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

Correct Answer: 4

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:

\[ -15x + 60 = 0 \]

\[ 15x = 60 \]

\[ x = 4 \]