Vectors in Space

Try this

Choose $u = A C_{1}$ . How are its coordinates related to the three edge lengths of the box?
Set all three edges to $2$ (a cube) and look at the angle between the two space diagonals $A C_{1}$ and $B D_{1}$ .
Try several pairs of vectors until $u \cdot v = 0$ . Do they look perpendicular in the picture? (Perspective can fool you; coordinates cannot.)
Turn the figure: do the readings of coordinates, dot product and angle change?

Take $A$ as the origin and $A B$ , $A D$ , $A A_{1}$ as the $x$ -, $y$ - and $z$ -axes. Every vertex of the box then has three coordinates, and the coordinates of each vector $P Q$ are “end minus start”. Everything from the plane carries over, with just one more coordinate:

u \cdot v = x_{1} x_{2} + y_{1} y_{2} + z_{1} z_{2}, ∣ u ∣ = x_{1}^{2} + y_{1}^{2} + z_{1}^{2} .

So all three kinds of angle in solid geometry can be calculated:

the angle between two lines: $cos θ = \frac{∣ u \cdot v ∣}{∣ u ∣∣ v ∣}$ ;
the angle between a line and a plane: first find a normal vector $n$ of the plane (perpendicular to two intersecting lines in the plane); then $sin φ = \frac{∣ u \cdot n ∣}{∣ u ∣∣ n ∣}$ ;
the dihedral angle: the angle between the normal vectors of the two planes, or its supplement (decide from the figure whether it is acute or obtuse).

Predict first

In the cube $A B C D - A_{1} B_{1} C_{1} D_{1}$ , what is the angle between the lines $A_{1} B$ and $B_{1} C$ , in degrees?

Choose an answer first; then you can see the result.

Why: proving a line perpendicular to a plane with coordinates

In a cube, the space diagonal AC₁ ⟂ plane A₁BD

Step 1 / 4

Set up axes and write coordinates
Take the edge to be $2$ : $A (0, 0, 0)$ , $B (2, 0, 0)$ , $D (0, 2, 0)$ , $A_{1} (0, 0, 2)$ , $C_{1} (2, 2, 2)$ .

$A C_{1} = (2, 2, 2)$
Perpendicular to the first line in the plane
Perpendicular to the second line
Line perpendicular to plane

To find a normal vector: put $n = (x, y, z)$ , require its dot products with two non-collinear vectors in the plane to be $0$ , and take any nonzero solution. (The widget uses the “cross product” of two edges; the result is the same up to a multiple.)

Remove one condition

Break it

Counterexample lab: remove one condition

Conclusion: $A C_{1} ⊥$ plane $A_{1} B D$

When the three edges are equal, $A C_{1} = (a, a, a)$ has dot product $0$ with both $A_{1} B = (a, 0, - a)$ and $A_{1} D = (0, a, - a)$ .

Practice

Exercise 1Basic

In a box, $A B = 3$ , $A D = 2$ , $A A_{1} = 1$ . Find $∣ A C_{1} ∣^{2}$ .

Exercise 2Basic

$a = (1, - 1, 2)$ , $b = (2, 1, x)$ , and $a ⊥ b$ . Find $x$ .

Exercise 3Advanced

In a cube with edge $1$ , the angle between the space diagonal $A C_{1}$ and the edge $A B$ is $θ$ . Find $cos^{2} θ$ .

Exercise 4Challenge

In a cube with edge $1$ , let $d$ be the distance from the point $C$ to the plane $A B_{1} D_{1}$ . Find $d^{2}$ .

Try this

Why: proving a line perpendicular to a plane with coordinates

In a cube, the space diagonal AC₁ ⟂ plane A₁BD

Remove one condition

Counterexample lab: remove one condition

Practice

Mistake Museum · Exhibits from this stop

Taking the angle between two lines to be obtuse

Using the cosine for the angle between a line and a plane