Question

Let \[ \vec{a} = \hat{i} + \hat{j} + \hat{k}, \quad \vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k}, \quad \text{and} \quad \vec{c} = x\hat{i} + 2\hat{j} + 3\hat{k}, \, x \in \mathbb{R}. \] If \( \vec{d} \) is the unit vector in the direction of \( \vec{b} + \vec{c} \) such that \( \vec{a} \cdot \vec{d} = 1 \), then \( (\vec{a} \times \vec{b}) \cdot \vec{c} \) is equal to:

• A. 9
• B. 6
• C. 3
• D. 11

Answer 1

The Correct Option is D

Step 1. \( \vec{d} = \lambda (\vec{b} + \vec{c}) \), where \( \lambda \) is a scalar constant.
Given \( \vec{a} \cdot \vec{d} = 1 \):

Substituting:
\( \vec{a} \cdot \vec{d} = \lambda (\vec{a} \cdot (\vec{b} + \vec{c})) \).
\( 1 = \lambda (\vec{a} \cdot (\vec{b} + \vec{c})) = \lambda (1 + x + 5) \).
Simplify:
\( 1 = \lambda (x + 6) \quad \dots \, (1). \)

Step 2. Since \( |\vec{d}| = 1 \):
\( |\vec{d}| = |\lambda (\vec{b} + \vec{c})| = 1 \).
Substituting \( \lambda = \frac{1}{x+6} \):
\( \left| \frac{1}{x+6} (\vec{b} + \vec{c}) \right| = 1 \).
Simplify:
\( |\vec{b} + \vec{c}|^2 = (x+6)^2 \).

Expand \( \vec{b} + \vec{c} = (2 + x)\hat{i} + 6\hat{j} - 2\hat{k} \):
\( |\vec{b} + \vec{c}|^2 = (x+2)^2 + 6^2 + (-2)^2 = x^2 + 4x + 4 + 36 + 4 \).
Equate:
\( x^2 + 4x + 44 = (x+6)^2 = x^2 + 12x + 36. \)
Simplify:
\( 8x = 8 \implies x = 1. \)

Step 3. Calculate \( (\vec{a} \times \vec{b}) \cdot \vec{c} \):
Expand:

\( \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 4 & -5 \end{vmatrix}. \)
Simplify:
\( \vec{a} \times \vec{b} = \hat{i}(1 \cdot -5 - 1 \cdot 4) - \hat{j}(1 \cdot -5 - 1 \cdot 2) + \hat{k}(1 \cdot 4 - 1 \cdot 2). \)
\( \vec{a} \times \vec{b} = -9\hat{i} + 3\hat{j} + 2\hat{k}. \)

Now:
\( (\vec{a} \times \vec{b}) \cdot \vec{c} = (-9)(1) + (3)(2) + (2)(3). \)
Simplify:
\( (\vec{a} \times \vec{b}) \cdot \vec{c} = 20 - 9 = 11. \)

Option (4) is correct.