Practice / Linear algebra

Vectors in 3-D: dot and cross products, lines and planes

Ten problems on vectors in three dimensions: lengths, the dot product and the angle between two vectors, work and the component along a direction, the cross product and the area of a parallelogram, a triangle's area and the plane through three points, the volume of a parallelepiped from a determinant, a line meeting a plane, the distance from a point to a plane, the angle and line of intersection of two planes, the distance between skew lines and a linear classifier's signed distances, with worked solutions and the mistakes of the middle sign of the cross product, dividing by squared lengths, forgetting to divide by |n|, a plane through the wrong point and the product of lengths as a volume.

Before you start

Two products do most of the geometry in three dimensions. The dot product $u \cdot v$ measures how much two vectors point the same way: it gives lengths, angles, projections and the work done by a force, and a zero dot product means perpendicular. The cross product $u \times v$ builds a third vector perpendicular to both, whose length is the area of the parallelogram they span. A plane is described by a point and a normal vector, a line by a point and a direction, and every distance between them is a dot product with a unit vector. The same formulas reach machine learning unchanged: cosine similarity is the angle formula, and a linear classifier's decision boundary $w \cdot x + b = 0$ is a plane with normal $w$ These ten problems compute angles, areas and volumes, write lines and planes, find where they meet and how far apart they are, and end with the signed distance of a point to a decision boundary.

For $u = (u_1, u_2, u_3)$ and $v = (v_1, v_2, v_3)$ $u \cdot v = u_1v_1 + u_2v_2 + u_3v_3$ and $\|u\| = \sqrt{u \cdot u}$
Angle. $u \cdot v = \|u\|\,\|v\|\cos\theta$ so $\cos\theta = \dfrac{u \cdot v}{\|u\|\,\|v\|}$ with $0 \le \theta \le \pi$ $u \cdot v = 0$ exactly when $u \perp v$ (or one of them is $0$
Component and projection. The component of $v$ along $u$ is $\dfrac{u \cdot v}{\|u\|}$ and the projection is $\dfrac{u \cdot v}{u \cdot u}\,u$
Cross product. $u \times v = (u_2v_3 - u_3v_2,\ u_3v_1 - u_1v_3,\ u_1v_2 - u_2v_1)$ It is perpendicular to $u$ and $v$ follows the right-hand rule, $v \times u = -(u \times v)$ and $\|u \times v\| = \|u\|\,\|v\|\sin\theta$ is the area of the parallelogram on $u$ and $v$
Triple product. $a \cdot (b \times c) = \det[a\ b\ c]$ the determinant with $a, b, c$ as rows, and $|a \cdot (b \times c)|$ is the volume of the parallelepiped on $a, b, c$
Lines and planes. The line through $P$ with direction $d$ is $r(t) = P + td$ The plane through $P_0$ with normal $n = (a, b, c)$ is $n \cdot (x - P_0) = 0$ that is $ax + by + cz = n \cdot P_0$ The distance from a point $Q$ to the plane $n \cdot x = k$ is $\dfrac{|n \cdot Q - k|}{\|n\|}$

Problems

·
For $u = (1, 2, 2)$ and $v = (3, 0, 4)$ find $\|u\|$ $\|v\|$ $u \cdot v$ and the angle between them.
·
A force $F = (2, -1, 3)$ moves an object along the displacement $d = (4, 2, 1)$ Find the work done, the component of $F$ along $d$ and the projection of $F$ onto $d$
·
For $u = (1, 2, 2)$ and $v = (3, 0, 4)$ find $u \times v$ check that it is perpendicular to both, and find the area of the parallelogram and of the triangle on $u$ and $v$
··
Find the area of the triangle with vertices $A(1, 0, 0)$ $B(0, 2, 0)$ $C(0, 0, 3)$ and the equation of the plane through them.
··
Find the volume of the parallelepiped on $a = (1, 0, 1)$ $b = (2, 1, 0)$ $c = (0, 1, 3)$ and the volume of the tetrahedron with the same three edges from one vertex.
··
Find the line through $P(1, 2, 3)$ and $Q(4, 0, 5)$ and the point where it meets the plane $x + y + z = 12$
··
Find the distance from $Q(2, 1, 3)$ to the plane $2x - y + 2z = 4$ and the point of the plane nearest to $Q$
··
Find the angle between the planes $x + y + z = 1$ and $x - y + 2z = 3$ and their line of intersection.
···
Show that the lines $L_1: (1, 0, 0) + s(1, 1, 0)$ and $L_2: (0, 0, 2) + t(0, 1, 1)$ do not meet and are not parallel, and find the distance between them.
···
A linear classifier predicts the positive class when $w \cdot x + b > 0$ with $w = (2, -1, 2)$ and $b = -3$ Find the signed distances of $x_1 = (2, 0, 1)$ and $x_2 = (0, 1, 0)$ to the decision boundary, and the point of the boundary nearest to $x_1$

Answers

$\|u\| = 3$ $\|v\| = 5$ $u \cdot v = 11$ $\theta \approx 42.83°$
$W = 9$ component $\tfrac{9}{\sqrt{21}} \approx 1.964$ projection $\big(\tfrac{12}{7}, \tfrac67, \tfrac37\big)$
$u \times v = (8, 2, -6)$ parallelogram area $2\sqrt{26} \approx 10.20$ triangle area $\sqrt{26} \approx 5.10$
Area $\tfrac72$ plane $6x + 3y + 2z = 6$
Parallelepiped $5$ tetrahedron $\tfrac56$
$r(t) = (1, 2, 3) + t(3, -2, 2)$ it meets the plane at $(7, -2, 7)$
Distance $\tfrac53$ nearest point $\big(\tfrac89, \tfrac{14}{9}, \tfrac{17}{9}\big)$
$\theta \approx 61.87°$ line $(2, -1, 0) + t(3, -1, -2)$
Distance $\dfrac{1}{\sqrt3} \approx 0.5774$
Signed distances $1$ and $-\tfrac43$ nearest boundary point $\big(\tfrac43, \tfrac13, \tfrac13\big)$

Worked solutions

Problem 1

For $u = (1, 2, 2)$ and $v = (3, 0, 4)$ find $\|u\|$ $\|v\|$ $u \cdot v$ and the angle between them.

$\|u\| = \sqrt{1 + 4 + 4} = 3$ and $\|v\| = \sqrt{9 + 0 + 16} = 5$ Square, add, take the root.
$u \cdot v = 3 + 0 + 8 = 11$ Multiply matching entries and add.
$\cos\theta = \dfrac{11}{3\cdot5} = \dfrac{11}{15} \approx 0.7333$ Divide by the product of the lengths, not their squares.
$\theta = \arccos(0.7333) \approx 0.7476$ radA positive dot product means an acute angle.
$\|u\| = 3$ $\|v\| = 5$ $u \cdot v = 11$ $\theta \approx 42.83°$ In machine learning $\tfrac{11}{15}$ would be the cosine similarity of $u$ and $v$ Dividing by $\|u\|^2\|v\|^2$ instead gives $87.2°$ (Mistake 2).

Problem 2

A force $F = (2, -1, 3)$ moves an object along the displacement $d = (4, 2, 1)$ Find the work done, the component of $F$ along $d$ and the projection of $F$ onto $d$

$W = F \cdot d = 8 - 2 + 3 = 9$ Work is force times displacement, counting only the part of the force along the motion.
$\|d\| = \sqrt{16 + 4 + 1} = \sqrt{21} \approx 4.583$ Needed to turn $d$ into a unit vector.
Component $= \dfrac{F \cdot d}{\|d\|} = \dfrac{9}{\sqrt{21}} \approx 1.964$ The dot product with the unit vector $d/\|d\|$
Projection $= \dfrac{F \cdot d}{d \cdot d}\,d = \dfrac{9}{21}(4, 2, 1) = \dfrac37(4, 2, 1)$ The component times the unit vector, which divides by $\|d\|$ twice.
$W = 9$ component $\tfrac{9}{\sqrt{21}} \approx 1.964$ projection $\big(\tfrac{12}{7}, \tfrac67, \tfrac37\big)$ Check: $F$ minus the projection is $\big(\tfrac27, -\tfrac{13}{7}, \tfrac{18}{7}\big)$ and its dot product with $d$ is $\tfrac{8 - 26 + 18}{7} = 0$

Problem 3

For $u = (1, 2, 2)$ and $v = (3, 0, 4)$ find $u \times v$ check that it is perpendicular to both, and find the area of the parallelogram and of the triangle on $u$ and $v$

First entry: $u_2v_3 - u_3v_2 = 2\cdot4 - 2\cdot0 = 8$ Cover the first column and take the $2\times2$ determinant of what is left.
Second entry: $u_3v_1 - u_1v_3 = 2\cdot3 - 1\cdot4 = 2$ third: $u_1v_2 - u_2v_1 = 0 - 6 = -6$ The middle entry has the order reversed, which is the minus sign of the cofactor expansion (Mistake 1).
$(8, 2, -6)\cdot(1, 2, 2) = 8 + 4 - 12 = 0$ and $(8, 2, -6)\cdot(3, 0, 4) = 24 + 0 - 24 = 0$ The check catches almost every slip.
$\|u \times v\| = \sqrt{64 + 4 + 36} = \sqrt{104} = 2\sqrt{26} \approx 10.198$ Check with Lagrange's identity: $\|u\|^2\|v\|^2 - (u \cdot v)^2 = 225 - 121 = 104$
$u \times v = (8, 2, -6)$ parallelogram area $2\sqrt{26} \approx 10.20$ triangle area $\sqrt{26} \approx 5.10$ The triangle is half the parallelogram.

Problem 4

Find the area of the triangle with vertices $A(1, 0, 0)$ $B(0, 2, 0)$ $C(0, 0, 3)$ and the equation of the plane through them.

$\vec{AB} = (-1, 2, 0)$ and $\vec{AC} = (-1, 0, 3)$ Two edges from the same vertex.
$n = \vec{AB}\times\vec{AC} = (2\cdot3 - 0\cdot0,\ 0\cdot(-1) - (-1)\cdot3,\ (-1)\cdot0 - 2\cdot(-1)) = (6, 3, 2)$ Perpendicular to both edges, so normal to the plane.
Area $= \tfrac12\|n\| = \tfrac12\sqrt{36 + 9 + 4} = \tfrac72$ Half the parallelogram on the two edges.
$6x + 3y + 2z = n \cdot A = 6$ The constant comes from a point on the plane. Check $B$ $6$ $C$ $6$
Area $\tfrac72$ plane $6x + 3y + 2z = 6$ Dividing by $6$ gives the intercept form $\tfrac x1 + \tfrac y2 + \tfrac z3 = 1$ Taking the constant from the vector $\vec{AB}$ instead of a point gives a parallel plane through the origin (Mistake 4).

Problem 5

Find the volume of the parallelepiped on $a = (1, 0, 1)$ $b = (2, 1, 0)$ $c = (0, 1, 3)$ and the volume of the tetrahedron with the same three edges from one vertex.

$\det\begin{bmatrix}1&0&1\\2&1&0\\0&1&3\end{bmatrix} = 1(1\cdot3 - 0\cdot1) - 0(2\cdot3 - 0\cdot0) + 1(2\cdot1 - 1\cdot0)$ Expand along the first row.
$= 3 - 0 + 2 = 5$ The triple product $a \cdot (b \times c)$
Check: $b \times c = (3, -6, 2)$ and $a \cdot (3, -6, 2) = 3 + 0 + 2 = 5$ The same number computed the other way.
The tetrahedron is $\tfrac16$ of the parallelepipedHalf for the triangular base, a third for the pyramid.
Parallelepiped $5$ tetrahedron $\tfrac56$ The sign of the determinant is positive, so $a, b, c$ are a right-handed triple. The product of the lengths, $\sqrt2\cdot\sqrt5\cdot\sqrt{10} = 10$ is the volume only when the edges are perpendicular (Mistake 5).

Problem 6

Find the line through $P(1, 2, 3)$ and $Q(4, 0, 5)$ and the point where it meets the plane $x + y + z = 12$

$d = Q - P = (3, -2, 2)$ The direction is the difference of two points on the line.
$r(t) = (1 + 3t,\ 2 - 2t,\ 3 + 2t)$ $t = 0$ gives $P$ and $t = 1$ gives $Q$
$(1 + 3t) + (2 - 2t) + (3 + 2t) = 6 + 3t = 12$ Substitute the line into the plane: one equation in $t$
$t = 2$ So the meeting point is beyond $Q$
$r(t) = (1, 2, 3) + t(3, -2, 2)$ it meets the plane at $(7, -2, 7)$ Check: $7 - 2 + 7 = 12$ If $n \cdot d$ had been $0$ the line would be parallel to the plane, and the equation in step 3 would have no $t$ in it.

Problem 7

Find the distance from $Q(2, 1, 3)$ to the plane $2x - y + 2z = 4$ and the point of the plane nearest to $Q$

$n = (2, -1, 2)$ and $\|n\| = 3$ The normal is read off the coefficients.
$n \cdot Q - 4 = 4 - 1 + 6 - 4 = 5$ How far the equation is from being satisfied, in units of $\|n\|$
Distance $= \dfrac{|5|}{3} = \dfrac53$ Dividing by $\|n\|$ turns it into a length.
Nearest point: $Q - \dfrac{5}{\|n\|^2}\,n = (2, 1, 3) - \dfrac59(2, -1, 2) = \Big(\dfrac89, \dfrac{14}{9}, \dfrac{17}{9}\Big)$ Move from $Q$ along the normal by the right amount. Check: $\tfrac{16 - 14 + 34}{9} = 4$
Distance $\tfrac53$ nearest point $\big(\tfrac89, \tfrac{14}{9}, \tfrac{17}{9}\big)$ The distance from $Q$ to that point is $\tfrac59\|n\| = \tfrac53$ Forgetting to divide by $\|n\|$ gives $5$ (Mistake 3).

Problem 8

Find the angle between the planes $x + y + z = 1$ and $x - y + 2z = 3$ and their line of intersection.

$n_1 = (1, 1, 1)$ $n_2 = (1, -1, 2)$ $n_1 \cdot n_2 = 1 - 1 + 2 = 2$ The angle between two planes is the angle between their normals.
$\cos\theta = \dfrac{2}{\sqrt3\sqrt6} = \dfrac{\sqrt2}{3} \approx 0.4714$ so $\theta \approx 61.87°$ $\sqrt{18} = 3\sqrt2$
$d = n_1 \times n_2 = (1\cdot2 - 1\cdot(-1),\ 1\cdot1 - 1\cdot2,\ 1\cdot(-1) - 1\cdot1) = (3, -1, -2)$ The line lies in both planes, so it is perpendicular to both normals.
With $z = 0$ $x + y = 1$ and $x - y = 3$ so $(x, y) = (2, -1)$ Any point on both planes will do.
$\theta \approx 61.87°$ line $(2, -1, 0) + t(3, -1, -2)$ Check a second point, $t = 1$ $(5, -2, -2)$ gives $5 - 2 - 2 = 1$ and $5 + 2 - 4 = 3$

Problem 9

Show that the lines $L_1: (1, 0, 0) + s(1, 1, 0)$ and $L_2: (0, 0, 2) + t(0, 1, 1)$ do not meet and are not parallel, and find the distance between them.

The directions $(1, 1, 0)$ and $(0, 1, 1)$ are not multiples of each otherSo the lines are not parallel.
Meeting would need $1 + s = 0$ $s = t$ $0 = 2 + t$ $s = -1$ from the first, $t = -2$ from the third, and then $s \ne t$ The three equations have no common solution: the lines are skew.
$n = (1, 1, 0)\times(0, 1, 1) = (1, -1, 1)$ Perpendicular to both lines.
$P_2 - P_1 = (-1, 0, 2)$ and $(P_2 - P_1)\cdot n = -1 + 0 + 2 = 1$ Project the vector between any two points of the lines onto the common normal.
Distance $\dfrac{1}{\sqrt3} \approx 0.5774$ Divide by $\|n\| = \sqrt3$ The closest points are at $s = -\tfrac43$ and $t = -\tfrac53$ $\big(-\tfrac13, -\tfrac43, 0\big)$ on $L_1$ and $\big(0, -\tfrac53, \tfrac13\big)$ on $L_2$ Their difference is $\tfrac13(1, -1, 1)$ a multiple of $n$ of length $\tfrac{1}{\sqrt3}$

Problem 10

A linear classifier predicts the positive class when $w \cdot x + b > 0$ with $w = (2, -1, 2)$ and $b = -3$ Find the signed distances of $x_1 = (2, 0, 1)$ and $x_2 = (0, 1, 0)$ to the decision boundary, and the point of the boundary nearest to $x_1$

The boundary is the plane $w \cdot x = 3$ with normal $w$ and $\|w\| = 3$ The weight vector is perpendicular to the decision boundary.
$w \cdot x_1 + b = 4 + 0 + 2 - 3 = 3$ so the signed distance is $\tfrac33 = 1$ Positive: $x_1$ is on the positive side.
$w \cdot x_2 + b = 0 - 1 + 0 - 3 = -4$ so the signed distance is $-\tfrac43$ Negative: $x_2$ is classified negative, $\tfrac43$ from the boundary.
Nearest point to $x_1$ $x_1 - 1\cdot\dfrac{w}{\|w\|} = (2, 0, 1) - \dfrac13(2, -1, 2) = \Big(\dfrac43, \dfrac13, \dfrac13\Big)$ Step back along the unit normal by the signed distance. Check: $\tfrac83 - \tfrac13 + \tfrac23 - 3 = 0$
Signed distances $1$ and $-\tfrac43$ nearest boundary point $\big(\tfrac43, \tfrac13, \tfrac13\big)$ The raw score $w \cdot x + b$ is the signed distance times $\|w\|$ so scaling $w$ and $b$ by $10$ multiplies every score by $10$ without moving the boundary. That is why a margin is measured as $\tfrac{w \cdot x + b}{\|w\|}$

Where this goes wrong

1. The middle sign of the cross product

The three entries are computed with the same pattern, $u_iv_j - u_jv_i$ in index order.

$u = (1, 2, 2)$ $v = (3, 0, 4)$ first entry $u_2v_3 - u_3v_2 = 8$ Right so far: Problem 3.
“Second entry: $u_1v_3 - u_3v_1 = 4 - 6 = -2$ ”The step that causes the mistake: the middle entry is $u_3v_1 - u_1v_3$ the minus sign of the cofactor expansion.
$u \times v = (8, -2, -6)$ The check catches it: $(8, -2, -6)\cdot(1, 2, 2) = 8 - 4 - 12 = -8 \ne 0$ so this vector is not perpendicular to $u$ The cross product is $(8, 2, -6)$

2. Dividing by the squared lengths

The angle formula is remembered with $u \cdot u$ and $v \cdot v$ in the denominator.

$u \cdot v = 11$ $\|u\|^2 = 9$ $\|v\|^2 = 25$ Right so far: Problem 1.
“ $\cos\theta = \dfrac{u \cdot v}{\|u\|^2\|v\|^2}$ ”The squared lengths belong to the projection coefficient $\tfrac{u \cdot v}{u \cdot u}$ not to the cosine.
$\cos\theta = \tfrac{11}{225} \approx 0.0489$ so $\theta \approx 87.2°$ Nearly perpendicular, for two vectors whose dot product is large and positive. A cosine must not change when $u$ is doubled; this one halves. With $\|u\|\,\|v\| = 15$ $\theta \approx 42.83°$

3. The distance without dividing by |n|

The value of the plane's equation at the point is taken as the distance.

$n \cdot Q - 4 = 5$ for $Q = (2, 1, 3)$ and the plane $2x - y + 2z = 4$ Right so far: Problem 7.
“The distance is $|n \cdot Q - 4|$ ”That number depends on how the equation is written: $4x - 2y + 4z = 8$ is the same plane and would give $10$
Distance $= 5$ $5$ is the distance measured in units of $\|n\| = 3$ Divide: the distance is $\tfrac53$ Check with the nearest point $\big(\tfrac89, \tfrac{14}{9}, \tfrac{17}{9}\big)$ which is $\tfrac53$ from $Q$

4. The plane through the wrong point

The constant on the right of the plane's equation is computed from an edge vector.

$n = (6, 3, 2)$ is normal to the plane through $A$ $B$ $C$ Right so far: Problem 4.
“ $k = n \cdot \vec{AB} = -6 + 6 + 0 = 0$ ” $\vec{AB}$ lies in the plane's direction but is not a point of it; $n$ is perpendicular to every edge, so this always gives $0$
The plane is $6x + 3y + 2z = 0$ That plane passes through the origin, parallel to the right one. $A$ does not satisfy it: $6 \ne 0$ Use a point: $k = n \cdot A = 6$

5. The product of the lengths as a volume

The volume of a box is length times width times height, and the same is done for a slanted one.

$\|a\| = \sqrt2$ $\|b\| = \sqrt5$ $\|c\| = \sqrt{10}$ Right so far: the edges of Problem 5.
“Volume $= \|a\|\,\|b\|\,\|c\|$ ”True only when the edges are perpendicular; here $a \cdot b = 2$ and $a \cdot c = 3$
Volume $= \sqrt{2\cdot5\cdot10} = 10$ Twice the true volume. The product of the lengths is an upper bound (Hadamard's inequality), reached only by a rectangular box. The triple product gives $5$

Print this set: vectors-dot-and-cross-products-lines-and-planes.pdf (problems, answers, and worked solutions on separate pages).