Scalar triple product and their applications Questions in English

(A) The volume of a parallelepiped is given by the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$.
First, calculate the cross product $\vec{b} \times \vec{c}$:
$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 4 & -1 \\ 1 & 1 & 3 \end{vmatrix} = \hat{i}(12 - (-1)) - \hat{j}(6 - (-1)) + \hat{k}(2 - 4) = 13\hat{i} - 7\hat{j} - 2\hat{k}$.
The volume $V = |\vec{a} \cdot (\vec{b} \times \vec{c})| = |(1, 1, 1) \cdot (13, -7, -2)| = |13 - 7 - 2| = 4 \text{ cubic units}$.
The base area $A$ is the magnitude of the cross product of the base vectors $\vec{a}$ and $\vec{b}$:
$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 4 & -1 \end{vmatrix} = \hat{i}(-1 - 4) - \hat{j}(-1 - 2) + \hat{k}(4 - 2) = -5\hat{i} + 3\hat{j} + 2\hat{k}$.
$A = |\vec{a} \times \vec{b}| = \sqrt{(-5)^2 + 3^2 + 2^2} = \sqrt{25 + 9 + 4} = \sqrt{38}$.
The altitude $h = V / A = 4 / \sqrt{38} = 4 / (\sqrt{2} \cdot \sqrt{19}) = 2\sqrt{2} / \sqrt{19} \text{ units}$.

(A) Three vectors $\vec{a}$, $\vec{b}$, and $\vec{c}$ are coplanar if their scalar triple product is zero, i.e., $[\vec{a} \ \vec{b} \ \vec{c}] = 0$.
This is equivalent to the determinant of the matrix formed by their components being zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ 1 & 2 & c_3 \end{vmatrix} = 0$.
Expanding along the second row:
$-1 \begin{vmatrix} 1 & 1 \\ 2 & c_3 \end{vmatrix} = 0$.
$-1(c_3 - 2) = 0$.
$c_3 - 2 = 0$.
$c_3 = 2$.

(C) In parallelogram $OBDC$, diagonals are $\vec{d_1} = \vec{OD} = \hat{i} + 2\hat{j} + 6\hat{k}$ and $\vec{d_2} = \vec{CB} = \vec{OB} - \vec{OC} = -3\hat{i} - 2\hat{k}$.
We know $\vec{OD} = \vec{OB} + \vec{OC}$.
Adding the two equations: $2\vec{OB} = (\hat{i} + 2\hat{j} + 6\hat{k}) + (-3\hat{i} - 2\hat{k}) = -2\hat{i} + 2\hat{j} + 4\hat{k} \implies \vec{OB} = -\hat{i} + \hat{j} + 2\hat{k}$.
Subtracting the two equations: $2\vec{OC} = (\hat{i} + 2\hat{j} + 6\hat{k}) - (-3\hat{i} - 2\hat{k}) = 4\hat{i} + 2\hat{j} + 8\hat{k} \implies \vec{OC} = 2\hat{i} + \hat{j} + 4\hat{k}$.
The volume of the parallelepiped is given by the scalar triple product $|\vec{OA} \cdot (\vec{OB} \times \vec{OC})|$.
$\vec{OA} \cdot (\vec{OB} \times \vec{OC}) = \begin{vmatrix} 1 & 2 & 3 \\ -1 & 1 & 2 \\ 2 & 1 & 4 \end{vmatrix}$.
$= 1(4 - 2) - 2(-4 - 4) + 3(-1 - 2) = 1(2) - 2(-8) + 3(-3) = 2 + 16 - 9 = 9$.
Thus, the volume is $9$ cubic units.

(D) Let $V = [\vec{a} \vec{b} \vec{c}]$. By definition, $\vec{a} \cdot \vec{p} = \vec{a} \cdot \frac{\vec{b} \times \vec{c}}{V} = \frac{[\vec{a} \vec{b} \vec{c}]}{V} = 1$.
Similarly, $\vec{b} \cdot \vec{q} = 1$ and $\vec{c} \cdot \vec{r} = 1$.
Also, $\vec{b} \cdot \vec{p} = 0$, $\vec{c} \cdot \vec{p} = 0$, $\vec{c} \cdot \vec{q} = 0$, $\vec{a} \cdot \vec{q} = 0$, $\vec{a} \cdot \vec{r} = 0$, and $\vec{b} \cdot \vec{r} = 0$ because the dot product of a vector with a cross product containing itself is zero.
Now, expand the expression: $(\vec{a} + \vec{b}) \cdot \vec{p} + (\vec{b} + \vec{c}) \cdot \vec{q} + (\vec{c} + \vec{a}) \cdot \vec{r} = (\vec{a} \cdot \vec{p} + \vec{b} \cdot \vec{p}) + (\vec{b} \cdot \vec{q} + \vec{c} \cdot \vec{q}) + (\vec{c} \cdot \vec{r} + \vec{a} \cdot \vec{r})$.
Substituting the values: $(1 + 0) + (1 + 0) + (1 + 0) = 3$.

(B) Three vectors are coplanar if their scalar triple product is zero.
Let the vectors be $\vec{u} = a\hat{i} + a\hat{j} + c\hat{k}$, $\vec{v} = \hat{i} + 0\hat{j} + \hat{k}$, and $\vec{w} = c\hat{i} + c\hat{j} + b\hat{k}$.
The condition for coplanarity is $\det(\vec{u}, \vec{v}, \vec{w}) = 0$.
$\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$.
Expanding along the second row: $-1(ab - c^2) + 0 - 1(ac - ac) = 0$.
$-(ab - c^2) = 0$.
$c^2 = ab$.
$c = \sqrt{ab}$.
Thus, $c$ is the geometric mean of $a$ and $b$.

(A) Step $1$: Calculate the volume of the parallelepiped using the scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$.
$\vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 3 & -2 \\ -3 & 1 & 4 \end{vmatrix} = \hat{i}(12 - (-2)) - \hat{j}(4 - 6) + \hat{k}(1 - (-9)) = 14\hat{i} + 2\hat{j} + 10\hat{k}$.
Volume $V = |\vec{a} \cdot (\vec{b} \times \vec{c})| = |(-2)(14) + (5)(2) + (3)(10)| = |-28 + 10 + 30| = 12$.
Step $2$: Calculate the area of the base formed by $\vec{b}$ and $\vec{c}$, which is $|\vec{b} \times \vec{c}|$.
$|\vec{b} \times \vec{c}| = \sqrt{14^2 + 2^2 + 10^2} = \sqrt{196 + 4 + 100} = \sqrt{300} = 10\sqrt{3}$.
Step $3$: The altitude $h$ is given by $V / \text{Area} = 12 / (10\sqrt{3}) = 6 / (5\sqrt{3}) = (6\sqrt{3}) / 15 = 2\sqrt{3} / 5$.

(D) The function $f(x)$ is the scalar triple product $[\vec{a} \vec{b} \vec{c}]$, given by the determinant:
$f(x) = \begin{vmatrix} x & -2 & 3 \\ -2 & x & -1 \\ 7 & -2 & x \end{vmatrix}$
Expanding along the first row:
$f(x) = x(x^2 - 2) + 2(-2x + 7) + 3(4 - 7x)$
$f(x) = x^3 - 2x - 4x + 14 + 12 - 21x = x^3 - 27x + 26$
To find local maxima, find $f'(x) = 3x^2 - 27$. Setting $f'(x) = 0$ gives $x^2 = 9$, so $x = \pm 3$.
$f''(x) = 6x$. For local maxima, $f''(x) < 0$, so $x_0 = -3$.
Now calculate $\vec{a} \cdot \vec{c}$ at $x = -3$:
$\vec{a} = -3\hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{c} = 7\hat{i} - 2\hat{j} - 3\hat{k}$.
$\vec{a} \cdot \vec{c} = (-3)(7) + (-2)(-2) + (3)(-3) = -21 + 4 - 9 = -26$.

(B) The scalar triple product is defined as $[\vec{a} \quad \vec{a} + \vec{b} \quad \vec{a} \times \vec{b}] = \vec{a} \cdot ((\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b}))$.
Using the vector triple product identity $\vec{u} \times (\vec{v} \times \vec{w}) = (\vec{u} \cdot \vec{w})\vec{v} - (\vec{u} \cdot \vec{v})\vec{w}$, we expand $(\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b})$:
$(\vec{a} + \vec{b}) \times (\vec{a} \times \vec{b}) = \vec{a} \times (\vec{a} \times \vec{b}) + \vec{b} \times (\vec{a} \times \vec{b})$
$= ((\vec{a} \cdot \vec{b})\vec{a} - (\vec{a} \cdot \vec{a})\vec{b}) + ((\vec{b} \cdot \vec{b})\vec{a} - (\vec{b} \cdot \vec{a})\vec{b})$
$= (\vec{a} \cdot \vec{b})\vec{a} - |\vec{a}|^2 \vec{b} + |\vec{b}|^2 \vec{a} - (\vec{a} \cdot \vec{b})\vec{b}$
Now, take the dot product with $\vec{a}$:
$[\vec{a} \quad \vec{a} + \vec{b} \quad \vec{a} \times \vec{b}] = \vec{a} \cdot [(\vec{a} \cdot \vec{b})\vec{a} - |\vec{a}|^2 \vec{b} + |\vec{b}|^2 \vec{a} - (\vec{a} \cdot \vec{b})\vec{b}]$
$= (\vec{a} \cdot \vec{b})(\vec{a} \cdot \vec{a}) - |\vec{a}|^2 (\vec{a} \cdot \vec{b}) + |\vec{b}|^2 (\vec{a} \cdot \vec{a}) - (\vec{a} \cdot \vec{b})(\vec{a} \cdot \vec{b})$
$= (8)(16) - (16)(8) + (9)(16) - (8)(8)$
$= 0 + 144 - 64 = 80$.

(D) Since $\vec{c}$ lies in the plane of $\vec{a}$ and $\vec{b}$, the scalar triple product $[\vec{a} \ \vec{b} \ \vec{c}] = 0$.
This is equivalent to the determinant of the components being zero:
$\begin{vmatrix} 1 & 1 & 1 \\ 1 & -1 & 2 \\ x & x-2 & -1 \end{vmatrix} = 0$.
Expanding along the first row:
$1((-1)(-1) - (2)(x-2)) - 1((1)(-1) - (2)(x)) + 1((1)(x-2) - (-1)(x)) = 0$.
$1(1 - 2x + 4) - 1(-1 - 2x) + 1(x - 2 + x) = 0$.
$(5 - 2x) + (1 + 2x) + (2x - 2) = 0$.
$4 + 2x = 0$.
$2x = -4$.
$x = -2$.

(A) The volume of a tetrahedron with vertices $\vec{a}, \vec{b}, \vec{c}, \vec{d}$ is given by $V = \frac{1}{6} |(\vec{b}-\vec{a}) \cdot ((\vec{c}-\vec{a}) \times (\vec{d}-\vec{a}))|$.
Let $\vec{a} = (5, -1, 1)$, $\vec{b} = (7, -4, p)$, $\vec{c} = (1, -6, 10)$, $\vec{d} = (-1, -3, 7)$.
$\vec{b}-\vec{a} = (2, -3, p-1)$, $\vec{c}-\vec{a} = (-4, -5, 9)$, $\vec{d}-\vec{a} = (-6, -2, 6)$.
Volume $V = \frac{1}{6} |\det \begin{pmatrix} 2 & -3 & p-1 \\ -4 & -5 & 9 \\ -6 & -2 & 6 \end{pmatrix}| = 11$.
$|2(-30 + 18) + 3(-24 + 54) + (p-1)(8 - 30)| = 66$.
$|2(-12) + 3(30) + (p-1)(-22)| = 66$.
$|-24 + 90 - 22p + 22| = 66$.
$|88 - 22p| = 66$.
Case $1$: $88 - 22p = 66 \implies 22p = 22 \implies p = 1$.
Case $2$: $88 - 22p = -66 \implies 22p = 154 \implies p = 7$.
Thus, one of the values of $p$ is $1$.

(A) Given, the volume of the parallelepiped is $|[\vec{a} \vec{b} \vec{c}]| = 3$.
The volume of a tetrahedron with coterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by $V = \frac{1}{6} |[\vec{u} \vec{v} \vec{w}]|$.
Here, $\vec{u} = \vec{a} \times \vec{b}$, $\vec{v} = 2(\vec{a} \times \vec{c})$, $\vec{w} = 2(\vec{b} \times \vec{c})$.
Thus, $[\vec{u} \vec{v} \vec{w}] = [(\vec{a} \times \vec{b}) \ (2(\vec{a} \times \vec{c})) \ (2(\vec{b} \times \vec{c}))] = 4 [(\vec{a} \times \vec{b}) \ (\vec{a} \times \vec{c}) \ (\vec{b} \times \vec{c})]$.
Using the identity $[(\vec{a} \times \vec{b}) \ (\vec{b} \times \vec{c}) \ (\vec{c} \times \vec{a})] = [\vec{a} \vec{b} \vec{c}]^2$, we have $[(\vec{a} \times \vec{b}) \ (\vec{a} \times \vec{c}) \ (\vec{b} \times \vec{c})] = [\vec{a} \vec{b} \vec{c}]^2$.
So, $[\vec{u} \vec{v} \vec{w}] = 4 [\vec{a} \vec{b} \vec{c}]^2 = 4 \times (3)^2 = 4 \times 9 = 36$.
The volume of the tetrahedron is $V = \frac{1}{6} \times 36 = 6 \text{ cubic units}$.

(A) The determinant $\left| \begin{matrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{matrix} \right|$ represents the scalar triple product $\vec{a} \cdot (\vec{b} \times \vec{c})$.
Since $\vec{a}$ is a unit vector perpendicular to both $\vec{b}$ and $\vec{c}$, $\vec{a} = \pm \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|}$.
Thus, $\vec{a} \cdot (\vec{b} \times \vec{c}) = \pm \frac{\vec{b} \times \vec{c}}{|\vec{b} \times \vec{c}|} \cdot (\vec{b} \times \vec{c}) = \pm |\vec{b} \times \vec{c}|$.
Squaring both sides, we get the value as $|\vec{b} \times \vec{c}|^2$.
Using the formula $|\vec{b} \times \vec{c}| = |\vec{b}||\vec{c}| \sin(\theta)$, where $\theta = \frac{\pi}{3}$.
$|\vec{b} \times \vec{c}|^2 = |\vec{b}|^2|\vec{c}|^2 \sin^2(\frac{\pi}{3}) = |\vec{b}|^2|\vec{c}|^2 (\frac{\sqrt{3}}{2})^2 = \frac{3}{4}|\vec{b}|^2|\vec{c}|^2$.