Question

For three vectors $\vec{A} = (-xi - 6j - 2k),$ $\vec{B} = (-i + 4j + 3k)$ and $\vec{C} = (-8i - j + 3k),$ if $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0,$ then value of x is ___________.

Answer 1

First, compute $\vec{B} \times \vec{C}$:

$\vec{B} \times \vec{C} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\\ -1 & 4 & 3 \\\ -8 & -1 & 3 \end{vmatrix}$

$\vec{B} \times \vec{C} = \hat{i}(12 + 3) - \hat{j}(-3 + 24) + \hat{k}(-1 - (-32))$

$\vec{B} \times \vec{C} = 15\hat{i} - 21\hat{j} + 33\hat{k}$

Now, compute $\vec{A} \cdot (\vec{B} \times \vec{C})$:

$\vec{A} \cdot (\vec{B} \times \vec{C}) = (-x\hat{i} - 6\hat{j} - 2\hat{k}) \cdot (15\hat{i} - 21\hat{j} + 33\hat{k})$

$\vec{A} \cdot (\vec{B} \times \vec{C}) = (-x)(15) + (-6)(-21) + (-2)(33)$

$\vec{A} \cdot (\vec{B} \times \vec{C}) = -15x + 126 - 66$

$\vec{A} \cdot (\vec{B} \times \vec{C}) = -15x + 60$

Since $\vec{A} \cdot (\vec{B} \times \vec{C}) = 0$, we get:

$-15x + 60 = 0$

$15x = 60$

$x = 4$