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.

Answer 2

Step 1: Find \( \vec{a} \times \vec{b} \)
Given: \( \vec{a} = \hat{i} + \hat{j} + \hat{k} \), \( \vec{b} = 2\hat{i} + 4\hat{j} - 5\hat{k} \)
Cross product:
\[ \vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 1 & 1 \\ 2 & 4 & -5 \end{vmatrix} = \hat{i}[(1)(-5)-(1)(4)] - \hat{j}[1(-5)-1(2)] + \hat{k}[1(4)-1(2)] = \hat{i}(-5 - 4) - \hat{j}(-5 - 2) + \hat{k}(4 - 2) = -9\hat{i} + 7\hat{j} + 2\hat{k} \]
Step 2: Dot product with \( \vec{c} \)
\( \vec{c} = x\hat{i} + 2\hat{j} + 3\hat{k} \)
\[ (\vec{a} \times \vec{b}) \cdot \vec{c} = (-9)x + 7 \times 2 + 2 \times 3 = -9x + 14 + 6 = -9x + 20 \]
Step 3: Find x using vector condition
Given: Unit vector \( \vec{d} \) in the direction of \( \vec{b} + \vec{c} \) such that \( \vec{a} \cdot \vec{d} = 1 \)
\( \vec{b} + \vec{c} = (2 + x)\hat{i} + 6\hat{j} - 2\hat{k} \)
Magnitude: \( \sqrt{(2+x)^2 + 36 + 4} = \sqrt{(x+2)^2 + 40} \)
Unit vector \( \vec{d} = \frac{(x+2)\hat{i} + 6\hat{j} - 2\hat{k}}{\sqrt{(x+2)^2 + 40}} \)
Dot with \( \vec{a} \):
\( \vec{a} \cdot \vec{d} = \frac{(x+2) + 6 + (-2)}{\sqrt{(x+2)^2 + 40}} = \frac{x + 6}{\sqrt{(x+2)^2 + 40}} = 1 \)
\( x + 6 = \sqrt{(x+2)^2 + 40} \)
Square both sides:
\( (x + 6)^2 = (x + 2)^2 + 40 \\ x^2 + 12x + 36 = x^2 + 4x + 4 + 40 \\ 12x + 36 = 4x + 44 \\ 8x = 8 \implies x = 1 \)
Step 4: Get final value
Plug \( x = 1 \) in Step 2:
\( (\vec{a} \times \vec{b}) \cdot \vec{c} = -9(1) + 20 = 11 \)
Final Answer: 11