Corrected exercises: cross product, areas and volumes (MATH 133)

a) Expanding along the first row, each unit vector multiplies the minor obtained by deleting its row and column, with the signs $+, -, +$. For $i$: $\det\begin{pmatrix} -1 & 3 \\ 4 & -2 \end{pmatrix} = (-1)(-2) - (3)(4) = 2 - 12 = -10$. For $j$, with its MINUS sign: $-\det\begin{pmatrix} 2 & 3 \\ 1 & -2 \end{pmatrix} = -\big((2)(-2) - (3)(1)\big) = -(-7) = 7$. For $k$: $\det\begin{pmatrix} 2 & -1 \\ 1 & 4 \end{pmatrix} = 8 - (-1) = 9$. So $u \times v = (-10, 7, 9)$. Writing the three minors on the page is not decoration: in MATH 133 the method marks are given on them, and a single wrong entry becomes visible to the marker instead of being buried in one line.

b) The result must be perpendicular to both factors. $u \cdot (-10, 7, 9) = -20 - 7 + 27 = 0$ and $v \cdot (-10, 7, 9) = -10 + 28 - 18 = 0$. Both vanish, so the vector is right. This check is worth its ten seconds because the most common error of the chapter is forgetting the minus sign in front of the $j$ minor. Had we written $(-10, -7, 9)$, the first dot product would give $-20 + 7 + 27 = 14 \neq 0$ and the mistake would be caught before it contaminated a plane equation, an area or a distance further down the page.

c) The cross product is ANTICOMMUTATIVE: $v \times u = -(u \times v) = (10, -7, -9)$. In the determinant, exchanging $u$ and $v$ swaps two rows, and a row swap multiplies a determinant by $-1$. Geometrically the new vector has the same length (same parallelogram) and the opposite direction (the right hand now curls from $v$ to $u$). The order of the factors is therefore never a detail: it decides which side of a plane a normal points to and the sign of a torque.

d) $i \times j = k$, $j \times k = i$, $k \times i = j$: going around the cycle $i \to j \to k \to i$ gives the next vector. Against the cycle the sign flips, so $j \times i = -k$. On the figure, curl the fingers of the right hand from the $x$-axis towards the $y$-axis through the smaller angle: the thumb points along $+z$. The frame is called right-handed precisely because $i \times j = k$ and not $-k$, and all the signs in the component formula of a) come from that convention.

e) Directly: $u + v = (3, 3, 1)$ and $u - v = (1, -5, 5)$, so the product is $\big((3)(5) - (1)(-5),\ (1)(1) - (3)(5),\ (3)(-5) - (3)(1)\big) = (20, -14, -18)$. By the rules: $(u + v) \times (u - v) = u \times u - u \times v + v \times u - v \times v = 0 - u \times v - u \times v - 0 = -2(u \times v)$, and $-2(-10, 7, 9) = (20, -14, -18)$: the two ways agree. The tempting $u \times u - v \times v$ copies the identity $(a + b)(a - b) = a^2 - b^2$, which relies on $ab = ba$. Here the cross terms do NOT cancel, they add, because $v \times u = -u \times v$.

a) $u \times v = \big((2)(4) - (2)(0),\ (2)(3) - (1)(4),\ (1)(0) - (2)(3)\big) = (8, 2, -6)$. Check: $u \cdot (8, 2, -6) = 8 + 4 - 12 = 0$ and $v \cdot (8, 2, -6) = 24 + 0 - 24 = 0$. The area is the length: $\|u \times v\| = \sqrt{64 + 4 + 36} = \sqrt{104} = 2\sqrt{26}$. The vector itself is not the area. An answer written as $(8, 2, -6)$ to a question that asks for an area answers with the wrong half of the product and loses the mark, even though every number in it is right.

b) $\|u\|^2 = 1 + 4 + 4 = 9$, $\|v\|^2 = 9 + 0 + 16 = 25$ and $u \cdot v = 3 + 0 + 8 = 11$. Lagrange gives $9 \times 25 - 11^2 = 225 - 121 = 104$, the same $104$ as in a). The identity is nothing but $\sin^2\theta = 1 - \cos^2\theta$ multiplied by $\|u\|^2\|v\|^2$, and it is the fastest independent check of a cross product length: it uses only dot products, so it shares no arithmetic with the determinant.

c) $\cos\theta = \dfrac{u \cdot v}{\|u\|\|v\|} = \dfrac{11}{3 \times 5} = \dfrac{11}{15}$, and $\sin\theta = \dfrac{\|u \times v\|}{\|u\|\|v\|} = \dfrac{2\sqrt{26}}{15}$. Then $\sin^2\theta + \cos^2\theta = \dfrac{104 + 121}{225} = \dfrac{225}{225} = 1$. Note that $\sin\theta$ comes out positive with no choice of sign: the angle between two vectors lies in $[0, \pi]$, where the sine is never negative, which is why a LENGTH can be written as $\|u\|\|v\|\sin\theta$ without an absolute value.

d) Two vectors are parallel exactly when their cross product is the zero vector, since the parallelogram then collapses to a segment. $u \times w = \big((2)(-4) - (2)(-4),\ (2)(-2) - (1)(-4),\ (1)(-4) - (2)(-2)\big) = (0, 0, 0)$: indeed $w = -2u$. $u \times t = \big((2)(5) - (2)(4),\ (2)(2) - (1)(5),\ (1)(4) - (2)(2)\big) = (2, -1, 0) \neq 0$: $t$ looks almost like $2u = (2, 4, 4)$, but one entry differs and that is enough. Comparing ratios entry by entry works too, but it breaks down as soon as a component is zero, whereas the cross product test never does.

e) The sides are $\frac{1}{2}(p + q) = (3, 0, 4)$ and $\frac{1}{2}(p - q) = (1, 2, 2)$, the very $v$ and $u$ of a). With $p = u + v$ and $q = v - u$, $p \times q = u \times v - v \times u = 2(u \times v)$ by anticommutativity, so the area is $\frac{1}{2}\|p \times q\|$. Here $p \times q = \big((2)(2) - (6)(-2),\ (6)(2) - (4)(2),\ (4)(-2) - (2)(2)\big) = (16, 4, -12)$, of length $4\sqrt{26}$, and the area is $2\sqrt{26}$, as in a). Taking $\|p \times q\|$ directly doubles the area: the cross product measures the parallelogram built on the two vectors you feed it, and the diagonals build a parallelogram twice as large.

a) $\overrightarrow{AB} = B - A = (2, 1, -1)$ and $\overrightarrow{AC} = C - A = (1, 3, 2)$. Then $\overrightarrow{AB} \times \overrightarrow{AC} = \big((1)(2) - (-1)(3),\ (-1)(1) - (2)(2),\ (2)(3) - (1)(1)\big) = (5, -5, 5)$. Check: $(2, 1, -1) \cdot (5, -5, 5) = 10 - 5 - 5 = 0$ and $(1, 3, 2) \cdot (5, -5, 5) = 5 - 15 + 10 = 0$. Its length is $\sqrt{75} = 5\sqrt{3}$, which is the area of the PARALLELOGRAM, so the triangle has area $\frac{5\sqrt{3}}{2}$. Forgetting the half is the cheapest mark to lose in the chapter and one of the most frequent.

b) The area of the triangle is also $\frac{1}{2} \times$ base $\times$ height with base $\|\overrightarrow{AB}\| = \sqrt{4 + 1 + 1} = \sqrt{6}$. So $h = \dfrac{\|\overrightarrow{AB} \times \overrightarrow{AC}\|}{\|\overrightarrow{AB}\|} = \dfrac{5\sqrt{3}}{\sqrt{6}} = \dfrac{5}{\sqrt{2}} = \dfrac{5\sqrt{2}}{2}$. This is exactly the formula for the distance from a point to a line that exercise 7 uses: the length of the cross product is the area, and dividing by the base leaves the height. Check: $\frac{1}{2}\sqrt{6} \times \frac{5}{\sqrt{2}} = \frac{5\sqrt{3}}{2}$, the area of a).

c) $\overrightarrow{OA} \times \overrightarrow{OC} = (1, 0, 2) \times (2, 3, 4) = \big((0)(4) - (2)(3),\ (2)(2) - (1)(4),\ (1)(3) - (0)(2)\big) = (-6, 0, 3)$, of length $\sqrt{45} = 3\sqrt{5}$. Half of it, $\frac{3\sqrt{5}}{2}$, is the area of the triangle $OAC$, a different triangle with a vertex at the origin. The proof that it cannot be the area of $ABC$: an area does not change when the whole figure is translated, but position vectors do. Move the origin and $\overrightarrow{OA}$, $\overrightarrow{OC}$ change while $ABC$ does not. Edge vectors are differences of points, and a difference is unchanged by a translation.

d) In the parallelogram $ABCD$, $\overrightarrow{AD} = \overrightarrow{BC}$, so $D = A + (C - B) = (1, 0, 2) + (-1, 2, 3) = (0, 2, 5)$. Check: $\overrightarrow{AD} = (-1, 2, 3) = \overrightarrow{BC}$. The diagonal $AC$ cuts the parallelogram into two copies of triangle $ABC$, so its area is $2 \times \frac{5\sqrt{3}}{2} = 5\sqrt{3}$. The trap here is the order of the letters: $D = A + B - C$ gives a point that makes $ADBC$ a parallelogram, a different quadrilateral with the same area, and the wrong point for the question asked.

e) $\overrightarrow{BA} = (-2, -1, 1)$ and $\overrightarrow{BC} = (-1, 2, 3)$, so $\overrightarrow{BA} \times \overrightarrow{BC} = \big((-1)(3) - (1)(2),\ (1)(-1) - (-2)(3),\ (-2)(2) - (-1)(-1)\big) = (-5, 5, -5)$. It is $-(5, -5, 5)$: same line, opposite direction, same length, hence the same area $\frac{5\sqrt{3}}{2}$. Any vertex works, provided both edges leave from it. What changes with the vertex and the order is at most the SIGN of the vector, and the area only uses its length.

a) $\overrightarrow{PQ} = (2, -2, 1)$ and $\overrightarrow{PR} = (-1, -1, 2)$. Then $\overrightarrow{PQ} \times \overrightarrow{PR} = \big((-2)(2) - (1)(-1),\ (1)(-1) - (2)(2),\ (2)(-1) - (-2)(-1)\big) = (-3, -5, -4)$, and any non-zero multiple is a normal, so take $n = (3, 5, 4)$. The plane is $n \cdot (X - P) = 0$: $3(x - 1) + 5(y - 2) + 4z = 0$, that is $3x + 5y + 4z = 13$. Check with the three points: $3 + 10 + 0 = 13$, $9 + 0 + 4 = 13$, $0 + 5 + 8 = 13$. Checking only $P$ proves nothing, since $P$ was used to build the constant; $Q$ and $R$ are the real test.

b) $\overrightarrow{AB} = (1, 2, -1)$ and $\overrightarrow{AC} = (3, 6, -3) = 3\overrightarrow{AB}$, so $\overrightarrow{AB} \times \overrightarrow{AC} = \big((2)(-3) - (-1)(6),\ (-1)(3) - (1)(-3),\ (1)(6) - (2)(3)\big) = (0, 0, 0)$. The zero vector is not a normal: plugging it in gives $0x + 0y + 0z = 0$, which is all of space. The three points are COLLINEAR, and infinitely many planes contain the line through them, like the pages of a book around its spine. The right answer is that sentence, not an equation. A cross product that comes out zero is information, never a computing accident to be patched.

c) First make sure $P$ is not on $L$: $x = 2$ forces $t = 1$, which gives $(2, 1, 1) \neq (2, 3, 1)$. The plane contains the direction $d = (1, 1, -1)$ and the vector from a point of $L$ to $P$, $\overrightarrow{AP} = (2, 3, 1) - (1, 0, 2) = (1, 3, -1)$. Then $d \times \overrightarrow{AP} = \big((1)(-1) - (-1)(3),\ (-1)(1) - (1)(-1),\ (1)(3) - (1)(1)\big) = (2, 0, 2)$, so $n = (1, 0, 1)$ and the plane is $x + z = 1 + 2 = 3$. Check: $P$ gives $2 + 1 = 3$ and $n \cdot d = 1 - 1 = 0$, so the whole line lies in it. The student's $\overrightarrow{OP} = (2, 3, 1)$ goes from the ORIGIN to $P$, and the origin is not on $L$: $d \times \overrightarrow{OP} = (4, -3, 1)$ gives the plane $4x - 3y + z = 0$ through $P$, and the point $(1, 0, 2)$ of $L$ gives $4 + 2 = 6 \neq 0$. Both vectors fed to the cross product must lie IN the plane.

d) $\overrightarrow{AB} = (-2, 3, 0)$ and $\overrightarrow{AC} = (-2, 0, 6)$, so the cross product is $\big((3)(6) - (0)(0),\ (0)(-2) - (-2)(6),\ (-2)(0) - (3)(-2)\big) = (18, 12, 6) = 6(3, 2, 1)$. With $n = (3, 2, 1)$ through $(2, 0, 0)$: $3x + 2y + z = 6$. Dividing by $6$ gives exactly $\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1$, the intercept form: each denominator is where the plane cuts an axis. Dividing the cross product by $6$ first was not required, but it keeps the numbers small, and small numbers are fewer chances of an arithmetic slip without a calculator.

e) The first classmate is right: $-3x - 5y - 4z = -13$ is $3x + 5y + 4z = 13$ multiplied by $-1$, the same set of points. A normal is only defined up to a non-zero factor, and the order of the two edge vectors, $\overrightarrow{PQ} \times \overrightarrow{PR}$ or $\overrightarrow{PR} \times \overrightarrow{PQ}$, only flips its sign. The second classmate kept the right normal but forgot to put a point in: $3x + 5y + 4z = 0$ is the parallel plane through the origin, and $P$ gives $13 \neq 0$. The normal fixes the ORIENTATION of the plane, the point fixes its POSITION, and the question needs both.

a) $v \times w = \big((2)(3) - (1)(0),\ (1)(1) - (0)(3),\ (0)(0) - (2)(1)\big) = (6, 1, -2)$, and $u \cdot (6, 1, -2) = 6 + 1 + 0 = 7$. As a determinant with rows $u$, $v$, $w$, expanded along the first row because it holds a zero: $1\det\begin{pmatrix} 2 & 1 \\ 0 & 3 \end{pmatrix} - 1\det\begin{pmatrix} 0 & 1 \\ 1 & 3 \end{pmatrix} + 0 = 1(6) - 1(0 - 1) = 7$. The two computations are the same computation: the $i$, $j$, $k$ row of the symbolic determinant has simply been replaced by $u$. The volume is $7$.

b) $u \times w = \big((1)(3) - (0)(0),\ (0)(1) - (1)(3),\ (1)(0) - (1)(1)\big) = (3, -3, -1)$ and $v \cdot (3, -3, -1) = 0 - 6 - 1 = -7$. The triple $(v, u, w)$ is $(u, v, w)$ with two vectors exchanged, which swaps two rows of the determinant and flips the sign. The cyclic orders $u, v, w$ and $v, w, u$ and $w, u, v$ all give $+7$; the others give $-7$. The sign records the ORIENTATION of the triple (right-handed or not), which a volume does not care about, so a volume is always $|u \cdot (v \times w)|$.

c) The tetrahedron on $O$, $u$, $v$, $w$ has volume $\frac{1}{3} \times$ base $\times$ height. Its base is the TRIANGLE on $v$ and $w$, half of the parallelogram, and its height is the height of the box. So it is $\frac{1}{3} \times \frac{1}{2} = \frac{1}{6}$ of the parallelepiped: $\frac{7}{6}$. The two factors come from two different places, the pyramid giving a third and the triangular base giving a half. Writing $\frac{7}{3}$ keeps the first and forgets the second, and it is the most common wrong answer to this question.

d) Four points are coplanar exactly when the three edges from one of them span zero volume. $\overrightarrow{AB} = (1, 1, 2)$, $\overrightarrow{AC} = (2, -1, 1)$, $\overrightarrow{AD} = (3, 2, 5)$. $\overrightarrow{AB} \times \overrightarrow{AC} = \big((1)(1) - (2)(-1),\ (2)(2) - (1)(1),\ (1)(-1) - (1)(2)\big) = (3, 3, -3)$ and $\overrightarrow{AD} \cdot (3, 3, -3) = 9 + 6 - 15 = 0$: the points ARE coplanar. The normal $(3, 3, -3)$, or $(1, 1, -1)$, through $A$ gives $x + y - z = 0$, and all four points check: $1 + 0 - 1$, $2 + 1 - 3$, $3 - 1 - 2$ and $4 + 2 - 6$ are all $0$.

e) $\overrightarrow{AD'} = (3, 2, 6)$, so $\overrightarrow{AD'} \cdot (3, 3, -3) = 9 + 6 - 18 = -3$. The points are no longer coplanar, and the tetrahedron has volume $\frac{|-3|}{6} = \frac{1}{2}$. A useful check that costs nothing: when all the vertices have integer coordinates, six times the volume of a tetrahedron is always an INTEGER, because the triple product of integer vectors is an integer. An answer like $\frac{2}{5}$ for such a tetrahedron is wrong on sight.

a) $n_1 = (1, -1, 1)$ and $n_2 = (2, 1, -1)$. Then $n_1 \times n_2 = \big((-1)(-1) - (1)(1),\ (1)(2) - (1)(-1),\ (1)(1) - (-1)(2)\big) = (0, 3, 3)$, and dividing by $3$ gives $d = (0, 1, 1)$. Check: $n_1 \cdot d = 0 - 1 + 1 = 0$ and $n_2 \cdot d = 0 + 1 - 1 = 0$. The reasoning deserves one sentence on the copy: a vector lies in a plane exactly when it is perpendicular to its normal, so a direction common to both planes is perpendicular to both normals, and $n_1 \times n_2$ is built to be that. The alternative, row reducing the $2 \times 3$ system and reading the free variable, gives the same line and takes longer.

b) With $x = 0$ the equations become $-y + z = 2$ and $y - z = 1$; adding them gives $0 = 3$, a contradiction. No point of the line has $x = 0$: adding the original equations gives $3x = 3$, so $x = 1$ at EVERY point of the line. The direction vector said it in advance, since its first component is $0$: moving along the line never changes $x$. The rule to keep is to set to zero a variable whose component in $d$ is NOT zero. Here take $z = 0$: $x - y = 2$ and $2x + y = 1$ give $3x = 3$, $x = 1$, $y = -1$. The point is $(1, -1, 0)$, and it checks in both: $1 + 1 + 0 = 2$ and $2 - 1 - 0 = 1$.

c) $(x, y, z) = (1, -1, 0) + t(0, 1, 1) = (1, -1 + t, t)$. Substituting a general point: $\Pi_1$ gives $1 - (-1 + t) + t = 2$ and $\Pi_2$ gives $2 + (-1 + t) - t = 1$, for EVERY $t$, the $t$ cancelling each time. Checking with a general $t$ rather than one value of $t$ is what proves the whole line lies in both planes; a single point would only show that the line touches them.

d) This is the plane through a line and a point, the gesture of exercise 4 c). The plane contains $d = (0, 1, 1)$ and the vector from $A = (1, -1, 0)$ on the line to the origin, $\overrightarrow{AO} = (-1, 1, 0)$. Then $d \times \overrightarrow{AO} = \big((1)(0) - (1)(1),\ (1)(-1) - (0)(0),\ (0)(1) - (1)(-1)\big) = (-1, -1, 1)$, so $n = (1, 1, -1)$ and, through the origin, $x + y - z = 0$. Check: $A$ gives $1 - 1 - 0 = 0$, and $n \cdot d = 0 + 1 - 1 = 0$, so the line lies in it.