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The vector product and the scalar triple product questions
The vector product of two vectors is a third vector perpendicular to both, whose magnitude is the area of the parallelogram they span. Taking the scalar product of that with a third vector gives the volume of the parallelepiped, which turns several geometry problems into a single calculation.
7 original questions · 25 marks · the vector product and the scalar triple product notes · Further Pure 1
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State what |a × b| measures geometrically, and what a·(b × c) measures.
Worked 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.Find a × b for a = (2, −1, 3) and b = (1, 4, −2), and verify that it is perpendicular to a.
Worked 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. The commonest slip is the sign of the middle component, and this check catches it.Find the area of the triangle with vertices A(1, 0, 1), B(3, −1, 4) and C(2, 4, −1).
Worked answer
Work with edge vectors from a common vertex: 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. Crossing the position vectors instead of the differences is the usual error and loses the first method mark.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).
Worked 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.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.
Worked 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.Explain what a·(b × c) = 0 tells you about three non-zero vectors, and how this compares with a × b = 0.
Worked 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. Both are degeneracy tests, one dimension apart.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.
Worked 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.
Two marks hang on the modulus. 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. One triple product does both jobs here: non-zero proves the lines are skew, and its size gives the gap.
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