Scalar Triple Product - Multivariable Calculus

1. Let $\vec{u} = \langle 2, 1, -3 \rangle$, $\vec{v} = \langle 1, -2, 1 \rangle$, and $\vec{w} = \langle -1, 3, 2 \rangle$. If $\vec{n}$ is a unit vector such that the scalar triple product $(\vec{u} + t\vec{n}) \cdot (\vec{v} \times \vec{w}) = 0$ for some real number $t$, which of the following could be $\vec{n}$?
   Answer: $\frac{1}{\sqrt{14}}\langle -1, 2, -3 \rangle$ First, compute $\vec{v} \times \vec{w} = \langle 1,-2,1 \rangle \times \langle -1,3,2 \rangle = \langle (-2)(2)-(1)(3), (1)(-1)-(1)(2), (1)(3)-(-2)(-1) \rangle = \langle -7, -3, -1 \rangle$. The condition $(\vec{u} + t\vec{n}) \cdot (\vec{v} \times \vec{w}) = 0$ becomes $\vec{u} \cdot (\vec{v} \times \vec{w}) + t\vec{n} \cdot (\vec{v} \times \vec{w}) = 0$. We have $\vec{u} \cdot (\vec{v} \times \vec{w}) = \langle 2,1,-3 \rangle \cdot \langle -7,-3,-1 \rangle = 2(-7) + 1(-3) + (-3)(-1) = -14 - 3 + 3 = -14$. So we need $-14 + t\vec{n} \cdot \langle -7,-3,-1 \rangle = 0$, which means $t\vec{n} \cdot \langle -7,-3,-1 \rangle = 14$. For this to have a solution, $\vec{n}$ must not be perpendicular to $\langle -7,-3,-1 \rangle$. Let's check each option: (A) $\frac{1}{\sqrt{14}}\langle 1,2,3 \rangle \cdot \langle -7,-3,-1 \rangle = \frac{1}{\sqrt{14}}(-7-6-3) = \frac{-16}{\sqrt{14}}$. (B) $\frac{1}{\sqrt{14}}\langle 3,2,1 \rangle \cdot \langle -7,-3,-1 \rangle = \frac{1}{\sqrt{14}}(-21-6-1) = \frac{-28}{\sqrt{14}} = -2\sqrt{14}$. (C) $\frac{1}{\sqrt{14}}\langle -1,2,-3 \rangle \cdot \langle -7,-3,-1 \rangle = \frac{1}{\sqrt{14}}(7-6+3) = \frac{4}{\sqrt{14}}$. (D) $\frac{1}{\sqrt{14}}\langle -3,2,1 \rangle \cdot \langle -7,-3,-1 \rangle = \frac{1}{\sqrt{14}}(21-6-1) = \frac{14}{\sqrt{14}} = \sqrt{14}$. Any of these could work since none are zero, but we need $t \cdot \text{(dot product)} = 14$. For option (C), $t \cdot \frac{4}{\sqrt{14}} = 14$ gives $t = \frac{14\sqrt{14}}{4} = \frac{7\sqrt{14}}{2}$. This is valid.
2. Let $\vec{x} = \langle 2, 1, -1 \rangle$, $\vec{y} = \langle 1, -1, 2 \rangle$, and $\vec{z} = \langle -1, 2, 1 \rangle$. If the vectors $\vec{u} = \vec{x} + k\vec{y}$, $\vec{v} = \vec{y} + k\vec{z}$, and $\vec{w} = \vec{z} + k\vec{x}$ are coplanar for some value of $k$, what is the value of $k$?
   Answer: $-1$ Three vectors are coplanar if and only if their scalar triple product is zero. We need $\vec{u} \cdot (\vec{v} \times \vec{w}) = 0$. First, express the vectors in terms of $k$: $\vec{u} = \langle 2+k, 1-k, -1+2k \rangle$, $\vec{v} = \langle 1-k, -1+2k, 2+k \rangle$, $\vec{w} = \langle -1+2k, 2+k, 1-k \rangle$. The scalar triple product is the determinant: $\begin{vmatrix} 2+k & 1-k & -1+2k \\ 1-k & -1+2k & 2+k \\ -1+2k & 2+k & 1-k \end{vmatrix} = 0$. 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: $\vec{w} = \alpha\vec{u} + \beta\vec{v}$ for some scalars $\alpha$ and $\beta$. Substituting: $\vec{z} + k\vec{x} = \alpha(\vec{x} + k\vec{y}) + \beta(\vec{y} + k\vec{z})$. Rearranging: $\vec{z} + k\vec{x} = \alpha\vec{x} + \alpha k\vec{y} + \beta\vec{y} + \beta k\vec{z}$. Collecting terms: $\vec{z} + k\vec{x} = \alpha\vec{x} + (\alpha k + \beta)\vec{y} + \beta k\vec{z}$. Since $\vec{x}$, $\vec{y}$, and $\vec{z}$ are linearly independent (we can verify this by checking that their scalar triple product is non-zero), we must have: coefficient of $\vec{x}$: $k = \alpha$, coefficient of $\vec{y}$: $0 = \alpha k + \beta = k^2 + \beta$, coefficient of $\vec{z}$: $1 = \beta k$. From the third equation, $\beta = \frac{1}{k}$ (assuming $k \neq 0$). Substituting into the second equation: $0 = k^2 + \frac{1}{k}$, so $k^3 + 1 = 0$, giving $k^3 = -1$, thus $k = -1$. Let's verify: if $k = -1$, then $\alpha = -1$ and $\beta = \frac{1}{-1} = -1$. Check: $\alpha k + \beta = (-1)(-1) + (-1) = 1 - 1 = 0$ ✓. Also, $\vec{u} = \vec{x} - \vec{y} = \langle 2-1, 1+1, -1-2 \rangle = \langle 1, 2, -3 \rangle$, $\vec{v} = \vec{y} - \vec{z} = \langle 1+1, -1-2, 2-1 \rangle = \langle 2, -3, 1 \rangle$, $\vec{w} = \vec{z} - \vec{x} = \langle -1-2, 2-1, 1+1 \rangle = \langle -3, 1, 2 \rangle$. Verify coplanarity: $\vec{w} = -\vec{u} - \vec{v}$ means $\langle -3, 1, 2 \rangle = -\langle 1, 2, -3 \rangle - \langle 2, -3, 1 \rangle = \langle -1, -2, 3 \rangle + \langle -2, 3, -1 \rangle = \langle -3, 1, 2 \rangle$ ✓.
3. Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ satisfy $\vec{a} \cdot (\vec{b} \times \vec{c}) = 15$. If $\vec{a}' = 2\vec{a} + \vec{b}$, $\vec{b}' = \vec{a} - \vec{b}$, and $\vec{c}' = \vec{c} + 3\vec{a}$, what is the value of $\vec{a}' \cdot (\vec{b}' \times \vec{c}')$?
   Answer: $-45$ We need to compute $\vec{a}' \cdot (\vec{b}' \times \vec{c}')$ where $\vec{a}' = 2\vec{a} + \vec{b}$, $\vec{b}' = \vec{a} - \vec{b}$, and $\vec{c}' = \vec{c} + 3\vec{a}$. 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 $\vec{a}'$: coefficient of $\vec{a}$ is 2, $\vec{b}$ is 1, $\vec{c}$ is 0. For $\vec{b}'$: coefficient of $\vec{a}$ is 1, $\vec{b}$ is -1, $\vec{c}$ is 0. For $\vec{c}'$: coefficient of $\vec{a}$ is 3, $\vec{b}$ is 0, $\vec{c}$ is 1. The transformation matrix is $\begin{pmatrix}2&1&3\\1&-1&0\\0&0&1\end{pmatrix}$. Its determinant is $2[(-1)(1) - (0)(0)] - 1[(1)(1) - (0)(0)] + 3[(1)(0) - (-1)(0)] = 2(-1) - 1(1) + 3(0) = -2 - 1 = -3$. Therefore, $\vec{a}' \cdot (\vec{b}' \times \vec{c}') = -3 \times 15 = -45$.

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