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Scalar Triple Product - Calculus 3

  1. The vectors a⃗=⟨2,3,1⟩\vec{a} = \langle 2, 3, 1 \rangle, b⃗=⟨1,0,2⟩\vec{b} = \langle 1, 0, 2 \rangle, and c⃗=⟨4,6,k⟩\vec{c} = \langle 4, 6, k \rangle are given. For what value of kk will the volume of the parallelepiped formed by these vectors be minimized?
    Answer: k=2k = 2, and the minimum volume is 00 The volume of the parallelepiped is ∣a⃗⋅(b⃗×c⃗)∣|\vec{a} \cdot (\vec{b} \times \vec{c})|. Notice that c⃗=⟨4,6,k⟩=2⟨2,3,k2⟩\vec{c} = \langle 4, 6, k \rangle = 2\langle 2, 3, \frac{k}{2} \rangle. If k=2k = 2, then c⃗=2⟨2,3,1⟩=2a⃗\vec{c} = 2\langle 2, 3, 1 \rangle = 2\vec{a}. When c⃗\vec{c} is a scalar multiple of a⃗\vec{a}, the three vectors become coplanar, making the scalar triple product zero and thus minimizing the volume to 0. Let's verify: b⃗×c⃗=⟨1,0,2⟩×⟨4,6,k⟩=⟨0⋅k−2⋅6,2⋅4−1⋅k,1⋅6−0⋅4⟩=⟨−12,8−k,6⟩\vec{b} \times \vec{c} = \langle 1, 0, 2 \rangle \times \langle 4, 6, k \rangle = \langle 0 \cdot k - 2 \cdot 6, 2 \cdot 4 - 1 \cdot k, 1 \cdot 6 - 0 \cdot 4 \rangle = \langle -12, 8-k, 6 \rangle. Then a⃗⋅(b⃗×c⃗)=⟨2,3,1⟩⋅⟨−12,8−k,6⟩=2(−12)+3(8−k)+1(6)=−24+24−3k+6=6−3k\vec{a} \cdot (\vec{b} \times \vec{c}) = \langle 2, 3, 1 \rangle \cdot \langle -12, 8-k, 6 \rangle = 2(-12) + 3(8-k) + 1(6) = -24 + 24 - 3k + 6 = 6 - 3k. The volume is ∣6−3k∣|6 - 3k|, which is minimized when 6−3k=06 - 3k = 0, giving k=2k = 2. At this value, the volume is 0, which is indeed the minimum possible volume.
  2. Let a⃗\vec{a}, b⃗\vec{b}, and c⃗\vec{c} be three vectors such that a⃗⋅(b⃗×c⃗)=12\vec{a} \cdot (\vec{b} \times \vec{c}) = 12. If we define new vectors a′⃗=2a⃗\vec{a'} = 2\vec{a}, b′⃗=b⃗+a⃗\vec{b'} = \vec{b} + \vec{a}, and c′⃗=c⃗−a⃗\vec{c'} = \vec{c} - \vec{a}, what is the value of a′⃗⋅(b′⃗×c′⃗)\vec{a'} \cdot (\vec{b'} \times \vec{c'})?
    Answer: a′⃗⋅(b′⃗×c′⃗)=24\vec{a'} \cdot (\vec{b'} \times \vec{c'}) = 24 We need to compute a′⃗⋅(b′⃗×c′⃗)\vec{a'} \cdot (\vec{b'} \times \vec{c'}) where a′⃗=2a⃗\vec{a'} = 2\vec{a}, b′⃗=b⃗+a⃗\vec{b'} = \vec{b} + \vec{a}, and c′⃗=c⃗−a⃗\vec{c'} = \vec{c} - \vec{a}. First, find b′⃗×c′⃗=(b⃗+a⃗)×(c⃗−a⃗)\vec{b'} \times \vec{c'} = (\vec{b} + \vec{a}) \times (\vec{c} - \vec{a}). Using the distributive property of cross products: b′⃗×c′⃗=b⃗×c⃗−b⃗×a⃗+a⃗×c⃗−a⃗×a⃗\vec{b'} \times \vec{c'} = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} + \vec{a} \times \vec{c} - \vec{a} \times \vec{a}. Since a⃗×a⃗=0⃗\vec{a} \times \vec{a} = \vec{0} and b⃗×a⃗=−(a⃗×b⃗)\vec{b} \times \vec{a} = -(\vec{a} \times \vec{b}): b′⃗×c′⃗=b⃗×c⃗+a⃗×b⃗+a⃗×c⃗\vec{b'} \times \vec{c'} = \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c}. Now compute the scalar triple product: a′⃗⋅(b′⃗×c′⃗)=2a⃗⋅(b⃗×c⃗+a⃗×b⃗+a⃗×c⃗)=2a⃗⋅(b⃗×c⃗)+2a⃗⋅(a⃗×b⃗)+2a⃗⋅(a⃗×c⃗)\vec{a'} \cdot (\vec{b'} \times \vec{c'}) = 2\vec{a} \cdot (\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{a} \times \vec{c}) = 2\vec{a} \cdot (\vec{b} \times \vec{c}) + 2\vec{a} \cdot (\vec{a} \times \vec{b}) + 2\vec{a} \cdot (\vec{a} \times \vec{c}). Since a⃗⋅(a⃗×b⃗)=0\vec{a} \cdot (\vec{a} \times \vec{b}) = 0 and a⃗⋅(a⃗×c⃗)=0\vec{a} \cdot (\vec{a} \times \vec{c}) = 0 (a vector is always perpendicular to any cross product involving itself): a′⃗⋅(b′⃗×c′⃗)=2a⃗⋅(b⃗×c⃗)=2(12)=24\vec{a'} \cdot (\vec{b'} \times \vec{c'}) = 2\vec{a} \cdot (\vec{b} \times \vec{c}) = 2(12) = 24.
  3. A tetrahedron has vertices at A(1,0,0)A(1, 0, 0), B(0,2,0)B(0, 2, 0), C(0,0,3)C(0, 0, 3), and D(2,1,1)D(2, 1, 1). What is the volume of this tetrahedron?
    Answer: The volume is 116\frac{11}{6} cubic units The volume of a tetrahedron with vertices at AA, BB, CC, and DD is 16∣AB→⋅(AC→×AD→)∣\frac{1}{6}|\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD})|. First, find the vectors: AB→=B−A=(0,2,0)−(1,0,0)=(−1,2,0)\overrightarrow{AB} = B - A = (0, 2, 0) - (1, 0, 0) = (-1, 2, 0), AC→=C−A=(0,0,3)−(1,0,0)=(−1,0,3)\overrightarrow{AC} = C - A = (0, 0, 3) - (1, 0, 0) = (-1, 0, 3), and AD→=D−A=(2,1,1)−(1,0,0)=(1,1,1)\overrightarrow{AD} = D - A = (2, 1, 1) - (1, 0, 0) = (1, 1, 1). Next, compute the cross product: AC→×AD→=⟨−1,0,3⟩×⟨1,1,1⟩=⟨0⋅1−3⋅1,3⋅1−(−1)⋅1,(−1)⋅1−0⋅1⟩=⟨−3,4,−1⟩\overrightarrow{AC} \times \overrightarrow{AD} = \langle -1, 0, 3 \rangle \times \langle 1, 1, 1 \rangle = \langle 0 \cdot 1 - 3 \cdot 1, 3 \cdot 1 - (-1) \cdot 1, (-1) \cdot 1 - 0 \cdot 1 \rangle = \langle -3, 4, -1 \rangle. Then the scalar triple product is: AB→⋅(AC→×AD→)=⟨−1,2,0⟩⋅⟨−3,4,−1⟩=(−1)(−3)+(2)(4)+(0)(−1)=3+8+0=11\overrightarrow{AB} \cdot (\overrightarrow{AC} \times \overrightarrow{AD}) = \langle -1, 2, 0 \rangle \cdot \langle -3, 4, -1 \rangle = (-1)(-3) + (2)(4) + (0)(-1) = 3 + 8 + 0 = 11. Therefore, the volume is 16∣11∣=116\frac{1}{6}|11| = \frac{11}{6}.
  4. Consider three vectors u⃗\vec{u}, v⃗\vec{v}, and w⃗\vec{w} such that u⃗⋅(v⃗×w⃗)=−15\vec{u} \cdot (\vec{v} \times \vec{w}) = -15. What is the value of w⃗⋅(u⃗×v⃗)\vec{w} \cdot (\vec{u} \times \vec{v})?
    Answer: w⃗⋅(u⃗×v⃗)=−15\vec{w} \cdot (\vec{u} \times \vec{v}) = -15 This tests the cyclic property of scalar triple products. The scalar triple product is invariant under cyclic permutations: a⃗⋅(b⃗×c⃗)=b⃗⋅(c⃗×a⃗)=c⃗⋅(a⃗×b⃗)\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{c} \cdot (\vec{a} \times \vec{b}). Starting with u⃗⋅(v⃗×w⃗)=−15\vec{u} \cdot (\vec{v} \times \vec{w}) = -15, we can apply the cyclic property: u⃗⋅(v⃗×w⃗)=v⃗⋅(w⃗×u⃗)=w⃗⋅(u⃗×v⃗)\vec{u} \cdot (\vec{v} \times \vec{w}) = \vec{v} \cdot (\vec{w} \times \vec{u}) = \vec{w} \cdot (\vec{u} \times \vec{v}). Therefore, w⃗⋅(u⃗×v⃗)=−15\vec{w} \cdot (\vec{u} \times \vec{v}) = -15. The key insight is recognizing that we go from u⃗⋅(v⃗×w⃗)\vec{u} \cdot (\vec{v} \times \vec{w}) to w⃗⋅(u⃗×v⃗)\vec{w} \cdot (\vec{u} \times \vec{v}) by two cyclic permutations: first u→v→w→uu \to v \to w \to u gives v⃗⋅(w⃗×u⃗)\vec{v} \cdot (\vec{w} \times \vec{u}), then v→w→u→vv \to w \to u \to v gives w⃗⋅(u⃗×v⃗)\vec{w} \cdot (\vec{u} \times \vec{v}). Since each cyclic permutation preserves the value, the result is unchanged.
  5. A parallelepiped is formed by vectors u⃗=⟨1,0,1⟩\vec{u} = \langle 1, 0, 1 \rangle, v⃗=⟨0,1,1⟩\vec{v} = \langle 0, 1, 1 \rangle, and w⃗=⟨1,1,0⟩\vec{w} = \langle 1, 1, 0 \rangle. If we want to construct a new parallelepiped with twice the volume using vectors u⃗\vec{u}, v⃗\vec{v}, and kw⃗k\vec{w} for some scalar kk, what are the possible values of kk?
    Answer: k=2k = 2 or k=−2k = -2 The original volume is ∣u⃗⋅(v⃗×w⃗)∣|\vec{u} \cdot (\vec{v} \times \vec{w})|. First, compute v⃗×w⃗=⟨0,1,1⟩×⟨1,1,0⟩=⟨(1)(0)−(1)(1),(1)(1)−(0)(0),(0)(1)−(1)(1)⟩=⟨−1,1,−1⟩\vec{v} \times \vec{w} = \langle 0, 1, 1 \rangle \times \langle 1, 1, 0 \rangle = \langle (1)(0) - (1)(1), (1)(1) - (0)(0), (0)(1) - (1)(1) \rangle = \langle -1, 1, -1 \rangle. Then u⃗⋅(v⃗×w⃗)=⟨1,0,1⟩⋅⟨−1,1,−1⟩=1(−1)+0(1)+1(−1)=−2\vec{u} \cdot (\vec{v} \times \vec{w}) = \langle 1, 0, 1 \rangle \cdot \langle -1, 1, -1 \rangle = 1(-1) + 0(1) + 1(-1) = -2. The original volume is ∣−2∣=2|-2| = 2. For the new parallelepiped with vectors u⃗\vec{u}, v⃗\vec{v}, and kw⃗k\vec{w}, the volume is ∣u⃗⋅(v⃗×kw⃗)∣=∣u⃗⋅k(v⃗×w⃗)∣=∣k∣∣u⃗⋅(v⃗×w⃗)∣=∣k∣⋅2|\vec{u} \cdot (\vec{v} \times k\vec{w})| = |\vec{u} \cdot k(\vec{v} \times \vec{w})| = |k||\vec{u} \cdot (\vec{v} \times \vec{w})| = |k| \cdot 2. We want this to equal twice the original volume: ∣k∣⋅2=2⋅2=4|k| \cdot 2 = 2 \cdot 2 = 4. Therefore, ∣k∣=2|k| = 2, which gives k=2k = 2 or k=−2k = -2. Both values work because the volume depends on the absolute value of the scalar triple product.
  6. Given vectors p⃗=⟨3,1,−2⟩\vec{p} = \langle 3, 1, -2 \rangle, q⃗=⟨1,−1,1⟩\vec{q} = \langle 1, -1, 1 \rangle, and r⃗=⟨2,3,−1⟩\vec{r} = \langle 2, 3, -1 \rangle, determine the relationship between the scalar triple products p⃗⋅(q⃗×r⃗)\vec{p} \cdot (\vec{q} \times \vec{r}) and r⃗⋅(p⃗×q⃗)\vec{r} \cdot (\vec{p} \times \vec{q}).
    Answer: r⃗⋅(p⃗×q⃗)=p⃗⋅(q⃗×r⃗)\vec{r} \cdot (\vec{p} \times \vec{q}) = \vec{p} \cdot (\vec{q} \times \vec{r}) This question tests the cyclic property of scalar triple products. The scalar triple product is invariant under cyclic permutations: a⃗⋅(b⃗×c⃗)=b⃗⋅(c⃗×a⃗)=c⃗⋅(a⃗×b⃗)\vec{a} \cdot (\vec{b} \times \vec{c}) = \vec{b} \cdot (\vec{c} \times \vec{a}) = \vec{c} \cdot (\vec{a} \times \vec{b}). Applying this property: p⃗⋅(q⃗×r⃗)=q⃗⋅(r⃗×p⃗)=r⃗⋅(p⃗×q⃗)\vec{p} \cdot (\vec{q} \times \vec{r}) = \vec{q} \cdot (\vec{r} \times \vec{p}) = \vec{r} \cdot (\vec{p} \times \vec{q}). Therefore, r⃗⋅(p⃗×q⃗)=p⃗⋅(q⃗×r⃗)\vec{r} \cdot (\vec{p} \times \vec{q}) = \vec{p} \cdot (\vec{q} \times \vec{r}). We can verify this by calculation: q⃗×r⃗=⟨1,−1,1⟩×⟨2,3,−1⟩=⟨(−1)(−1)−(1)(3),(1)(2)−(1)(−1),(1)(3)−(−1)(2)⟩=⟨−2,3,5⟩\vec{q} \times \vec{r} = \langle 1, -1, 1 \rangle \times \langle 2, 3, -1 \rangle = \langle (-1)(-1) - (1)(3), (1)(2) - (1)(-1), (1)(3) - (-1)(2) \rangle = \langle -2, 3, 5 \rangle. Then p⃗⋅(q⃗×r⃗)=⟨3,1,−2⟩⋅⟨−2,3,5⟩=3(−2)+1(3)+(−2)(5)=−6+3−10=−13\vec{p} \cdot (\vec{q} \times \vec{r}) = \langle 3, 1, -2 \rangle \cdot \langle -2, 3, 5 \rangle = 3(-2) + 1(3) + (-2)(5) = -6 + 3 - 10 = -13. For the other expression: p⃗×q⃗=⟨3,1,−2⟩×⟨1,−1,1⟩=⟨(1)(1)−(−2)(−1),(−2)(1)−(3)(1),(3)(−1)−(1)(1)⟩=⟨−1,−5,−4⟩\vec{p} \times \vec{q} = \langle 3, 1, -2 \rangle \times \langle 1, -1, 1 \rangle = \langle (1)(1) - (-2)(-1), (-2)(1) - (3)(1), (3)(-1) - (1)(1) \rangle = \langle -1, -5, -4 \rangle. Then r⃗⋅(p⃗×q⃗)=⟨2,3,−1⟩⋅⟨−1,−5,−4⟩=2(−1)+3(−5)+(−1)(−4)=−2−15+4=−13\vec{r} \cdot (\vec{p} \times \vec{q}) = \langle 2, 3, -1 \rangle \cdot \langle -1, -5, -4 \rangle = 2(-1) + 3(-5) + (-1)(-4) = -2 - 15 + 4 = -13. Both expressions equal −13-13, confirming they are equal.
  7. Three vectors a⃗=⟨1,2,3⟩\vec{a} = \langle 1, 2, 3 \rangle, b⃗=⟨2,−1,1⟩\vec{b} = \langle 2, -1, 1 \rangle, and c⃗=⟨3,0,−1⟩\vec{c} = \langle 3, 0, -1 \rangle form a parallelepiped. If this parallelepiped is projected onto a plane perpendicular to the vector n⃗=⟨1,1,1⟩\vec{n} = \langle 1, 1, 1 \rangle, what happens to the volume?
    Answer: The projection creates a 2D figure with zero volume When a 3D parallelepiped is projected onto a plane, the result is a 2D parallelogram, which has area but zero volume. The question is testing whether students understand that "volume" refers specifically to 3D measurements. The original parallelepiped has volume ∣a⃗⋅(b⃗×c⃗)∣|\vec{a} \cdot (\vec{b} \times \vec{c})|. Let's calculate: b⃗×c⃗=⟨2,−1,1⟩×⟨3,0,−1⟩=⟨(−1)(−1)−(1)(0),(1)(3)−(2)(−1),(2)(0)−(−1)(3)⟩=⟨1,5,3⟩\vec{b} \times \vec{c} = \langle 2, -1, 1 \rangle \times \langle 3, 0, -1 \rangle = \langle (-1)(-1) - (1)(0), (1)(3) - (2)(-1), (2)(0) - (-1)(3) \rangle = \langle 1, 5, 3 \rangle. Then a⃗⋅(b⃗×c⃗)=⟨1,2,3⟩⋅⟨1,5,3⟩=1+10+9=20\vec{a} \cdot (\vec{b} \times \vec{c}) = \langle 1, 2, 3 \rangle \cdot \langle 1, 5, 3 \rangle = 1 + 10 + 9 = 20. So the original volume is 20 cubic units. However, when projected onto any plane, the result is a 2D figure with area but no volume. The other answer choices incorrectly suggest that a projection can have volume, which reflects a common misconception about projecting 3D objects onto planes.

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