Questions on the vector product and the scalar triple product

Question 1

State what |a × b| measures geometrically, and what a·(b × c) measures.

Accepted Answer

|a × b| is the area of the parallelogram spanned by a and b, equal to |a||b| sin θ. The scalar triple product a·(b × c) is the volume of the parallelepiped on the three vectors, read as a modulus, since the triple product itself can come out negative. B1 for the area of the parallelogram, B1 for the volume of the parallelepiped.

Question 2

Find a × b for a = (2, −1, 3) and b = (1, 4, −2), and verify that it is perpendicular to a.

Accepted Answer

a × b = ((−1)(−2) − (3)(4), (3)(1) − (2)(−2), (2)(4) − (−1)(1)) = (−10, 7, 9). Dot with a: −20 − 7 + 27 = 0, so the product is perpendicular to a, as it must be. M1 for the vector product, A1 for (−10, 7, 9), B1 for the zero dot product. Dotting with b gives 0 as well.

Question 3

Write down i × j, j × k and k × i, and also j × i and i × i, stating the rule that fixes the sign of each product.

Accepted Answer

i × j = k, j × k = i and k × i = j: in the cyclic order i, j, k, the product of two consecutive unit vectors is the third. B1 for the three products. Reversing the order reverses the sign, so j × i = −k, and any vector crossed with itself is the zero vector, so i × i = 0. B1 for j × i = −k and i × i = 0 with the rule. The signs follow the right-handed triple: with i and j in the plane of the page, k points out of it, and swapping the factors sends the product back into the page.

Question 4

Two vectors a and b have |a| = 3 and |b| = 4, and the angle between them is 30 degrees. Find |a × b| and a·b.

Accepted Answer

|a × b| = |a||b| sin θ = 3 × 4 × ½ = 6, and a·b = |a||b| cos θ = 12 × √3/2 = 6√3, about 10.39. B1 B1, one for each product. The two products share the same |a||b| and split it between the sine and the cosine of the angle, which is why their squares add to (|a||b|)² = 144.

Question 5

Find the area of the triangle with vertices A(1, 0, 1), B(3, −1, 4) and C(2, 4, −1).

Accepted Answer

Form both edge vectors from the same vertex before taking their vector product: AB = (2, −1, 3) and AC = (1, 4, −2). Their vector product is (−10, 7, 9), of magnitude √(100 + 49 + 81) = √230 ≈ 15.17, the area of the parallelogram on AB and AC. The triangle is half of it, so its area is √230/2 ≈ 7.58. M1 for the edge vectors, M1 for the vector product, A1 for √230, A1 for half of it.

Question 6

Find the volume of the tetrahedron with edges from the origin along a = (1, 1, 1), b = (2, 3, 1) and c = (0, 2, 4).

Accepted Answer

b × c = ((3)(4) − (1)(2), (1)(0) − (2)(4), (2)(2) − (3)(0)) = (10, −8, 4). Then a·(b × c) = 10 − 8 + 4 = 6.
The parallelepiped's volume is the modulus of that, |6| = 6, and the tetrahedron takes a sixth of it, so its volume is 1. M1 for b × c, A1 for (10, −8, 4), M1 for the scalar triple product, A1 for the volume. Take the modulus even when the answer is already positive. Listing the same three vectors in the other order gives −6, and a volume is never negative.

Question 7

The line l passes through A(2, −1, 4) and is parallel to b = i + 2j + 2k. Write down an equation for l in the form (r − a) × b = 0, and find the direction cosines of l.

Accepted Answer

Taking a as the position vector of A, the equation is (r − (2i − j + 4k)) × (i + 2j + 2k) = 0. Every point of l gives a displacement from A parallel to b, and the vector product of parallel vectors is zero.
|b| = √(1 + 4 + 4) = 3, so the direction cosines are 1/3, 2/3 and 2/3. B1 for the vector form of the line, M1 for |b| = 3, A1 for the direction cosines. Check: (1 + 4 + 4)/9 = 1, which the cosines of the angles a direction makes with the three axes must always satisfy.

Question 8

Explain what a·(b × c) = 0 tells you about three non-zero vectors, and how this compares with a × b = 0.

Accepted Answer

A zero triple product means the parallelepiped has no volume, so the three vectors are coplanar: one lies in the plane of the other two. A zero vector product means the parallelogram has no area, so those two vectors are parallel. B1 for coplanar, B1 for parallel, B1 for the comparison.

Question 9

Given a = (1, 2, 0) and b = (0, 1, 3), use the distributive rule to express (a + 2b) × (3a − b) in terms of a × b, and hence find it as a vector.

Accepted Answer

Expanding as with brackets, but keeping the order of every product: (a + 2b) × (3a − b) = 3(a × a) − (a × b) + 6(b × a) − 2(b × b). M1 for the four-term expansion in order. Now a × a = b × b = 0 and b × a = −(a × b), so this is −(a × b) − 6(a × b) = −7(a × b). A1.
a × b = ((2)(3) − (0)(1), (0)(0) − (1)(3), (1)(1) − (2)(0)) = (6, −3, 1). M1. So the product is (−42, 21, −7). A1. Direct calculation confirms it: a + 2b = (1, 4, 6) and 3a − b = (3, 5, −3), whose vector product is (−12 − 30, 18 + 3, 5 − 12) = (−42, 21, −7). The expansion is shorter, provided b × a is written as −a × b rather than a × b.

Question 10

Find a × b for a = (3, −2, 1) and b = (1, 1, −4). Check that your answer is perpendicular to both a and b, and hence write down a unit vector perpendicular to both.

Accepted Answer

a × b = ((−2)(−4) − (1)(1), (1)(1) − (3)(−4), (3)(1) − (−2)(1)) = (7, 13, 5). M1 for the vector product, A1.
Check: (7, 13, 5)·(3, −2, 1) = 21 − 26 + 5 = 0 and (7, 13, 5)·(1, 1, −4) = 7 + 13 − 20 = 0. B1 for both scalar products being zero.
|(7, 13, 5)| = √(49 + 169 + 25) = √243 = 9√3, so a unit vector is (7, 13, 5)/(9√3), or its negative. A1. The two checks are worth the thirty seconds: a sign slip in the middle component turns 1 + 12 into 1 − 12 and the second scalar product then comes out as −24 instead of 0, which is how the slip is caught.

Question 11

Show that the vectors a = (1, 2, 3), b = (2, −1, 1) and c = (4, 3, 7) are coplanar, and express c in terms of a and b.

Accepted Answer

b × c = ((−1)(7) − (1)(3), (1)(4) − (2)(7), (2)(3) − (−1)(4)) = (−10, −10, 10), and a·(b × c) = −10 − 20 + 30 = 0. M1 for the scalar triple product, A1 for zero with the conclusion that the parallelepiped has no volume, so the three vectors lie in one plane. Then c = 2a + b, since 2(1, 2, 3) + (2, −1, 1) = (4, 3, 7). B1.

Question 12

The line l₁ has equation r = i + 2k + λ(2i + j − k) and the line l₂ has equation r = 3i − j + μ(i + 2j + k). Show that l₁ and l₂ are skew, and find the shortest distance between them, giving your answer in exact form.

Accepted Answer

The direction vectors are d₁ = (2, 1, −1) and d₂ = (1, 2, 1). Neither is a multiple of the other, so the lines are not parallel.
d₁ × d₂ = ((1)(1) − (−1)(2), (−1)(1) − (2)(1), (2)(2) − (1)(1)) = (3, −3, 3), of magnitude √(9 + 9 + 9) = 3√3.
Join a point of each line: a₂ − a₁ = (3, −1, 0) − (1, 0, 2) = (2, −1, −2). Then (a₂ − a₁)·(d₁ × d₂) = 6 + 3 − 6 = 3. That is non-zero, so the lines do not intersect. Not parallel and not intersecting makes them skew.
The shortest distance is the component of a₂ − a₁ along the common perpendicular: |(a₂ − a₁)·(d₁ × d₂)|/|d₁ × d₂| = 3/(3√3) = 1/√3 = √3/3, about 0.577.
B1 for the directions not being parallel, M1 for d₁ × d₂, A1 for (3, −3, 3), M1 for the scalar triple product, A1 for the skew conclusion, A1 for the distance.
The modulus matters here. The triple product carries the sign of the orientation, so naming the lines the other way round turns 3 into −3, and a distance is never negative.

Question 13

Starting from the definitions |a × b| = |a||b| sin θ and a·b = |a||b| cos θ, prove that |a × b|² = |a|²|b|² − (a·b)². Two vectors have |a| = 5, |b| = 3 and a·b = 9. Find the area of the parallelogram they span.

Accepted Answer

Squaring the two definitions: |a × b|² = |a|²|b|² sin² θ and (a·b)² = |a|²|b|² cos² θ. M1 for squaring both. Adding, |a × b|² + (a·b)² = |a|²|b|²(sin² θ + cos² θ) = |a|²|b|², so |a × b|² = |a|²|b|² − (a·b)². M1 for adding and using sin² θ + cos² θ = 1, A1 for the identity.
With the numbers: |a × b|² = 25 × 9 − 81 = 144, so |a × b| = 12 and the parallelogram has area 12. M1 for substituting, A1. The angle was never needed: cos θ = 9/15 = 3/5 gives sin θ = 4/5 and 15 × 4/5 = 12 by the longer road. The identity is what lets the area be found from a scalar product alone, and it can be evaluated from components as well, which makes it a useful check on any computed vector product.

Question 14

The line l has equation r = (1, 0, 1) + λ(2, 2, 1), and P is the point (3, 1, 2). By considering the area of the parallelogram spanned by AP and the direction vector d, where A is the point (1, 0, 1), show that the perpendicular distance from P to l is |AP × d|/|d|, and find that distance in exact form. Check your answer by finding the foot of the perpendicular from P to l.

Accepted Answer

The parallelogram on AP and d has area |AP × d| = |AP||d| sin θ, where θ is the angle between them. Its base is |d| and its height is |AP| sin θ, the perpendicular distance from P to the line. So distance × |d| = |AP × d|, giving distance = |AP × d|/|d|. M1 for area as base × height, A1 for the formula.
AP = (3, 1, 2) − (1, 0, 1) = (2, 1, 1), and AP × d = ((1)(1) − (1)(2), (1)(2) − (2)(1), (2)(2) − (1)(2)) = (−1, 0, 2), of magnitude √5, while |d| = √9 = 3. M1 for AP × d, A1 for (−1, 0, 2). The distance is √5/3, about 0.745. A1.
Check: the foot F is (1 + 2λ, 2λ, 1 + λ) with PF·d = 0: (2λ − 2)(2) + (2λ − 1)(2) + (λ − 1)(1) = 9λ − 7 = 0, so λ = 7/9 and PF = (−4/9, 5/9, −2/9), of length √(45)/9 = √5/3. B1 for the foot and the matching length. The vector product route needs no unknown and no equation to solve; the foot is the longer road, but it is the one to take when the question asks where the perpendicular lands.

Question 15

Let a = a₁i + a₂j + a₃k and b = b₁i + b₂j + b₃k. Using only the products i × j = k, j × k = i, k × i = j, the rule that reversing the order of a product changes its sign, and the distributive law, derive the component formula for a × b.

Accepted Answer

Expanding by the distributive law gives nine products, in order: a₁b₁(i × i) + a₁b₂(i × j) + a₁b₃(i × k) + a₂b₁(j × i) + a₂b₂(j × j) + a₂b₃(j × k) + a₃b₁(k × i) + a₃b₂(k × j) + a₃b₃(k × k). M1 for the nine-term expansion with the order of each factor kept.
The three products of a vector with itself are zero. B1. The three reversed products are i × k = −j, j × i = −k and k × j = −i, by the sign rule. M1 for converting the three reversed products.
Collecting by unit vector: the i terms are a₂b₃ − a₃b₂, the j terms are a₃b₁ − a₁b₃, and the k terms are a₁b₂ − a₂b₁. A1 for the collection. So a × b = (a₂b₃ − a₃b₂)i + (a₃b₁ − a₁b₃)j + (a₁b₂ − a₂b₁)k. A1. Every sign in the formula comes from the right-handed order of i, j, k, which is why the middle component reads a₃b₁ − a₁b₃ rather than a₁b₃ − a₃b₁: the cyclic order after k is i, not j. The same expansion, done with the factors in the other order, produces every term with the opposite sign, which is b × a = −a × b again.