Practice: Cross Products and Geometry in Space

Recognition · Error diagnosis

A learner needs the normal to the plane of $u = (1, 2, 3)$ and $v = (4, 5, 6)$ . They compute $v \times u = (3, - 6, 3)$ but had intended $u \times v$ . What is the consequence?

2 hints available, least help first.

Hint 1: Retrieval cue

What happens to a determinant when two of its rows are swapped?

Hint 2: Concept cue

Compare $(3, - 6, 3)$ with $(- 3, 6, - 3)$ component by component, and compare their lengths.

Direct application

Let $u = (1, 2, 3)$ and $v = (4, 5, 6)$ .

Compute $u \times v$ and give its second component.

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

The second component omits the second coordinate and uses the third and first.

Hint 2: Concept cue

Compute $3 \times 4 - 1 \times 6$ .

Direct application

For $u = (1, 2, 3)$ and $v = (4, 5, 6)$ , the cross product is $u \times v = (- 3, 6, - 3)$ .

What is $‖ u \times v ‖^{2}$ , the squared area of the parallelogram they span?

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

The squared length of a vector is the sum of the squares of its components.

Hint 2: Concept cue

Add $9$ , $36$ and $9$ .

Direct application

Let $u = (1, 2, 3)$ , $v = (4, 5, 6)$ and $w = (2, - 1, 1)$ , with $v \times w = (11, 8, - 14)$ .

Compute the scalar triple product $u \cdot (v \times w)$ .

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

The triple product is a dot product of $u$ with the vector already given.

Hint 2: Concept cue

$11 + 16 - 42$ .

Direct application

A plane passes through $P_{0} = (1, 0, 0)$ with normal $n = (2, - 1, 2)$ . Find the distance from $Q = (3, 4, 5)$ to the plane.

Give your answer as a decimal.

Your answer

Enter the value. It is checked against the answer and the precision this task asks for.

2 hints available, least help first.

Hint 1: Retrieval cue

Form the displacement $Q - P_{0}$ first.

Hint 2: Concept cue

$(2, 4, 5) \cdot (2, - 1, 2) = 10$ , and $‖ n ‖ = 3$ .

Construction · Direct application · Explanation

Let $P = (1, 0, 2)$ , $Q = (2, 1, 0)$ , $R = (0, 3, 1)$ and $S = (4, 4, 4)$ .

(a) Find a normal to the plane through $P$ , $Q$ and $R$ , and verify it is perpendicular to both edge vectors. Give the plane's equation and check all three points satisfy it.

(b) Find the area of triangle $P Q R$ .

(c) Find the volume of the tetrahedron $P Q R S$ , and state what the sign of your triple product means.

(d) Find the distance from $S$ to the plane of $P Q R$ . Explain why the number appearing in your numerator also appeared in part (c).

(e) A colleague computes the normal as $P R \times P Q$ instead. Say precisely which of your answers change and which do not, and why.

Write your answer

Write your answer, then compare it with the worked solution.

3 hints available, least help first.

Hint 1: Retrieval cue

Form both edge vectors from the same starting point before taking any cross product.

Hint 2: Concept cue

Once you have the normal, part (c) and part (d) can both use $P S \cdot n$ rather than a second cross product.

Hint 3: Strategy cue

In part (e), sort your answers into those defined by a magnitude and those defined by a direction.

Compare with the worked solution

Comparing does not record a result. Judging your own written answer cannot show that you can do this without help.

(a) The plane. Edge vectors from $P$ :

P Q = (1, 1, - 2), P R = (- 1, 3, - 1) .

Cross product:

n = P Q \times P R = (1 (- 1) - (- 2) (3), (- 2) (- 1) - 1 (- 1), 1 (3) - 1 (- 1)) = (5, 3, 4) .

Perpendicularity check. $n \cdot P Q = 5 + 3 - 8 = 0$ ✓; $n \cdot P R = - 5 + 9 - 4 = 0$ ✓. Equation. $n \cdot (X - P) = 0$ gives $5 (x - 1) + 3 (y - 0) + 4 (z - 2) = 0$ , so

5 x + 3 y + 4 z = 13 .

Check all three. $P$ : $5 (1) + 0 + 4 (2) = 13$ ✓. $Q$ : $5 (2) + 3 (1) + 0 = 13$ ✓. $R$ : $0 + 3 (3) + 4 (1) = 13$ ✓. (b) Triangle area. $‖ n ‖^{2} = 25 + 9 + 16 = 50$ , so the parallelogram on $P Q$ and $P R$ has area $\sqrt{50} = 5 \sqrt{2} \approx 7.071$ and

area (P Q R) = \frac{1}{2} \sqrt{50} = \frac{5 \sqrt{2}}{2} \approx 3.536 .

(c) Tetrahedron volume. The third edge is $P S = S - P = (3, 4, 2)$ . The scalar triple product is

P Q \cdot (P R \times P S) = 35,

which equals $det$ of the matrix with rows $P Q$ , $P R$ , $P S$ . Equivalently and more simply, $P S \cdot n = 15 + 12 + 8 = 35$ using the normal already computed. The parallelepiped on the three edges has volume $| 35 | = 35$ , so

volume (P Q R S) = \frac{35}{6} \approx 5.833 .

The sign. It is positive, so $P Q$ , $P R$ , $P S$ form a right-handed frame in that order. A negative value would mean a left-handed one; either way the volume takes the absolute value, so the sign carries orientation rather than size. (d) Distance from $S$ to the plane.

dist = \frac{| P S \cdot n |}{‖ n ‖} = \frac{35}{\sqrt{50}} = \frac{35}{5 \sqrt{2}} = \frac{7}{\sqrt{2}} \approx 4.950 .

Why 35 appears twice. The triple product is the parallelepiped's volume, and volume equals base area times height. The base is the parallelogram on $P Q$ and $P R$ , with area $‖ n ‖ = \sqrt{50}$ , and the height is exactly the distance from $S$ to the plane containing that base. So

35 = \underset{base area}{\underset{⏟}{\sqrt{50}}} \times \underset{height}{\underset{⏟}{\frac{35}{\sqrt{50}}}},

and dividing the triple product by $‖ n ‖$ is precisely the step that converts a volume into a height. The two parts are one computation read two ways. (e) Reversing the cross product. $P R \times P Q = - (P Q \times P R) = (- 5, - 3, - 4)$ . What changes. The normal vector itself, which now points to the other side of the plane. The triple product changes sign, from $35$ to $- 35$ , so the frame is reported as left-handed instead of right-handed. Any statement about which side of the plane $S$ lies on flips. What does not change. The plane's equation, since $- 5 x - 3 y - 4 z = - 13$ describes the same set of points. The length $‖ n ‖ = \sqrt{50}$ , so the triangle area is unchanged. The tetrahedron volume, since it takes an absolute value. And the distance, for the same reason. The numerator is $| P S \cdot n | = 35$ either way. The general rule. Reversing the order negates the cross product, which negates every orientation-carrying quantity and leaves every magnitude alone. Areas, volumes and distances survive; normals, handedness and side-of-plane verdicts do not. A note on degeneracy. Had $P$ , $Q$ and $R$ been collinear, $P Q$ and $P R$ would be parallel and the cross product would be zero. No normal, no plane, no area. Similarly a zero triple product in part (c) would mean $S$ lies in the plane of $P Q R$ , giving zero volume and zero distance. In both cases the vanishing result is the informative answer rather than a failed computation.

A complete answer does each of these:

computes cross product
reads length as area
computes triple product
applies distance formulas
uses the anticommutativity
interprets a zero result

Practice: Cross Products and Geometry in Space

Recognition · Error diagnosis

Direct application

Direct application

Direct application

Direct application

Construction · Direct application · Explanation

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