Question

The vector a coplanar with the vectors i and j, perpendicular to the vector $\mathbf{b} = 4\mathbf{i} - 3\mathbf{j} + 5\mathbf{k}$ such that $|\mathbf{a}| = |\mathbf{b}|$ is

• A. $\sqrt{2}(3\mathbf{i} + 4\mathbf{j})$ or $- \sqrt{2}(3\mathbf{i} + 4\mathbf{j})$
• B. $\sqrt{2}(4\mathbf{i} + 3\mathbf{j})$ or $- \sqrt{2}(4\mathbf{i} + 3\mathbf{j})$
• C. $\sqrt{3}(4\mathbf{i} + 5\mathbf{j})$ or $- \sqrt{3}(4\mathbf{i} + 5\mathbf{j})$
• D. $\sqrt{3}(5\mathbf{i} + 4\mathbf{j})$ or $- \sqrt{3}(5\mathbf{i} + 4\mathbf{j})$

Answer 1

Answer: (A)

$\sqrt{2}(3\mathbf{i} + 4\mathbf{j})$ or $- \sqrt{2}(3\mathbf{i} + 4\mathbf{j})$

Answer 2

Let $\mathbf{a} = x\mathbf{i} + y\mathbf{j},$ then $\mathbf{a}.\mathbf{b} = 0$

$\Rightarrow 4x - 3y = 0 \Rightarrow \frac{x}{3} = \frac{y}{4} \Rightarrow x = 3\lambda,y = 4\lambda,\lambda \in R.$

Now $|\mathbf{a}| = |\mathbf{b}| \Rightarrow x^{2} + y^{2} = 16 + 9 + 25$

$= 9\lambda^{2} + 16\lambda^{2} = 50$

$\Rightarrow \lambda = \pm \sqrt{2} \Rightarrow x = \pm 3\sqrt{2},y = \pm 4\sqrt{2}$

Hence, $\mathbf{a} = \pm \sqrt{2}(3\mathbf{i} + 4\mathbf{j})$.