Multivariable Calculus Quiz: Scalar Triple Product
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Scalar Triple ProductQuestion 1 of 3

Three vectors a⃗\vec{a}, b⃗\vec{b}, and c⃗\vec{c} satisfy a⃗⋅(b⃗×c⃗)=15\vec{a} \cdot (\vec{b} \times \vec{c}) = 15. If a⃗′=2a⃗+b⃗\vec{a}' = 2\vec{a} + \vec{b}, b⃗′=a⃗−b⃗\vec{b}' = \vec{a} - \vec{b}, and c⃗′=c⃗+3a⃗\vec{c}' = \vec{c} + 3\vec{a}, what is the value of a⃗′⋅(b⃗′×c⃗′)\vec{a}' \cdot (\vec{b}' \times \vec{c}')?

−45-45
−30-30
3030
4545
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Multivariable Calculus Quiz

Multivariable Calculus Quiz: Scalar Triple Product

Practice Scalar Triple Product in Multivariable Calculus with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Scalar Triple Product, giving you a quick way to practice the rules, question types, and explanations that matter most for Multivariable Calculus.

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Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

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Question 1

Three vectors a⃗\vec{a}, b⃗\vec{b}, and c⃗\vec{c} satisfy a⃗⋅(b⃗×c⃗)=15\vec{a} \cdot (\vec{b} \times \vec{c}) = 15. If a⃗′=2a⃗+b⃗\vec{a}' = 2\vec{a} + \vec{b}, b⃗′=a⃗−b⃗\vec{b}' = \vec{a} - \vec{b}, and c⃗′=c⃗+3a⃗\vec{c}' = \vec{c} + 3\vec{a}, what is the value of a⃗′⋅(b⃗′×c⃗′)\vec{a}' \cdot (\vec{b}' \times \vec{c}')?

  1. −45-45 (correct answer)
  2. −30-30
  3. 3030
  4. 4545
Explanation: We need to compute a⃗′⋅(b⃗′×c⃗′)\vec{a}' \cdot (\vec{b}' \times \vec{c}') where a⃗′=2a⃗+b⃗\vec{a}' = 2\vec{a} + \vec{b}, b⃗′=a⃗−b⃗\vec{b}' = \vec{a} - \vec{b}, and c⃗′=c⃗+3a⃗\vec{c}' = \vec{c} + 3\vec{a}. This scalar triple product can be expressed as the determinant of the transformation matrix times the original scalar triple product. The transformation matrix has columns representing how each new vector is expressed in terms of the old vectors. For a⃗′\vec{a}': coefficient of a⃗\vec{a} is 2, b⃗\vec{b} is 1, c⃗\vec{c} is 0. For b⃗′\vec{b}': coefficient of a⃗\vec{a} is 1, b⃗\vec{b} is -1, c⃗\vec{c} is 0. For c⃗′\vec{c}': coefficient of a⃗\vec{a} is 3, b⃗\vec{b} is 0, c⃗\vec{c} is 1. The transformation matrix is (2131−10001)\begin{pmatrix}2&1&3\\1&-1&0\\0&0&1\end{pmatrix}. Its determinant is 2[(−1)(1)−(0)(0)]−1[(1)(1)−(0)(0)]+3[(1)(0)−(−1)(0)]=2(−1)−1(1)+3(0)=−2−1=−32[(-1)(1) - (0)(0)] - 1[(1)(1) - (0)(0)] + 3[(1)(0) - (-1)(0)] = 2(-1) - 1(1) + 3(0) = -2 - 1 = -3. Therefore, a⃗′⋅(b⃗′×c⃗′)=−3×15=−45\vec{a}' \cdot (\vec{b}' \times \vec{c}') = -3 \times 15 = -45.

Question 2

Let u⃗=⟨2,1,−3⟩\vec{u} = \langle 2, 1, -3 \rangle, v⃗=⟨1,−2,1⟩\vec{v} = \langle 1, -2, 1 \rangle, and w⃗=⟨−1,3,2⟩\vec{w} = \langle -1, 3, 2 \rangle. If n⃗\vec{n} is a unit vector such that the scalar triple product (u⃗+tn⃗)⋅(v⃗×w⃗)=0(\vec{u} + t\vec{n}) \cdot (\vec{v} \times \vec{w}) = 0 for some real number tt, which of the following could be n⃗\vec{n}?

  1. 114⟨1,2,3⟩\frac{1}{\sqrt{14}}\langle 1, 2, 3 \rangle
  2. 114⟨3,2,1⟩\frac{1}{\sqrt{14}}\langle 3, 2, 1 \rangle
  3. 114⟨−1,2,−3⟩\frac{1}{\sqrt{14}}\langle -1, 2, -3 \rangle (correct answer)
  4. 114⟨−3,2,1⟩\frac{1}{\sqrt{14}}\langle -3, 2, 1 \rangle
Explanation: First, compute v⃗×w⃗=⟨1,−2,1⟩×⟨−1,3,2⟩=⟨(−2)(2)−(1)(3),(1)(−1)−(1)(2),(1)(3)−(−2)(−1)⟩=⟨−7,−3,−1⟩\vec{v} \times \vec{w} = \langle 1,-2,1 \rangle \times \langle -1,3,2 \rangle = \langle (-2)(2)-(1)(3), (1)(-1)-(1)(2), (1)(3)-(-2)(-1) \rangle = \langle -7, -3, -1 \rangle. The condition (u⃗+tn⃗)⋅(v⃗×w⃗)=0(\vec{u} + t\vec{n}) \cdot (\vec{v} \times \vec{w}) = 0 becomes u⃗⋅(v⃗×w⃗)+tn⃗⋅(v⃗×w⃗)=0\vec{u} \cdot (\vec{v} \times \vec{w}) + t\vec{n} \cdot (\vec{v} \times \vec{w}) = 0. We have u⃗⋅(v⃗×w⃗)=⟨2,1,−3⟩⋅⟨−7,−3,−1⟩=2(−7)+1(−3)+(−3)(−1)=−14−3+3=−14\vec{u} \cdot (\vec{v} \times \vec{w}) = \langle 2,1,-3 \rangle \cdot \langle -7,-3,-1 \rangle = 2(-7) + 1(-3) + (-3)(-1) = -14 - 3 + 3 = -14. So we need −14+tn⃗⋅⟨−7,−3,−1⟩=0-14 + t\vec{n} \cdot \langle -7,-3,-1 \rangle = 0, which means tn⃗⋅⟨−7,−3,−1⟩=14t\vec{n} \cdot \langle -7,-3,-1 \rangle = 14. For this to have a solution, n⃗\vec{n} must not be perpendicular to ⟨−7,−3,−1⟩\langle -7,-3,-1 \rangle. Let's check each option: (A) 114⟨1,2,3⟩⋅⟨−7,−3,−1⟩=114(−7−6−3)=−1614\frac{1}{\sqrt{14}}\langle 1,2,3 \rangle \cdot \langle -7,-3,-1 \rangle = \frac{1}{\sqrt{14}}(-7-6-3) = \frac{-16}{\sqrt{14}}. (B) 114⟨3,2,1⟩⋅⟨−7,−3,−1⟩=114(−21−6−1)=−2814=−214\frac{1}{\sqrt{14}}\langle 3,2,1 \rangle \cdot \langle -7,-3,-1 \rangle = \frac{1}{\sqrt{14}}(-21-6-1) = \frac{-28}{\sqrt{14}} = -2\sqrt{14}. (C) 114⟨−1,2,−3⟩⋅⟨−7,−3,−1⟩=114(7−6+3)=414\frac{1}{\sqrt{14}}\langle -1,2,-3 \rangle \cdot \langle -7,-3,-1 \rangle = \frac{1}{\sqrt{14}}(7-6+3) = \frac{4}{\sqrt{14}}. (D) 114⟨−3,2,1⟩⋅⟨−7,−3,−1⟩=114(21−6−1)=1414=14\frac{1}{\sqrt{14}}\langle -3,2,1 \rangle \cdot \langle -7,-3,-1 \rangle = \frac{1}{\sqrt{14}}(21-6-1) = \frac{14}{\sqrt{14}} = \sqrt{14}. Any of these could work since none are zero, but we need t⋅(dot product)=14t \cdot \text{(dot product)} = 14. For option (C), t⋅414=14t \cdot \frac{4}{\sqrt{14}} = 14 gives t=14144=7142t = \frac{14\sqrt{14}}{4} = \frac{7\sqrt{14}}{2}. This is valid.

Question 3

Let x⃗=⟨2,1,−1⟩\vec{x} = \langle 2, 1, -1 \rangle, y⃗=⟨1,−1,2⟩\vec{y} = \langle 1, -1, 2 \rangle, and z⃗=⟨−1,2,1⟩\vec{z} = \langle -1, 2, 1 \rangle. If the vectors u⃗=x⃗+ky⃗\vec{u} = \vec{x} + k\vec{y}, v⃗=y⃗+kz⃗\vec{v} = \vec{y} + k\vec{z}, and w⃗=z⃗+kx⃗\vec{w} = \vec{z} + k\vec{x} are coplanar for some value of kk, what is the value of kk?

  1. −1-1 (correct answer)
  2. −12-\frac{1}{2}
  3. 12\frac{1}{2}
  4. 11
Explanation: Three vectors are coplanar if and only if their scalar triple product is zero. We need u⃗⋅(v⃗×w⃗)=0\vec{u} \cdot (\vec{v} \times \vec{w}) = 0. First, express the vectors in terms of kk: u⃗=⟨2+k,1−k,−1+2k⟩\vec{u} = \langle 2+k, 1-k, -1+2k \rangle, v⃗=⟨1−k,−1+2k,2+k⟩\vec{v} = \langle 1-k, -1+2k, 2+k \rangle, w⃗=⟨−1+2k,2+k,1−k⟩\vec{w} = \langle -1+2k, 2+k, 1-k \rangle. The scalar triple product is the determinant: ∣2+k1−k−1+2k1−k−1+2k2+k−1+2k2+k1−k∣=0 \begin{vmatrix} 2+k & 1-k & -1+2k \\ 1-k & -1+2k & 2+k \\ -1+2k & 2+k & 1-k \end{vmatrix} = 0. This is a complex calculation, so let me try a different approach. Since the vectors are coplanar, we can write one as a linear combination of the other two: w⃗=αu⃗+βv⃗\vec{w} = \alpha\vec{u} + \beta\vec{v} for some scalars α\alpha and β\beta. Substituting: z⃗+kx⃗=α(x⃗+ky⃗)+β(y⃗+kz⃗)\vec{z} + k\vec{x} = \alpha(\vec{x} + k\vec{y}) + \beta(\vec{y} + k\vec{z}). Rearranging: z⃗+kx⃗=αx⃗+αky⃗+βy⃗+βkz⃗\vec{z} + k\vec{x} = \alpha\vec{x} + \alpha k\vec{y} + \beta\vec{y} + \beta k\vec{z}. Collecting terms: z⃗+kx⃗=αx⃗+(αk+β)y⃗+βkz⃗\vec{z} + k\vec{x} = \alpha\vec{x} + (\alpha k + \beta)\vec{y} + \beta k\vec{z}. Since x⃗\vec{x}, y⃗\vec{y}, and z⃗\vec{z} are linearly independent (we can verify this by checking that their scalar triple product is non-zero), we must have: coefficient of x⃗\vec{x}: k=αk = \alpha, coefficient of y⃗\vec{y}: 0=αk+β=k2+β0 = \alpha k + \beta = k^2 + \beta, coefficient of z⃗\vec{z}: 1=βk1 = \beta k. From the third equation, β=1k\beta = \frac{1}{k} (assuming k≠0k \neq 0). Substituting into the second equation: 0=k2+1k0 = k^2 + \frac{1}{k}, so k3+1=0k^3 + 1 = 0, giving k3=−1k^3 = -1, thus k=−1k = -1. Let's verify: if k=−1k = -1, then α=−1\alpha = -1 and β=1−1=−1\beta = \frac{1}{-1} = -1. Check: αk+β=(−1)(−1)+(−1)=1−1=0\alpha k + \beta = (-1)(-1) + (-1) = 1 - 1 = 0 ✓. Also, u⃗=x⃗−y⃗=⟨2−1,1+1,−1−2⟩=⟨1,2,−3⟩\vec{u} = \vec{x} - \vec{y} = \langle 2-1, 1+1, -1-2 \rangle = \langle 1, 2, -3 \rangle, v⃗=y⃗−z⃗=⟨1+1,−1−2,2−1⟩=⟨2,−3,1⟩\vec{v} = \vec{y} - \vec{z} = \langle 1+1, -1-2, 2-1 \rangle = \langle 2, -3, 1 \rangle, w⃗=z⃗−x⃗=⟨−1−2,2−1,1+1⟩=⟨−3,1,2⟩\vec{w} = \vec{z} - \vec{x} = \langle -1-2, 2-1, 1+1 \rangle = \langle -3, 1, 2 \rangle. Verify coplanarity: w⃗=−u⃗−v⃗\vec{w} = -\vec{u} - \vec{v} means ⟨−3,1,2⟩=−⟨1,2,−3⟩−⟨2,−3,1⟩=⟨−1,−2,3⟩+⟨−2,3,−1⟩=⟨−3,1,2⟩\langle -3, 1, 2 \rangle = -\langle 1, 2, -3 \rangle - \langle 2, -3, 1 \rangle = \langle -1, -2, 3 \rangle + \langle -2, 3, -1 \rangle = \langle -3, 1, 2 \rangle ✓.