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

Answer: (D)

11

Answer 2

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.