Scalar triple product and their applications Questions in English

(A) Three vectors $\vec{a}, \vec{b},$ and $\vec{c}$ are collinear (or coplanar with a common point) if the scalar triple product of the vectors formed by their differences is zero,or more simply,if the determinant of the matrix formed by their components is zero.
Let the vectors be $\vec{a} = i + 2j + 3k$,$\vec{b} = \lambda i + 4j + 7k$,and $\vec{c} = -3i - 2j - 5k$.
For these vectors to be collinear,the determinant of the matrix formed by their components must be zero:
$\begin{vmatrix} 1 & 2 & 3 \\ \lambda & 4 & 7 \\ -3 & -2 & -5 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(4(-5) - 7(-2)) - 2(\lambda(-5) - 7(-3)) + 3(\lambda(-2) - 4(-3)) = 0$
$1(-20 + 14) - 2(-5\lambda + 21) + 3(-2\lambda + 12) = 0$
$-6 + 10\lambda - 42 - 6\lambda + 36 = 0$
$4\lambda - 12 = 0$
$4\lambda = 12$
$\lambda = 3$.

(A) Given $d = \lambda (a \times b) + \mu (b \times c) + \nu (c \times a)$ and $[a, b, c] = \frac{1}{8}$.
Taking the dot product of $d$ with $c$:
$d \cdot c = \lambda (a \times b) \cdot c + \mu (b \times c) \cdot c + \nu (c \times a) \cdot c$
$d \cdot c = \lambda [a, b, c] + 0 + 0 = \lambda \left(\frac{1}{8}\right) \implies \lambda = 8(d \cdot c)$.
Similarly,taking the dot product of $d$ with $a$:
$d \cdot a = \lambda (a \times b) \cdot a + \mu (b \times c) \cdot a + \nu (c \times a) \cdot a$
$d \cdot a = 0 + \mu [b, c, a] + 0 = \mu \left(\frac{1}{8}\right) \implies \mu = 8(d \cdot a)$.
Similarly,taking the dot product of $d$ with $b$:
$d \cdot b = \lambda (a \times b) \cdot b + \mu (b \times c) \cdot b + \nu (c \times a) \cdot b$
$d \cdot b = 0 + 0 + \nu [c, a, b] = \nu \left(\frac{1}{8}\right) \implies \nu = 8(d \cdot b)$.
Therefore,$\lambda + \mu + \nu = 8(d \cdot c) + 8(d \cdot a) + 8(d \cdot b) = 8d \cdot (a + b + c)$.

(D) The given determinant is the Gram determinant $G(\vec{r}_{1}, \vec{r}_{2}, \vec{r}_{3})$,which is equal to the square of the scalar triple product: $G(\vec{r}_{1}, \vec{r}_{2}, \vec{r}_{3}) = [\vec{r}_{1} \vec{r}_{2} \vec{r}_{3}]^2$.
Given that the determinant is $0$,it implies that $[\vec{r}_{1} \vec{r}_{2} \vec{r}_{3}] = 0$.
This condition means that the three vectors are coplanar,which implies they are linearly dependent.
If vectors are linearly dependent,they lie in the same plane.
Option $D$ states that all three vectors are perpendicular to each other,which would imply they are linearly independent (forming an orthogonal basis in $3D$ space),so the determinant would be non-zero.
Therefore,the statement in option $D$ is false.

(D) The scalar triple product is defined as $|(a \times b) \cdot c| = |a| |b| |c| |\sin \theta| |\cos \alpha|$,where $\theta$ is the angle between $a$ and $b$,and $\alpha$ is the angle between $(a \times b)$ and $c$.
Given $|(a \times b) \cdot c| = |a| |b| |c|$,we have $|\sin \theta| |\cos \alpha| = 1$.
Since the maximum value of $|\sin \theta|$ and $|\cos \alpha|$ is $1$,this equality holds only if $|\sin \theta| = 1$ and $|\cos \alpha| = 1$.
$|\sin \theta| = 1 \Rightarrow \theta = \frac{\pi}{2}$,which means $a \perp b$ $(a \cdot b = 0)$.
$|\cos \alpha| = 1 \Rightarrow \alpha = 0$ or $\pi$,which means $c$ is parallel to the vector $(a \times b)$.
Since $(a \times b)$ is perpendicular to both $a$ and $b$,$c$ must also be perpendicular to both $a$ and $b$.
Therefore,$c \perp a$ $(c \cdot a = 0)$ and $c \perp b$ $(c \cdot b = 0)$.
Thus,$a, b, c$ are mutually perpendicular,implying $a \cdot b = b \cdot c = c \cdot a = 0$.

(A) The scalar triple product is defined as $[a, b, c] = a \cdot (b \times c)$.
Given that $a, b, c$ are non-coplanar,$[a, b, c] \neq 0$.
We know the properties of scalar triple products: $[c, a, b] = [a, b, c]$ and $[b, a, c] = -[a, b, c]$.
Substituting these into the expression:
$\frac{a \cdot (b \times c)}{c \times a \cdot b} + \frac{b \cdot (a \times c)}{c \cdot (a \times b)} = \frac{[a, b, c]}{[c, a, b]} + \frac{[b, a, c]}{[c, a, b]}$
$= \frac{[a, b, c]}{[a, b, c]} + \frac{-[a, b, c]}{[a, b, c]}$
$= 1 - 1 = 0$.

(B) The scalar triple product is defined as $[a + b, b + c, c + a] = (a + b) \cdot \{(b + c) \times (c + a)\}$.
Expanding the cross product: $(b + c) \times (c + a) = b \times c + b \times a + c \times c + c \times a$.
Since $c \times c = 0$,this simplifies to $b \times c + b \times a + c \times a$.
Now,compute the dot product: $(a + b) \cdot (b \times c + b \times a + c \times a) = a \cdot (b \times c) + a \cdot (b \times a) + a \cdot (c \times a) + b \cdot (b \times c) + b \cdot (b \times a) + b \cdot (c \times a)$.
Using the properties of scalar triple products: $a \cdot (b \times a) = 0$,$a \cdot (c \times a) = 0$,$b \cdot (b \times c) = 0$,and $b \cdot (b \times a) = 0$.
This leaves: $a \cdot (b \times c) + b \cdot (c \times a) = [a, b, c] + [b, c, a]$.
Since $[a, b, c] = [b, c, a]$,the expression becomes $[a, b, c] + [a, b, c] = 2[a, b, c]$.

(C) Let the vectors be $\vec{a} = 2i - 3j + 0k$,$\vec{b} = i + j - k$,and $\vec{c} = 3i + 0j - k$.
The volume of a parallelepiped with concurrent edges $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|[\vec{a} \, \vec{b} \, \vec{c}]|$.
This is calculated as the determinant of the matrix formed by the components of the vectors:
$V = |\det \begin{bmatrix} 2 & -3 & 0 \\ 1 & 1 & -1 \\ 3 & 0 & -1 \end{bmatrix}|$
Expanding the determinant along the first row:
$V = |2(1(-1) - (-1)(0)) - (-3)(1(-1) - (-1)(3)) + 0|$
$V = |2(-1) + 3(-1 + 3)|$
$V = |-2 + 3(2)|$
$V = |-2 + 6| = |4| = 4$.
Thus,the volume of the parallelepiped is $4$ cubic units.

(C) The scalar triple product $[a, b, c]$ is defined as $a \cdot (b \times c)$.
If three vectors $a, b, c$ are coplanar,they lie in the same plane.
The vector $(b \times c)$ is perpendicular to the plane containing vectors $b$ and $c$.
Since $a$ is also in the same plane,$a$ is perpendicular to $(b \times c)$.
Therefore,the dot product of $a$ and $(b \times c)$ must be zero,i.e.,$a \cdot (b \times c) = 0$.
By the property of the scalar triple product,$[a, b, c] = [b, c, a] = [c, a, b] = (a \times b) \cdot c = 0$.

(A) The scalar triple product is defined as $[a \ c \ b] = a \cdot (c \times b)$.
Using the properties of the scalar triple product,we can rearrange the cyclic order: $[a \ c \ b] = c \cdot (b \times a)$.
Since $a$ and $b$ are parallel vectors,their cross product $b \times a = 0$.
Therefore,$[a \ c \ b] = c \cdot 0 = 0$.

(D) Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{a} \, \vec{b} \, \vec{c}] = 0$.
The scalar triple product is given by the determinant of the components of the vectors:
$\begin{vmatrix} 2 & -1 & 1 \\ 1 & 2 & -3 \\ 3 & \lambda & 5 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(2 \times 5 - (-3) \times \lambda) - (-1)(1 \times 5 - (-3) \times 3) + 1(1 \times \lambda - 2 \times 3) = 0$
$2(10 + 3\lambda) + 1(5 + 9) + 1(\lambda - 6) = 0$
$20 + 6\lambda + 14 + \lambda - 6 = 0$
$7\lambda + 28 = 0$
$7\lambda = -28$
$\lambda = -4$
Therefore,the correct option is $D$.

(D) Given $p = \frac{b \times c}{[a, b, c]}$,$q = \frac{c \times a}{[a, b, c]}$,and $r = \frac{a \times b}{[a, b, c]}$.
We calculate each term separately:
$(a+b) \cdot p = a \cdot p + b \cdot p = a \cdot \frac{b \times c}{[a, b, c]} + b \cdot \frac{b \times c}{[a, b, c]} = \frac{[a, b, c]}{[a, b, c]} + 0 = 1$.
$(b+c) \cdot q = b \cdot q + c \cdot q = b \cdot \frac{c \times a}{[a, b, c]} + c \cdot \frac{c \times a}{[a, b, c]} = \frac{[b, c, a]}{[a, b, c]} + 0 = 1$.
$(c+a) \cdot r = c \cdot r + a \cdot r = c \cdot \frac{a \times b}{[a, b, c]} + a \cdot \frac{a \times b}{[a, b, c]} = \frac{[c, a, b]}{[a, b, c]} + 0 = 1$.
Summing these results: $1 + 1 + 1 = 3$.

(A) Let the position vectors of the points be $\vec{a} = 3i - 2j - k,$ $\vec{b} = 2i + 3j - 4k,$ $\vec{c} = -i + j + 2k,$ and $\vec{d} = 4i + 5j + \lambda k.$
Since the four points are coplanar,the scalar triple product of the vectors $(\vec{b}-\vec{a}),$ $(\vec{c}-\vec{a}),$ and $(\vec{d}-\vec{a})$ must be zero.
$\vec{b}-\vec{a} = (2-3)i + (3-(-2))j + (-4-(-1))k = -i + 5j - 3k$
$\vec{c}-\vec{a} = (-1-3)i + (1-(-2))j + (2-(-1))k = -4i + 3j + 3k$
$\vec{d}-\vec{a} = (4-3)i + (5-(-2))j + (\lambda-(-1))k = i + 7j + (\lambda+1)k$
For coplanarity,the determinant of these vectors is zero:
$\begin{vmatrix} -1 & 5 & -3 \\ -4 & 3 & 3 \\ 1 & 7 & \lambda+1 \end{vmatrix} = 0$
$-1(3(\lambda+1) - 21) - 5(-4(\lambda+1) - 3) - 3(-28 - 3) = 0$
$-1(3\lambda + 3 - 21) - 5(-4\lambda - 4 - 3) - 3(-31) = 0$
$-3\lambda + 18 + 20\lambda + 35 + 93 = 0$
$17\lambda + 146 = 0$
$\lambda = -\frac{146}{17}$

(A) Given that $p = \frac{b \times c}{[a, b, c]}$,$q = \frac{c \times a}{[a, b, c]}$,and $r = \frac{a \times b}{[a, b, c]}$.
Summing these vectors,we get $p + q + r = \frac{b \times c + c \times a + a \times b}{[a, b, c]}$.
Now,calculate the dot product $(a + b + c) \cdot (p + q + r)$:
$(a + b + c) \cdot \left( \frac{b \times c + c \times a + a \times b}{[a, b, c]} \right) = \frac{a \cdot (b \times c) + a \cdot (c \times a) + a \cdot (a \times b) + b \cdot (b \times c) + b \cdot (c \times a) + b \cdot (a \times b) + c \cdot (b \times c) + c \cdot (c \times a) + c \cdot (a \times b)}{[a, b, c]}$.
Using the property of scalar triple products $[a, b, c] = a \cdot (b \times c)$ and noting that any scalar triple product with repeated vectors is $0$:
$= \frac{[a, b, c] + 0 + 0 + 0 + [b, c, a] + 0 + 0 + 0 + [c, a, b]}{[a, b, c]}$.
Since $[a, b, c] = [b, c, a] = [c, a, b]$,we have:
$= \frac{[a, b, c] + [a, b, c] + [a, b, c]}{[a, b, c]} = \frac{3[a, b, c]}{[a, b, c]} = 3$.

(C) The volume of a parallelepiped with edges represented by vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ is given by the absolute value of the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$,which is equal to the determinant of the matrix formed by the components of the vectors.
Given vectors are $\vec{a} = -12i + 0j + \alpha k$,$\vec{b} = 0i + 3j - 1k$,and $\vec{c} = 2i + 1j - 15k$.
The volume is given by:
$|\det(\vec{a}, \vec{b}, \vec{c})| = 546$
$\left| \begin{matrix} -12 & 0 & \alpha \\ 0 & 3 & -1 \\ 2 & 1 & -15 \end{matrix} \right| = \pm 546$
Expanding the determinant along the first row:
$-12(3(-15) - (-1)(1)) - 0(...) + \alpha(0(1) - 3(2)) = \pm 546$
$-12(-45 + 1) + \alpha(-6) = \pm 546$
$-12(-44) - 6\alpha = \pm 546$
$528 - 6\alpha = \pm 546$
Case $1$: $528 - 6\alpha = 546 \Rightarrow -6\alpha = 18 \Rightarrow \alpha = -3$
Case $2$: $528 - 6\alpha = -546 \Rightarrow -6\alpha = -1074 \Rightarrow \alpha = 179$
Comparing with the given options,the correct value is $\alpha = -3$.

(B) Since the vectors are coplanar,their scalar triple product must be zero.
$\begin{vmatrix} a & a & c \\ 1 & 0 & 1 \\ c & c & b \end{vmatrix} = 0$
Applying the column operation $C_2 \to C_2 - C_1$:
$\begin{vmatrix} a & 0 & c \\ 1 & -1 & 1 \\ c & 0 & b \end{vmatrix} = 0$
Expanding along the second column:
$-(-1) \begin{vmatrix} a & c \\ c & b \end{vmatrix} = 0$
$ab - c^2 = 0 \Rightarrow c^2 = ab$
Thus,$c = \sqrt{ab}$,which is the geometric mean of $a$ and $b$.

(D) The reciprocal vectors are defined as:
$a^{-1} = \frac{b \times c}{[a, b, c]}, b^{-1} = \frac{c \times a}{[a, b, c]}, c^{-1} = \frac{a \times b}{[a, b, c]}$
Now,the scalar triple product is:
$[a^{-1}, b^{-1}, c^{-1}] = (a^{-1} \times b^{-1}) \cdot c^{-1}$
Substituting the expressions:
$[a^{-1}, b^{-1}, c^{-1}] = \left( \frac{b \times c}{[a, b, c]} \times \frac{c \times a}{[a, b, c]} \right) \cdot \frac{a \times b}{[a, b, c]}$
Using the property $(b \times c) \times (c \times a) = [b, c, a]c = [a, b, c]c$:
$[a^{-1}, b^{-1}, c^{-1}] = \frac{[a, b, c]c}{[a, b, c]^2} \cdot \frac{a \times b}{[a, b, c]} = \frac{c \cdot (a \times b)}{[a, b, c]^2} = \frac{[a, b, c]}{[a, b, c]^2} = \frac{1}{[a, b, c]}$
Since $[a, b, c] \neq 0$,the value is non-zero.

(A) Three vectors $a, b,$ and $c$ are coplanar if and only if their scalar triple product is zero,i.e.,$[a, b, c] = 0$.
The scalar triple product is given by the determinant of the matrix formed by the components of the vectors:
$\begin{vmatrix} 1 & -1 & 1 \\ 1 & 2 & -1 \\ 3 & p & 5 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(2 \times 5 - (-1) \times p) - (-1)(1 \times 5 - (-1) \times 3) + 1(1 \times p - 2 \times 3) = 0$
$1(10 + p) + 1(5 + 3) + 1(p - 6) = 0$
$10 + p + 8 + p - 6 = 0$
$2p + 12 = 0$
$2p = -12$
$p = -6$

(B) The scalar triple product $[i, k, j]$ is defined as $i \cdot (k \times j)$.
Since $i, j, k$ are mutually perpendicular unit vectors,we know that $k \times j = -i$.
Substituting this into the expression,we get $i \cdot (-i) = -(i \cdot i)$.
Since $i$ is a unit vector,$i \cdot i = |i|^2 = 1^2 = 1$.
Therefore,$[i, k, j] = -1$.

(D) The volume of a parallelepiped defined by vectors $\vec{a}, \vec{b},$ and $\vec{c}$ is given by the scalar triple product $|[\vec{a} \, \vec{b} \, \vec{c}]|$.
$V = |\vec{a} \cdot (\vec{b} \times \vec{c})| = \left| \det \begin{bmatrix} 12 & 4 & 3 \\ 8 & -12 & -9 \\ 33 & -4 & -24 \end{bmatrix} \right|$
Expanding the determinant along the first row:
$V = |12((-12)(-24) - (-9)(-4)) - 4((8)(-24) - (-9)(33)) + 3((8)(-4) - (-12)(33))|$
$V = |12(288 - 36) - 4(-192 + 297) + 3(-32 + 396)|$
$V = |12(252) - 4(105) + 3(364)|$
$V = |3024 - 420 + 1092|$
$V = |3696| = 3696 \text{ cubic units.}$
Since $3696$ is not among the options,the correct choice is $(d)$.

(A) The volume of a parallelepiped defined by three concurrent vectors $\vec{a}, \vec{b}, \vec{c}$ is given by the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$,which is equal to the absolute value of the determinant of the matrix formed by the components of the vectors.
Let $\vec{a} = 2i + j - k$,$\vec{b} = i + 2j + 3k$,and $\vec{c} = -3i - j + k$.
The volume $V = \left| \det \begin{bmatrix} 2 & 1 & -1 \\ 1 & 2 & 3 \\ -3 & -1 & 1 \end{bmatrix} \right|$.
Calculating the determinant:
$V = |2(2(1) - 3(-1)) - 1(1(1) - 3(-3)) + (-1)(1(-1) - 2(-3))|$
$V = |2(2 + 3) - 1(1 + 9) - 1(-1 + 6)|$
$V = |2(5) - 1(10) - 1(5)|$
$V = |10 - 10 - 5|$
$V = |-5| = 5$ cubic units.

(A) Given that $x$ is a non-zero vector such that $x \cdot a = 0, x \cdot b = 0$,and $x \cdot c = 0$.
This implies that the vector $x$ is perpendicular to each of the vectors $a, b$,and $c$.
If $x$ is a non-zero vector perpendicular to $a, b$,and $c$,then $a, b$,and $c$ must lie in a plane perpendicular to $x$.
Therefore,the vectors $a, b$,and $c$ are coplanar.
For any three coplanar vectors,their scalar triple product is zero,i.e.,$[a, b, c] = 0$.

(B) Three vectors are coplanar if their scalar triple product is zero,which is equivalent to the determinant of the matrix formed by these vectors being zero.
Let the determinant be $D = \begin{vmatrix} -bc & b^2 + bc & c^2 + bc \\ a^2 + ac & -ac & c^2 + ac \\ a^2 + ab & b^2 + ab & -ab \end{vmatrix} = 0$.
Dividing the rows by $a, b, c$ respectively (or factoring out $a, b, c$ from columns),we simplify the determinant.
Factoring $a$ from the first column,$b$ from the second,and $c$ from the third,we get:
$abc \begin{vmatrix} -c & b+c & c+b \\ a+c & -c & c+a \\ a+b & b+a & -b \end{vmatrix} = 0$.
After performing row and column operations,the determinant simplifies to $(ab + bc + ca)^3 = 0$.
Thus,$ab + bc + ca = 0$.

(D) The scalar triple product $[a + b, b + c, c + a]$ can be expanded as follows:
$[a + b, b + c, c + a] = (a + b) \cdot ((b + c) \times (c + a))$
$= (a + b) \cdot (b \times c + b \times a + c \times c + c \times a)$
Since $c \times c = 0$,this simplifies to:
$= (a + b) \cdot (b \times c + b \times a + c \times a)$
$= a \cdot (b \times c) + a \cdot (b \times a) + a \cdot (c \times a) + b \cdot (b \times c) + b \cdot (b \times a) + b \cdot (c \times a)$
Using the properties of scalar triple products,terms with repeated vectors are zero:
$= [a, b, c] + 0 + 0 + 0 + 0 + [b, c, a]$
$= [a, b, c] + [a, b, c] = 2[a, b, c]$
Since $a, b, c$ are coplanar,their scalar triple product $[a, b, c] = 0$.
Therefore,$2[a, b, c] = 2(0) = 0$.

(B) The scalar triple product of three vectors $a, b, c$ is defined as $[a, b, c] = (a \times b) \cdot c$.
Substituting $c = a \times b$ into the formula,we get:
$[a, b, a \times b] = (a \times b) \cdot (a \times b)$.
Using the property of the dot product $v \cdot v = |v|^2$,we have:
$(a \times b) \cdot (a \times b) = |a \times b|^2$.
Therefore,the correct option is $B$.

(A) Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if their scalar triple product is zero, i.e., $[\vec{a} \vec{b} \vec{c}] = 0$.
The scalar triple product is given by the determinant of the components of the vectors:
$\begin{vmatrix} 2 & -3 & 4 \\ 1 & 2 & -1 \\ x & -1 & 2 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$2(2(2) - (-1)(-1)) - (-3)(1(2) - (-1)(x)) + 4(1(-1) - 2(x)) = 0$
$2(4 - 1) + 3(2 + x) + 4(-1 - 2x) = 0$
$2(3) + 6 + 3x - 4 - 8x = 0$
$6 + 6 + 3x - 4 - 8x = 0$
$8 - 5x = 0$
$5x = 8$
$x = \frac{8}{5}$

(C) The volume $V$ of a parallelepiped with coterminous edges represented by vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ is given by the scalar triple product $|\vec{a} \cdot (\vec{b} \times \vec{c})|$,which is equal to the absolute value of the determinant of the matrix formed by the components of these vectors.
Let $\vec{a} = 2i - 3j + 4k$,$\vec{b} = i + 2j - 2k$,and $\vec{c} = 3i - j + k$.
The volume $V = \left| \det \begin{bmatrix} 2 & -3 & 4 \\ 1 & 2 & -2 \\ 3 & -1 & 1 \end{bmatrix} \right|$.
Calculating the determinant:
$V = |2(2(1) - (-2)(-1)) - (-3)(1(1) - (-2)(3)) + 4(1(-1) - 2(3))|$
$V = |2(2 - 2) + 3(1 + 6) + 4(-1 - 6)|$
$V = |2(0) + 3(7) + 4(-7)|$
$V = |0 + 21 - 28|$
$V = |-7| = 7$.
Thus,the volume is $7 \ cubic \ units$.

(B) We know that for unit vectors $i, j, k$ along the coordinate axes,the cross products are defined as:
$j \times k = i$
$k \times i = j$
$i \times j = k$
Substituting these into the expression:
$i \cdot (j \times k) + j \cdot (k \times i) + k \cdot (i \times j) = i \cdot i + j \cdot j + k \cdot k$
Since the dot product of a unit vector with itself is $1$ (i.e.,$i \cdot i = 1, j \cdot j = 1, k \cdot k = 1$):
$1 + 1 + 1 = 3$
Therefore,the correct option is $B$.

(C) The volume of a parallelepiped with coterminous edges $a$,$b$,and $c$ is given by the scalar triple product $|a \cdot (b \times c)|$,which is equal to the absolute value of the determinant of the matrix formed by the components of the vectors.
Volume $= \left| \det \begin{bmatrix} -3 & 7 & 5 \\ -3 & 7 & -3 \\ 7 & -5 & -3 \end{bmatrix} \right|$
Expanding the determinant along the first row:
$= | -3((7)(-3) - (-3)(-5)) - 7((-3)(-3) - (-3)(7)) + 5((-3)(-5) - (7)(7)) |$
$= | -3(-21 - 15) - 7(9 + 21) + 5(15 - 49) |$
$= | -3(-36) - 7(30) + 5(-34) |$
$= | 108 - 210 - 170 |$
$= | -272 |$
$= 272$ cubic units.

(C) The vector product $a \times b$ results in a vector that is perpendicular to both $a$ and $b$.
Since the dot product of two perpendicular vectors is always zero,we have $a \cdot (a \times b) = 0$.
This is a fundamental property of the scalar triple product $[a, a, b] = 0$ because two vectors are identical.

(D) The volume of a parallelepiped with conterminous edges $\vec{u}, \vec{v}, \vec{w}$ is given by the scalar triple product $[\vec{u} \vec{v} \vec{w}] = \vec{u} \cdot (\vec{v} \times \vec{w})$.
Here,the edges are $\vec{u} = \vec{a} - \vec{b}$,$\vec{v} = \vec{b} - \vec{c}$,and $\vec{w} = \vec{c} - \vec{a}$.
Volume $= (\vec{a} - \vec{b}) \cdot ((\vec{b} - \vec{c}) \times (\vec{c} - \vec{a}))$.
First,calculate the cross product: $(\vec{b} - \vec{c}) \times (\vec{c} - \vec{a}) = \vec{b} \times \vec{c} - \vec{b} \times \vec{a} - \vec{c} \times \vec{c} + \vec{c} \times \vec{a} = \vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a}$ (since $\vec{c} \times \vec{c} = 0$ and $-\vec{b} \times \vec{a} = \vec{a} \times \vec{b}$).
Now,take the dot product with $(\vec{a} - \vec{b})$:
$(\vec{a} - \vec{b}) \cdot (\vec{b} \times \vec{c} + \vec{a} \times \vec{b} + \vec{c} \times \vec{a})$
$= \vec{a} \cdot (\vec{b} \times \vec{c}) + \vec{a} \cdot (\vec{a} \times \vec{b}) + \vec{a} \cdot (\vec{c} \times \vec{a}) - \vec{b} \cdot (\vec{b} \times \vec{c}) - \vec{b} \cdot (\vec{a} \times \vec{b}) - \vec{b} \cdot (\vec{c} \times \vec{a})$.
Using the property that the scalar triple product is zero if any two vectors are identical,we have:
$\vec{a} \cdot (\vec{a} \times \vec{b}) = 0$,$\vec{a} \cdot (\vec{c} \times \vec{a}) = 0$,$\vec{b} \cdot (\vec{b} \times \vec{c}) = 0$,$\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$.
Thus,the expression simplifies to $\vec{a} \cdot (\vec{b} \times \vec{c}) - \vec{b} \cdot (\vec{c} \times \vec{a}) = [\vec{a} \vec{b} \vec{c}] - [\vec{b} \vec{c} \vec{a}] = [\vec{a} \vec{b} \vec{c}] - [\vec{a} \vec{b} \vec{c}] = 0$.

(C) The scalar triple product of three vectors $u, v, w$ is denoted by $[u, v, w] = u \cdot (v \times w)$.
By the cyclic property of the scalar triple product,$[u, v, w] = [v, w, u] = [w, u, v]$.
Option $(a)$ is $u \cdot (v \times w) = [u, v, w]$.
Option $(b)$ is $(v \times w) \cdot u = [v, w, u] = [u, v, w]$.
Option $(d)$ is $(u \times v) \cdot w = [u, v, w]$.
Option $(c)$ is $v \cdot (u \times w) = [v, u, w] = -[u, v, w]$.
Since $[u, v, w] \neq -[u, v, w]$ (in general),option $(c)$ is not equal to the others.

(A) Let $u$,$v$,and $w$ be vectors.
$1$. Consider $u \cdot (v \times w)$: The expression $(v \times w)$ results in a vector. The dot product of a vector $u$ with the resulting vector $(v \times w)$ is defined and results in a scalar. Thus,this expression is meaningful.
$2$. Consider $(u \cdot v) \cdot w$: The expression $(u \cdot v)$ results in a scalar. The dot product of a scalar with a vector $w$ is not defined. Thus,this expression is not meaningful.
$3$. Consider $(u \cdot v) \times w$: The expression $(u \cdot v)$ results in a scalar. The cross product of a scalar with a vector $w$ is not defined. Thus,this expression is not meaningful.
Therefore,only the first expression is meaningful.

(B) Given the vector equation $d = \lambda a + \mu b + \nu c$.
To find $\lambda$,take the scalar triple product of both sides with vectors $b$ and $c$.
Taking the dot product of both sides with $(b \times c)$:
$d \cdot (b \times c) = (\lambda a + \mu b + \nu c) \cdot (b \times c)$
Since the scalar triple product of vectors with repeated components is zero,we have:
$b \cdot (b \times c) = 0$ and $c \cdot (b \times c) = 0$.
Thus,$d \cdot (b \times c) = \lambda [a, b, c]$.
Therefore,$\lambda = \frac{[d, b, c]}{[a, b, c]}$.
Using the cyclic property of the scalar triple product,$[d, b, c] = [b, c, d]$ and $[a, b, c] = [b, c, a]$.
So,$\lambda = \frac{[b, c, d]}{[b, c, a]}$.

(B) For a set of vectors $\vec{A}$,$\vec{B}$,and $\vec{C}$ to form a left-handed system,the scalar triple product $[\vec{A} \, \vec{B} \, \vec{C}]$ must be negative,i.e.,$[\vec{A} \, \vec{B} \, \vec{C}] < 0$.
First,calculate the cross product $\vec{A} \times \vec{B}$:
$\vec{A} \times \vec{B} = \begin{vmatrix} i & j & k \\ 2 & 3 & 4 \\ 1 & 1 & 5 \end{vmatrix} = i(15 - 4) - j(10 - 4) + k(2 - 3) = 11i - 6j - k$.
For the system to be left-handed,we require $\vec{C} \cdot (\vec{A} \times \vec{B}) < 0$.
Testing option $B$: $\vec{C} = -11i + 6j + k$.
$\vec{C} \cdot (\vec{A} \times \vec{B}) = (-11)(11) + (6)(-6) + (1)(-1) = -121 - 36 - 1 = -158$.
Since $-158 < 0$,the vectors form a left-handed system.

(C) The volume of a parallelepiped with coterminous edges represented by vectors $a$,$b$,and $c$ is given by the scalar triple product $|[a, b, c]|$.
Given vectors:
$a = 1i - 1j + 1k$
$b = 1i - 3j + 4k$
$c = 2i - 5j + 3k$
The scalar triple product is the determinant of the matrix formed by these vectors:
$|[a, b, c]| = \begin{vmatrix} 1 & -1 & 1 \\ 1 & -3 & 4 \\ 2 & -5 & 3 \end{vmatrix}$
Expanding the determinant along the first row:
$= 1((-3)(3) - (4)(-5)) - (-1)((1)(3) - (4)(2)) + 1((1)(-5) - (-3)(2))$
$= 1(-9 + 20) + 1(3 - 8) + 1(-5 + 6)$
$= 1(11) + 1(-5) + 1(1)$
$= 11 - 5 + 1 = 7$
Thus,the volume of the parallelepiped is $7$ cubic units.

(B) Given that $a \cdot (b \times c) = \lambda \neq 0$.
We need to evaluate the expression $\frac{(b \times c) \cdot (a + b + c)}{\lambda}$.
Using the distributive property of the dot product:
$\frac{(b \times c) \cdot a + (b \times c) \cdot b + (b \times c) \cdot c}{\lambda}$
Since the scalar triple product of vectors with repeated components is zero,we have $(b \times c) \cdot b = 0$ and $(b \times c) \cdot c = 0$.
Thus,the expression simplifies to $\frac{(b \times c) \cdot a + 0 + 0}{\lambda}$.
Since the scalar triple product is cyclic,$(b \times c) \cdot a = a \cdot (b \times c) = \lambda$.
Substituting this into the expression,we get $\frac{\lambda}{\lambda} = 1$.

(A) The scalar triple product of three vectors $a, b,$ and $c$ is defined as $[a, b, c] = a \cdot (b \times c)$.
If three vectors are coplanar,they lie in the same plane.
For any three coplanar vectors,the volume of the parallelepiped formed by them is zero.
Since the scalar triple product represents the volume of the parallelepiped,it follows that $[a, b, c] = 0$ for any set of coplanar vectors.
Therefore,if $a, b,$ and $c$ are unit coplanar vectors,their scalar triple product is $0$.

(B) The vectors $\vec{a} = i + 3j - 2k$,$\vec{b} = 2i - j + 4k$,and $\vec{c} = 3i + 2j + xk$ are coplanar if and only if their scalar triple product is zero,i.e.,$[\vec{a} \, \vec{b} \, \vec{c}] = 0$.
This condition is equivalent to the determinant of the matrix formed by their components being zero:
$\left| \begin{array}{ccc} 1 & 3 & -2 \\ 2 & -1 & 4 \\ 3 & 2 & x \end{array} \right| = 0$
Expanding the determinant along the first row:
$1((-1)(x) - (4)(2)) - 3((2)(x) - (4)(3)) + (-2)((2)(2) - (-1)(3)) = 0$
$1(-x - 8) - 3(2x - 12) - 2(4 + 3) = 0$
$-x - 8 - 6x + 36 - 14 = 0$
$-7x + 14 = 0$
$7x = 14$
$x = 2$
Thus,the value of $x$ is $2$.

(A) The scalar triple product is defined as $[x, y, z] = (x \times y) \cdot z$.
We need to evaluate $[a - b, b - c, c - a]$.
Using the properties of the scalar triple product,we have:
$[a - b, b - c, c - a] = (a - b) \cdot ((b - c) \times (c - a))$
$= (a - b) \cdot (b \times c - b \times a - c \times c + c \times a)$
Since $c \times c = 0$,this simplifies to:
$= (a - b) \cdot (b \times c - b \times a + c \times a)$
$= a \cdot (b \times c) - a \cdot (b \times a) + a \cdot (c \times a) - b \cdot (b \times c) + b \cdot (b \times a) - b \cdot (c \times a)$
Using the property that the scalar triple product is zero if any two vectors are identical:
$a \cdot (b \times a) = [a, b, a] = 0$
$a \cdot (c \times a) = [a, c, a] = 0$
$b \cdot (b \times c) = [b, b, c] = 0$
$b \cdot (b \times a) = [b, b, a] = 0$
Thus,the expression simplifies to:
$= [a, b, c] - [b, c, a]$
Since $[a, b, c] = [b, c, a]$,we get:
$= [a, b, c] - [a, b, c] = 0$.

(D) Since the vectors $\vec{a}$,$\vec{b}$,and $\vec{c}$ are coplanar,their scalar triple product must be zero,i.e.,$[\vec{a} \, \vec{b} \, \vec{c}] = 0$.
This is given by the determinant:
$\begin{vmatrix} 1 & 1 & 1 \\ 2 & 0 & -4 \\ 1 & \lambda & 3 \end{vmatrix} = 0$
Expanding the determinant along the first row:
$1(0 - (-4\lambda)) - 1(6 - (-4)) + 1(2\lambda - 0) = 0$
$1(4\lambda) - 1(10) + 1(2\lambda) = 0$
$4\lambda - 10 + 2\lambda = 0$
$6\lambda = 10$
$\lambda = \frac{10}{6} = \frac{5}{3}$
Since $\frac{5}{3}$ is not among the given options,the correct choice is $(d)$.

(D) For three vectors $\overrightarrow{A}$,$\overrightarrow{B}$,and $\overrightarrow{C}$ to be coplanar,their scalar triple product must be zero,i.e.,$[\overrightarrow{A} \, \overrightarrow{B} \, \overrightarrow{C}] = 0$.
The scalar triple product is given by the determinant:
$[\overrightarrow{A} \, \overrightarrow{B} \, \overrightarrow{C}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ C_1 & C_2 & C_3 \end{vmatrix} = 0$.
Substituting the given values $C_2 = -1$ and $C_3 = 1$:
$[\overrightarrow{A} \, \overrightarrow{B} \, \overrightarrow{C}] = \begin{vmatrix} 1 & 1 & 1 \\ 1 & 0 & 0 \\ C_1 & -1 & 1 \end{vmatrix} = 0$.
Expanding along the second row:
$-1 \times \begin{vmatrix} 1 & 1 \\ -1 & 1 \end{vmatrix} = 0
\Rightarrow -1 \times (1 - (-1)) = 0
\Rightarrow -1 \times (2) = 0
\Rightarrow -2 = 0$.
Since $-2 \neq 0$,the scalar triple product is independent of $C_1$ and is never zero. Therefore,no value of $C_1$ can make these vectors coplanar.

(C) The scalar triple product $[a\,b\,c]$ is given by the determinant of the components of the vectors $a$, $b$, and $c$:
$[a\,b\,c] = \begin{vmatrix} 1 & 0 & -1 \\ x & 1 & 1 - x \\ y & x & 1 + x - y \end{vmatrix}$
Applying the column operation $C_3 \to C_3 + C_1$:
$[a\,b\,c] = \begin{vmatrix} 1 & 0 & 0 \\ x & 1 & 1 \\ y & x & 1 + x \end{vmatrix}$
Expanding along the first row:
$[a\,b\,c] = 1 \times \begin{vmatrix} 1 & 1 \\ x & 1 + x \end{vmatrix} - 0 + 0 = (1 + x) - x = 1$.
Since the result is a constant $1$, the value of $[a\,b\,c]$ does not depend on $x$ or $y$.

(B) The scalar triple product $a \cdot (b \times c)$ is given by the determinant of the matrix formed by the components of vectors $a$,$b$,and $c$:
$a \cdot (b \times c) = \begin{vmatrix} 3 & -2 & 2 \\ 6 & 4 & -2 \\ 3 & -2 & -4 \end{vmatrix}$
Expanding the determinant along the first row:
$= 3 \times [(4)(-4) - (-2)(-2)] - (-2) \times [(6)(-4) - (-2)(3)] + 2 \times [(6)(-2) - (4)(3)]$
$= 3 \times [-16 - 4] + 2 \times [-24 + 6] + 2 \times [-12 - 12]$
$= 3(-20) + 2(-18) + 2(-24)$
$= -60 - 36 - 48$
$= -144$

(B) We are given the expression $(a + b) \cdot ((b + c) \times (a + b + c))$.
First,simplify the cross product term: $(b + c) \times (a + b + c) = b \times a + b \times b + b \times c + c \times a + c \times b + c \times c$.
Since $b \times b = 0$ and $c \times c = 0$,and using the property $c \times b = -(b \times c)$,we get:
$(b + c) \times (a + b + c) = b \times a + b \times c + c \times a - b \times c = b \times a + c \times a$.
Now,take the dot product with $(a + b)$:
$(a + b) \cdot (b \times a + c \times a) = a \cdot (b \times a) + a \cdot (c \times a) + b \cdot (b \times a) + b \cdot (c \times a)$.
Using the properties of scalar triple products $[x y z] = x \cdot (y \times z)$:
$a \cdot (b \times a) = [a b a] = 0$ (since two vectors are identical).
$a \cdot (c \times a) = [a c a] = 0$.
$b \cdot (b \times a) = [b b a] = 0$.
$b \cdot (c \times a) = [b c a] = [a b c]$.
Thus,the expression simplifies to $[a b c]$.

(C) The scalar triple product is defined as $[a, b, c] = a \cdot (b \times c)$.
By the cyclic property of the scalar triple product,the value remains unchanged under cyclic permutation of the vectors.
Therefore,$a \cdot (b \times c) = b \cdot (c \times a) = c \cdot (a \times b)$.
Comparing this with the given options,$b \cdot (c \times a)$ is equivalent to $a \cdot (b \times c)$.

(A) We know that the scalar triple product of three vectors $x, y, z$ is given by $[x, y, z] = x \cdot (y \times z)$.
Let $x = a \times b$,$y = b \times c$,and $z = c \times a$.
Then $[a \times b, b \times c, c \times a] = (a \times b) \cdot ((b \times c) \times (c \times a))$.
Using the vector triple product identity $(p \times q) \times r = (p \cdot r)q - (q \cdot r)p$,we have:
$(b \times c) \times (c \times a) = ([b, c, a]c - [b, c, c]a)$.
Since $[b, c, c] = 0$ and $[b, c, a] = [a, b, c]$,we get:
$(b \times c) \times (c \times a) = [a, b, c]c$.
Substituting this back into the expression:
$[a \times b, b \times c, c \times a] = (a \times b) \cdot ([a, b, c]c) = [a, b, c] (a \times b) \cdot c = [a, b, c] [a, b, c]$.
Given $[a, b, c] = 4$,we have:
$[a \times b, b \times c, c \times a] = 4 \times 4 = 16$.