Compute $\vec{u}\times\vec{v}$ and deduce that $d_{1}$ and $d_{2}$ are not parallel.
b) By computing the scalar triple product $\vec{AC}\cdot(\vec{u}\times\vec{v})$ , show that $d_{1}$ and $d_{2}$ are not coplanar. Conclude that they are skew.
c) Explain why the distance between two skew lines is $\text{dist}=\dfrac{|\vec{AC}\cdot(\vec{u}\times\vec{v})|}{\|\vec{u}\times\vec{v}\|}$ , then compute it.
d) Replace the direction vector of $d_{2}$ by $\vec{v_{m}}=(2, -1, m)$ . Find the value of $m$ for which $d_{1}$ and $d_{2}$ become coplanar.

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Answers

a) $(3, 1, -5)$ : not parallel
b) $-12$ : skew
c) $\frac{12}{\sqrt{35}}\approx 2.03$
d) $m=-3$

a)

$\vec{u}\times\vec{v}=((2)(1)-(1)(-1), (1)(2)-(1)(1), (1)(-1)-(2)(2))=(3, 1, -5)$ . This vector is not zero, so $\vec{u}$ and $\vec{v}$ are not parallel: the lines are not parallel.

b)

$\vec{AC}=(0-1, 1-0, 2-0)=(-1, 1, 2)$ . The scalar triple product is $\vec{AC}\cdot(\vec{u}\times\vec{v})=(-1)(3)+(1)(1)+(2)(-5)=-3+1-10=-12\neq 0$ . The three vectors are not coplanar, so the lines do not meet. Being neither parallel nor intersecting, $d_{1}$ and $d_{2}$ are skew.

c)

The vector $\vec{u}\times\vec{v}$ is perpendicular to both $\vec{u}$ and $\vec{v}$ : it is the direction of the common perpendicular of the two lines. The required distance is the length of the projection of $\vec{AC}$ (which joins a point of each line) onto this common direction, that is $\dfrac{|\vec{AC}\cdot(\vec{u}\times\vec{v})|}{\|\vec{u}\times\vec{v}\|}$ . Numerically: $\|\vec{u}\times\vec{v}\|=\sqrt{3^{2}+1^{2}+(-5)^{2}}=\sqrt{35}$ , so $\text{dist}=\dfrac{|-12|}{\sqrt{35}}=\dfrac{12}{\sqrt{35}}=\dfrac{12\sqrt{35}}{35}\approx 2.03$ .

The figure says it in one picture: the two lines never meet, and the distance is measured along their common perpendicular.

d)

With $\vec{v_{m}}=(2, -1, m)$ : $\vec{u}\times\vec{v_{m}}=((2)(m)-(1)(-1), (1)(2)-(1)(m), (1)(-1)-(2)(2))=(2m+1, 2-m, -5)$ . The scalar triple product becomes $\vec{AC}\cdot(\vec{u}\times\vec{v_{m}})=(-1)(2m+1)+(1)(2-m)+(2)(-5)=-2m-1+2-m-10=-3m-9$ . The lines are coplanar when this triple product vanishes: $-3m-9=0$ , so $m=-3$ .

Exercise 7: Problem: the paths of two drones

Two drones fly in straight lines in space, with an orthonormal coordinate system graduated in kilometres. Time $t$ is measured in minutes from the same initial instant. The position of drone 1 is $\vec{r_{1}}(t)=(1, 2, 0)+t(2, 1, 1)$ and that of drone 2 is $\vec{r_{2}}(t)=(4, 6, -1)+t(1, -2, 3)$ .

The figure shows the two paths, with the position of each drone at the initial instant and its velocity.

a) Do the two paths, seen as lines, cross? If so, at what point in space?
b) Is there a collision, that is, do the two drones pass through the same place at the same instant? Justify.
c) Find the acute angle between the two paths, to the nearest tenth of a degree.

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Answers

a) Yes: $(5, 4, 2)$
b) No collision: times 2 and 1
c) $70.9^\circ$

a)

Look for parameters $t$ (drone 1) and $s$ (drone 2) that give the same point: $\begin{cases} 1+2t=4+s \\ 2+t=6-2s \\ t=-1+3s \end{cases}$ . The third equation gives $t=3s-1$ ; substituting into the first: $1+2(3s-1)=4+s$ , that is $6s-1=4+s$ , so $s=1$ and $t=2$ . Check the second: $2+2=4$ and $6-2(1)=4$ . The system is consistent: the paths cross. The common point is $\vec{r_{1}}(2)=(1+4, 2+2, 0+2)=(5, 4, 2)$ , and indeed $\vec{r_{2}}(1)=(4+1, 6-2, -1+3)=(5, 4, 2)$ .

b)

Drone 1 reaches the point $(5, 4, 2)$ at time $t=2$ min, while drone 2 passes through it at time $s=1$ min. The times are different, so there is NO collision: the paths cross in space, but the drones are never there at the same time. This is the key distinction of the problem: two paths can cross without the drones ever meeting.

c)

The direction vectors are $\vec{u_{1}}=(2, 1, 1)$ and $\vec{u_{2}}=(1, -2, 3)$ . We have $\vec{u_{1}}\cdot\vec{u_{2}}=(2)(1)+(1)(-2)+(1)(3)=3$ , $\|\vec{u_{1}}\|=\sqrt{6}$ and $\|\vec{u_{2}}\|=\sqrt{14}$ . So $\cos\theta=\dfrac{|3|}{\sqrt{6}\sqrt{14}}=\dfrac{3}{\sqrt{84}}\approx 0.3273$ , hence $\theta\approx 70.9^{\circ}$ .

Exercise 8: Orthogonal projection and reflection of a point: in a plane and in a line

Work in an orthonormal coordinate system of space. Consider the plane $\pi : x-y+z-3=0$ with normal vector $\vec{n}=(1, -1, 1)$ and the point $M(3, 1, 4)$ .

Consider also the line $\ell$ through $A(0, 1, 0)$ with direction vector $\vec{u}=(2, -1, 2)$ , and the point $P(1, 0, 3)$ .

a) Find the orthogonal projection $H$ of $M$ onto the plane $\pi$ , and the distance from $M$ to $\pi$ .
b) Find the reflection $M'$ of $M$ in the plane $\pi$ , and check that $H$ is the midpoint of $MM'$ .
c) Find the orthogonal projection $F$ of $P$ onto the line $\ell$ , and the distance from $P$ to $\ell$ .
d) Find the reflection $P'$ of $P$ in the line $\ell$ .

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Answers

a) $H(2, 2, 3)$ , distance $\sqrt3$
b) $M'(1, 3, 2)$
c) $F(2, 0, 2)$ , distance $\sqrt2$
d) $P'(3, 0, 1)$

a)

The orthogonal projection of $M$ onto $\pi$ is reached by moving from $M$ along the normal: $H=M-\dfrac{ax_{M}+by_{M}+cz_{M}+d}{a^{2}+b^{2}+c^{2}}\,\vec{n}$ . Here $ax_{M}+by_{M}+cz_{M}+d=3-1+4-3=3$ and $a^{2}+b^{2}+c^{2}=3$ , so the coefficient is $\dfrac{3}{3}=1$ and $H=M-1\cdot\vec{n}=(3, 1, 4)-(1, -1, 1)=(2, 2, 3)$ . Check that $H\in\pi$ : $2-2+3-3=0$ . The distance from $M$ to $\pi$ is $\dfrac{|3|}{\sqrt{3}}=\sqrt{3}\approx 1.73$ , which is also the length $MH$ .

b)

The reflection is reached by going twice as far along the normal: $M'=M-2\cdot 1\cdot\vec{n}=(3, 1, 4)-(2, -2, 2)=(1, 3, 2)$ . The midpoint of $MM'$ is $\left(\dfrac{3+1}{2}, \dfrac{1+3}{2}, \dfrac{4+2}{2}\right)=(2, 2, 3)=H$ , which confirms the computation: the projection is always the midpoint of a point and its reflection.

c)

The orthogonal projection of $P$ onto $\ell$ is the foot of the perpendicular: $F=A+t\,\vec{u}$ with $t=\dfrac{\vec{AP}\cdot\vec{u}}{\|\vec{u}\|^{2}}$ . We have $\vec{AP}=(1-0, 0-1, 3-0)=(1, -1, 3)$ , so $\vec{AP}\cdot\vec{u}=(1)(2)+(-1)(-1)+(3)(2)=2+1+6=9$ and $\|\vec{u}\|^{2}=4+1+4=9$ , hence $t=1$ . Thus $F=(0, 1, 0)+1\cdot(2, -1, 2)=(2, 0, 2)$ . Check the orthogonality: $\vec{PF}=(1, 0, -1)$ and $\vec{PF}\cdot\vec{u}=(1)(2)+(0)(-1)+(-1)(2)=0$ . The distance from $P$ to $\ell$ is $\|\vec{PF}\|=\sqrt{1+0+1}=\sqrt{2}\approx 1.41$ .

d)

As with the plane, the foot $F$ is the midpoint of $PP'$ , so $P'=2F-P=(4, 0, 4)-(1, 0, 3)=(3, 0, 1)$ . Check: the midpoint of $P(1, 0, 3)$ and $P'(3, 0, 1)$ is $(2, 0, 2)=F$ . The same principle governs reflection in a plane and in a line: find the foot of the perpendicular, then extend by the same length on the other side.

Exercise 9: Problem: the plane through a line and a point, and the pencil of planes

Consider the line $\ell$ through $A(1, 0, 2)$ with direction vector $\vec{u}=(2, 1, -1)$ , and the point $M(0, 3, 1)$ .

a) Show that $M$ does not lie on $\ell$ .
b) Find a normal vector, then a scalar equation of the plane $\pi$ that contains both the line $\ell$ and the point $M$ .
c) Check that $\vec{u}$ is indeed parallel to $\pi$ , and compute the distance from the point $S(4, 2, 0)$ to $\pi$ .
d) Explain why there are infinitely many planes containing the line $\ell$ , but only one that also contains the point $M$ . What role does the hypothesis “ $M$ is not on $\ell$ ” play here?

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Answers

a) No $t$ works
b) $2x+3y+7z-16=0$
c) $\vec u\cdot\vec n=0$ ; distance $\frac{2}{\sqrt{62}}\approx 0.25$
d) Pencil of planes with axis $\ell$ ; a single plane through $M$

a)

If $M$ were on $\ell$ , there would be a $t$ with $M=A+t\,\vec{u}$ . The first coordinate requires $0=1+2t$ , so $t=-\dfrac{1}{2}$ . But then the second coordinate would be $0+(-\tfrac{1}{2})(1)=-\dfrac{1}{2}\neq 3$ . No value of $t$ works, so $M\notin\ell$ .

b)

The required plane contains the direction $\vec{u}$ of the line and the vector $\vec{AM}=(0-1, 3-0, 1-2)=(-1, 3, -1)$ , which joins a point of the line to the point $M$ . A normal vector is their cross product: $\vec{n}=\vec{u}\times\vec{AM}=((1)(-1)-(-1)(3), (-1)(-1)-(2)(-1), (2)(3)-(1)(-1))=(2, 3, 7)$ . The plane passes through $A(1, 0, 2)$ : $2(x-1)+3(y-0)+7(z-2)=0$ , that is $2x+3y+7z-16=0$ . Check on $M$ : $2(0)+3(3)+7(1)-16=9+7-16=0$ .

c)

$\vec{u}$ is parallel to $\pi$ if and only if it is orthogonal to the normal: $\vec{u}\cdot\vec{n}=(2)(2)+(1)(3)+(-1)(7)=4+3-7=0$ . That is the case, which confirms that the whole line lies in the plane (it has one point there, $A$ , and a parallel direction). For the distance from $S(4, 2, 0)$ : $\dfrac{|2(4)+3(2)+7(0)-16|}{\sqrt{2^{2}+3^{2}+7^{2}}}=\dfrac{|8+6-16|}{\sqrt{62}}=\dfrac{2}{\sqrt{62}}\approx 0.25$ .

d)

Every plane containing $\ell$ must contain its direction $\vec{u}$ and one of its points: by rotating about the line, we get infinitely many of them, the pencil of planes with axis $\ell$ . To single out one plane, an extra condition is needed, and a point off the line is one: exactly one plane of the pencil passes through $M\notin\ell$ . The hypothesis “ $M$ is not on $\ell$ ” is essential: if $M$ were on the line, it would add no information and every plane of the pencil would contain it, without singling out any. The cross product $\vec{u}\times\vec{AM}$ would also be zero, and could not provide a normal.

Exercise 10: Problem: bisector planes and the set of points equidistant from two planes

Consider the two planes $\pi_{1} : 2x-y+2z=0$ and $\pi_{2} : x+2y-2z=0$ . Note first that their normal vectors have the same norm, which simplifies everything that follows.

a) Check that $\|\vec{n_{1}}\|=\|\vec{n_{2}}\|$ , then compute the distance from the point $Q(3, 1, 0)$ to each of the two planes. What do you notice?
b) Find the set of points in space equidistant from $\pi_{1}$ and $\pi_{2}$ . Show that it consists of two planes, and give their equations.
c) Prove that these two bisector planes are perpendicular to each other.
d) Which of the two bisector planes contains the point $Q$ from part a)? Interpret geometrically.

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Answers

a) Norms 3; $\frac53$ and $\frac53$
b) $x-3y+4z=0$ and $3x+y=0$
c) Orthogonal normals
d) $Q$ on $x-3y+4z=0$

a)

$\vec{n_{1}}=(2, -1, 2)$ and $\vec{n_{2}}=(1, 2, -2)$ have norms $\|\vec{n_{1}}\|=\sqrt{4+1+4}=3$ and $\|\vec{n_{2}}\|=\sqrt{1+4+4}=3$ : they are equal. For $Q(3, 1, 0)$ : $\text{dist}(Q,\pi_{1})=\dfrac{|2(3)-1+0|}{3}=\dfrac{5}{3}$ and $\text{dist}(Q,\pi_{2})=\dfrac{|3+2(1)-0|}{3}=\dfrac{5}{3}$ . The point $Q$ is equidistant from the two planes.

b)

A point $(x, y, z)$ is equidistant from the two planes when $\dfrac{|2x-y+2z|}{3}=\dfrac{|x+2y-2z|}{3}$ , that is $|2x-y+2z|=|x+2y-2z|$ . Since the denominators are equal, they cancel, and an equality of two absolute values gives two cases. Case $+$ : $2x-y+2z=x+2y-2z$ , that is $x-3y+4z=0$ . Case $-$ : $2x-y+2z=-(x+2y-2z)$ , that is $3x+y=0$ . The equidistant set is therefore the union of the two planes $x-3y+4z=0$ and $3x+y=0$ , called the bisector planes of the two original planes.

c)

The first bisector plane has normal $\vec{a}=(1, -3, 4)$ and the second $\vec{b}=(3, 1, 0)$ . Their dot product is $\vec{a}\cdot\vec{b}=(1)(3)+(-3)(1)+(4)(0)=3-3+0=0$ . The normals are orthogonal, so the two bisector planes are perpendicular. This is a general fact: the two bisector planes of two intersecting planes are always perpendicular to each other, exactly like the two angle bisectors of two intersecting lines in the plane.

d)

Test $Q(3, 1, 0)$ in both equations. First bisector plane: $3-3(1)+4(0)=0$ , satisfied. Second bisector plane: $3(3)+1=10\neq 0$ , not satisfied. The point $Q$ therefore lies on the bisector plane $x-3y+4z=0$ . Interpretation: this plane is the set of points inside one of the two dihedral angles formed by $\pi_{1}$ and $\pi_{2}$ , at equal distance from its two faces; the other bisector plane plays the same role for the supplementary dihedral angle.

Exercise 11: The angle between a line and a plane

Consider the line $d$ through the origin $O$ with direction vector $\vec{u}=(1, 2, 2)$ , and the plane $\pi : 2x-y+2z=9$ with normal vector $\vec{n}=(2, -1, 2)$ .

a) Compute the angle between $\vec{u}$ and $\vec{n}$ , to the nearest tenth of a degree.
b) Deduce the angle between the line $d$ and the plane $\pi$ .
c) Find the point of intersection of $d$ and $\pi$ .
d) Find a direction vector, with integer components, of the orthogonal projection of $d$ onto $\pi$ .
e) For which value of $k$ is the line with direction $(1, 2, k)$ parallel to $\pi$ ?

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Answers

a) $63.6^\circ$
b) $26.4^\circ$
c) $\left(\frac94, \frac92, \frac92\right)$
d) $(1, 22, 10)$
e) $k=0$

a)

$\vec{u}\cdot\vec{n}=2-2+4=4$ , $\|\vec{u}\|=\|\vec{n}\|=3$ , so $\cos\theta=\frac{4}{9}$ and $\theta\approx 63.6^{\circ}$ .

b)

The angle between a line and a plane is the angle between the line and its projection onto the plane: it is the COMPLEMENT of the angle with the normal. $\varphi=90^{\circ}-63.6^{\circ}\approx 26.4^{\circ}$ , or directly $\sin\varphi=\frac{|\vec{u}\cdot\vec{n}|}{\|\vec{u}\|\,\|\vec{n}\|}=\frac{4}{9}$ .

Pitfall: answering $63.6^{\circ}$ is the classic error, since it is the angle with the normal.

c)

The point $(t, 2t, 2t)$ satisfies $2t-2t+4t=9$ , so $t=\frac{9}{4}$ : the point of intersection is $\left(\frac{9}{4}, \frac{9}{2}, \frac{9}{2}\right)=(2.25, 4.5, 4.5)$ .

d)

Remove from $\vec{u}$ its component along the normal: $\vec{u}-\frac{\vec{u}\cdot\vec{n}}{\|\vec{n}\|^{2}}\vec{n}=(1, 2, 2)-\frac{4}{9}(2, -1, 2)=\left(\frac{1}{9}, \frac{22}{9}, \frac{10}{9}\right)$ , parallel to $(1, 22, 10)$ . Check: $(1, 22, 10)\cdot\vec{n}=2-22+20=0$ , so the direction does lie in the plane.

e)

Parallel to the plane means orthogonal to the normal: $2-2+2k=0$ , so $k=0$ .

The thread: between a line and a plane, compute with the normal, then take the complement; the sine replaces the cosine.

Exercise 12: Three planes: a point, a line or nothing

Consider the planes $\pi_{1} : x+y+z=6$ and $\pi_{2} : x-y+2z=5$ , and three possible planes to complete the system: $\pi_{3} : x+2y-z=2$ , $\pi_{3}' : 2x+3z=11$ and $\pi_{3}'' : 2x+3z=12$ .

a) Show that $\pi_{1}$ , $\pi_{2}$ and $\pi_{3}$ have exactly one point in common, and find it.
b) Compute the determinant of the normals of $\pi_{1}$ , $\pi_{2}$ and $\pi_{3}'$ . Show that these three planes share a common line, and give a direction vector of it.
c) Find the point of this common line that lies in the plane $z=0$ .
d) Do the planes $\pi_{1}$ , $\pi_{2}$ and $\pi_{3}''$ have a point in common? Describe their configuration.
e) For which values of $k$ do the planes $\pi_{1}$ , $\pi_{2}$ and $2x+3z=k$ have infinitely many points in common?

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Answers

a) Point $(1, 2, 3)$
b) Zero determinant; line with direction $(3, -1, -2)$
c) $(5.5, 0.5, 0)$
d) No point: prism
e) $k=11$

a)

$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ 1 & 2 & -1 \end{vmatrix}=1(1-4)-1(-1-2)+1(2+1)=3\neq 0$ : the normals are not coplanar, so the system has a unique solution. Adding $\pi_{1}$ and $\pi_{3}$ : $2x+3y=8$ ; subtracting $\pi_{2}$ from $\pi_{1}$ : $2y-z=1$ , so $z=2y-1$ , and $\pi_{1}$ gives $x=7-3y$ . Then $14-6y+3y=8$ , so $y=2$ , $x=1$ , $z=3$ . The common point is $(1, 2, 3)$ ; check in $\pi_{2}$ : $1-2+6=5$ .

b)

$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ 2 & 0 & 3 \end{vmatrix}=1(-3)-1(3-4)+1(0+2)=0$ . Indeed $\pi_{1}+\pi_{2}$ gives exactly $2x+3z=11$ : every point common to $\pi_{1}$ and $\pi_{2}$ lies on $\pi_{3}'$ . The three planes therefore contain the line $\pi_{1}\cap\pi_{2}$ , with direction vector $\vec{n_{1}}\times\vec{n_{2}}=(1\cdot 2-1\cdot(-1), 1\cdot 1-1\cdot 2, 1\cdot(-1)-1\cdot 1)=(3, -1, -2)$ .

c)

With $z=0$ : $x+y=6$ and $x-y=5$ , so $x=5.5$ and $y=0.5$ . The point is $(5.5, 0.5, 0)$ , and it does satisfy $2(5.5)+0=11$ .

d)

No: a point common to $\pi_{1}$ and $\pi_{2}$ satisfies $2x+3z=11$ , so never $2x+3z=12$ . There is no common point. Since the normals are the same as in b), the planes meet in pairs along three parallel lines with direction $(3, -1, -2)$ : the configuration is a PRISM.

e)

The equation must follow from the other two, so $k=11$ : it is the only value, the case of b). For any other value of $k$ , we are in the prism case, since the determinant is zero whatever $k$ is.

The thread: a non-zero determinant gives a point; when it is zero, it remains to see whether the third equation follows from the other two (a common line) or contradicts them (a prism).

Exercise 13: The common perpendicular of two skew lines

Consider the line $d_{1}$ given by $(t, t, 0)$ and the line $d_{2}$ given by $(1, s, 2+s)$ , for real $t$ and $s$ . Let $\vec{u}=(1, 1, 0)$ and $\vec{v}=(0, 1, 1)$ be their direction vectors.

a) Show that $d_{1}$ and $d_{2}$ are skew.
b) We look for the point $P$ of $d_{1}$ and the point $Q$ of $d_{2}$ such that $\vec{PQ}$ is orthogonal to $\vec{u}$ and to $\vec{v}$ . Write the two conditions and find $t$ and $s$ .
c) Give the coordinates of $P$ and $Q$ .
d) Compute $PQ$ to three decimal places, and check the result with the scalar triple product formula.
e) Show that $\vec{PQ}$ is parallel to $\vec{u}\times\vec{v}$ . Why is this expected?

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Answers

a) Scalar triple product 3: skew
b) $t=0$ , $s=-1$
c) $P(0, 0, 0)$ , $Q(1, -1, 1)$
d) $PQ=\sqrt3\approx 1.732$
e) $\vec{PQ}=\vec u\times\vec v$

a)

$\vec{u}$ and $\vec{v}$ are not parallel, so the lines are not parallel. With $A=(0, 0, 0)$ on $d_{1}$ and $C=(1, 0, 2)$ on $d_{2}$ : $\vec{u}\times\vec{v}=(1\cdot 1-0\cdot 1, 0\cdot 0-1\cdot 1, 1\cdot 1-1\cdot 0)=(1, -1, 1)$ and $\vec{AC}\cdot(\vec{u}\times\vec{v})=1+0+2=3\neq 0$ : they are not coplanar, hence skew.

b)

$\vec{PQ}=(1-t, s-t, 2+s)$ . $\vec{PQ}\cdot\vec{u}=1-t+s-t=1+s-2t=0$ and $\vec{PQ}\cdot\vec{v}=s-t+2+s=2s-t+2=0$ . The second gives $t=2s+2$ , and the first $1+s-4s-4=0$ , so $s=-1$ and then $t=0$ .

c)

$P=(0, 0, 0)$ and $Q=(1, -1, 1)$ .

d)

$\vec{PQ}=(1, -1, 1)$ and $PQ=\sqrt{3}\approx 1.732$ . With the scalar triple product: $\frac{|\vec{AC}\cdot(\vec{u}\times\vec{v})|}{\|\vec{u}\times\vec{v}\|}=\frac{3}{\sqrt{3}}=\sqrt{3}$ . The two computations agree.

e)

$\vec{PQ}=(1, -1, 1)=\vec{u}\times\vec{v}$ . This is expected: $\vec{PQ}$ is orthogonal to both directions, and in space the vectors orthogonal to two non-parallel vectors are exactly the multiples of their cross product. The common perpendicular is the only line that meets both lines at right angles, and $PQ$ is the shortest distance between them.

The thread: the distance between two skew lines comes from the scalar triple product; the common perpendicular is found by imposing two orthogonality conditions.

Exercise 14: Problem: the reflection of a light ray in a plane mirror

A mirror occupies the plane $\pi : x+y+z=3$ , with normal vector $\vec{n}=(1, 1, 1)$ . A light ray leaves the source $S(0, 0, 6)$ in the direction $\vec{d}=(1, 2, -5)$ .

On reflection, the component of the ray along the normal changes sign and the component parallel to the mirror is kept: $\vec{r}=\vec{d}-2\frac{\vec{d}\cdot\vec{n}}{\|\vec{n}\|^{2}}\vec{n}$ .

The figure shows the mirror, the source and the direction of the ray.

a) Find the point $I$ where the ray hits the mirror.
b) Compute the direction $\vec{r}$ of the reflected ray, then give it with integer components.
c) Check that the incident ray and the reflected ray make the same angle with the normal.
d) Find the reflection $S'$ of the source in the plane of the mirror.
e) Show that the vector $\vec{S'I}$ is parallel to the reflected direction, and interpret.

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Answers

a) $I(1.5, 3, -1.5)$
b) $(7, 10, -11)$
c) Same angle, about $77.8^\circ$
d) $S'(-2, -2, 4)$
e) $\vec{S'I}$ parallel to the reflected ray

a)

The point $(t, 2t, 6-5t)$ satisfies $t+2t+6-5t=3$ , that is $-2t=-3$ and $t=1.5$ : $I=(1.5, 3, -1.5)$ .

b)

$\vec{d}\cdot\vec{n}=1+2-5=-2$ and $\|\vec{n}\|^{2}=3$ , so $\vec{r}=(1, 2, -5)+\frac{4}{3}(1, 1, 1)=\left(\frac{7}{3}, \frac{10}{3}, -\frac{11}{3}\right)$ , parallel to $(7, 10, -11)$ .

c)

Incident: $\frac{|\vec{d}\cdot\vec{n}|}{\|\vec{d}\|\,\|\vec{n}\|}=\frac{2}{\sqrt{30}\sqrt{3}}\approx 0.211$ . Reflected, with $(7, 10, -11)$ : $\frac{|7+10-11|}{\sqrt{270}\sqrt{3}}=\frac{6}{\sqrt{810}}\approx 0.211$ . Same cosines, hence the same angle with the normal, about $77.8^{\circ}$ : this is the law of reflection. The incident ray travels towards the mirror, $\vec{d}\cdot\vec{n}<0$ , and the reflected ray travels away from it, $\vec{r}\cdot\vec{n}>0$ .

d)

$S$ gives $0+0+6-3=3$ , so $S'=S-2\cdot\frac{3}{3}\vec{n}=(0, 0, 6)-(2, 2, 2)=(-2, -2, 4)$ .

e)

$\vec{S'I}=(3.5, 5, -5.5)=\frac{1}{2}(7, 10, -11)$ : the reflected ray lies on the line $S'I$ . Everything happens as if the light came in a straight line from the image $S'$ of the source behind the mirror: that is why we see the object “in” the mirror, as far behind it as it is in front.

The thread: reflecting a vector in a plane means reversing its normal component; reflecting a point means extending the perpendicular by the same length.

Exercise 15: Problem: the line of steepest slope on a roof

A roof face is a rectangle with vertices $A(0, 0, 6)$ , $B(0, 10, 6)$ , $C(4, 0, 4)$ and $D(4, 10, 4)$ , in metres, with the $z$ -axis vertical. The ridge is the edge $AB$ , the gutter the edge $CD$ .

a) Find a scalar equation of the plane of the roof.
b) Compute the angle between the roof and the horizontal plane, to the nearest hundredth of a degree, and the slope as a percentage.
c) Find a direction vector of the horizontal lines of the roof, then one of the line of steepest slope pointing downwards.
d) A drop of water falls on the roof at the point $P(1, 5, 5.5)$ and follows the line of steepest slope. At what point does it reach the gutter?
e) Compute the area of the roof face, to the nearest tenth, and compare it with the area of its projection on the ground.

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Answers

a) $x+2z=12$
b) $26.57^\circ$ ; slope 50%
c) Horizontal lines $(0, 1, 0)$ ; slope $(2, 0, -1)$
d) $(4, 5, 4)$
e) $44.7$ m² versus 40 m²

a)

$\vec{AB}=(0, 10, 0)$ and $\vec{AC}=(4, 0, -2)$ , $\vec{AB}\times\vec{AC}=(10\cdot(-2)-0\cdot 0, 0\cdot 4-0\cdot(-2), 0\cdot 0-10\cdot 4)=(-20, 0, -40)$ , parallel to $\vec{n}=(1, 0, 2)$ . Through $A$ : $x+2(z-6)=0$ , that is $x+2z=12$ . Check on $D$ : $4+8=12$ .

b)

The angle between two planes is the angle between their normals: $\cos\theta=\frac{|\vec{n}\cdot\vec{k}|}{\|\vec{n}\|}=\frac{2}{\sqrt{5}}$ , so $\theta\approx 26.57^{\circ}$ . The slope is $\tan\theta=\frac{1}{2}$ : the roof drops $2\ \mathrm{m}$ over $4\ \mathrm{m}$ horizontally, a slope of $50\%$ .

c)

A horizontal line of the roof is orthogonal to $\vec{n}$ and to $\vec{k}=(0, 0, 1)$ : $\vec{n}\times\vec{k}=(0\cdot 1-2\cdot 0, 2\cdot 0-1\cdot 1, 0)=(0, -1, 0)$ , the direction of the ridge. The line of steepest slope lies in the roof and is orthogonal to the horizontal lines: $\vec{n}\times(0, 1, 0)=(0\cdot 0-2\cdot 1, 2\cdot 0-1\cdot 0, 1\cdot 1-0\cdot 0)=(-2, 0, 1)$ , which points up; downwards, $(2, 0, -1)$ .

d)

$P$ is indeed on the roof: $1+11=12$ . The drop follows $(1+2t, 5, 5.5-t)$ and reaches the gutter at $z=4$ , so $t=1.5$ : the point $(4, 5, 4)$ , in the middle of the gutter. It does not drift in $y$ : the steepest slope is perpendicular to the ridge.

e)

$AB=10$ and $AC=\sqrt{16+4}=\sqrt{20}\approx 4.47$ , so the area is $10\sqrt{20}\approx 44.7\ \mathrm{m^{2}}$ . The projection on the ground is a $10$ by $4$ rectangle, that is $40\ \mathrm{m^{2}}$ , and $44.7\times\cos\theta=44.7\times\frac{2}{\sqrt{5}}\approx 40$ : the projected area is the area multiplied by the cosine of the angle between the planes. It is the area of the roof, not that of the ground, that counts when ordering shingles.

The thread: the normal of a plane gives at once its tilt, its horizontal lines and its line of steepest slope.

Corrected exercises: lines and planes in space (201-SN4-RE)

Course recap

Calculations: equations, intersections, distances and angles

Exercise 0.1: Computing equations, intersections and distances in space: ten direct calculations

Part A: equations, intersections and distances (/50)

Exercise 1: The equations of a line in space

Exercise 2: The scalar equation of a plane

Exercise 3: Relative positions and the intersection of a line and a plane

Exercise 4: Distances in space

Exercise 5: The intersection of two planes and the angle between them

Part B: exam level (/50)

Exercise 6: Two skew lines and the distance between them

Exercise 7: Problem: the paths of two drones

Exercise 8: Orthogonal projection and reflection of a point: in a plane and in a line

Exercise 9: Problem: the plane through a line and a point, and the pencil of planes

Exercise 10: Problem: bisector planes and the set of points equidistant from two planes

Part C: the classics (/50)

Exercise 11: The angle between a line and a plane

Exercise 12: Three planes: a point, a line or nothing

Exercise 13: The common perpendicular of two skew lines

Exercise 14: Problem: the reflection of a light ray in a plane mirror

Exercise 15: Problem: the line of steepest slope on a roof

See also

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