Question

A vector whose modulus is $\sqrt{51}$ and makes the same angle with $\mathbf{a} = \frac{\mathbf{i} - 2\mathbf{j} + 2\mathbf{k}}{3},\mathbf{b} = \frac{- 4\mathbf{i} - 3\mathbf{k}}{5}$ and $\mathbf{c} = \mathbf{j},$ will be

• A. $5\mathbf{i} + 5\mathbf{j} + \mathbf{k}$
• B. $5\mathbf{i} + \mathbf{j} - 5\mathbf{k}$
• C. $5\mathbf{i} + \mathbf{j} + 5\mathbf{k}$
• D. $\pm (5\mathbf{i} - \mathbf{j} - 5\mathbf{k})$

Answer 1

Answer: (D)

$\pm (5\mathbf{i} - \mathbf{j} - 5\mathbf{k})$

Answer 2

Let the required vector be $\alpha = d_{1}\mathbf{i} + d_{2}\mathbf{j} + d_{3}\mathbf{k}$,

where $d_{1}^{2} + d_{2}^{2} + d_{3}^{2} = 51$, (given) .....(i)

Now, each of the given vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ is a unit vector $\cos\theta = \frac{\mathbf{d}.\mathbf{a}}{|\mathbf{d}||\mathbf{a}|} = \frac{\mathbf{d}.\mathbf{b}}{|\mathbf{d}||\mathbf{b}|} = \frac{\mathbf{d}.\mathbf{c}}{|\mathbf{d}||\mathbf{c}|}$ or $\mathbf{d}.\mathbf{a} = \mathbf{d}.\mathbf{b} = \mathbf{d}.\mathbf{c}$

$|\mathbf{d}| = \sqrt{51}$ cancels out and $|\mathbf{a}| = |\mathbf{b}| = |\mathbf{c}| = 1$

Hence, $\frac{1}{3}(d_{1} - 2d_{2} + 2d_{3}) = \frac{1}{5}( - 4d_{1} + 0d_{2} - 3d_{3}) = d_{2}$

$\Rightarrow d_{1} - 5d_{2} + 2d_{3} = 0$ and $4d_{1} + 5d_{2} + 3d_{3} = 0$

On solving, $\frac{d_{1}}{5} = \frac{d_{2}}{- 1} = \frac{d_{3}}{- 5} = \lambda$ (say)

Putting $d_{1},d_{2}$ and $d_{3}$ in (i), we get $\lambda = \pm 1$

Hence the required vectors are $\pm (5\mathbf{i} - \mathbf{j} - 5\mathbf{k}).$

Trick : Check it with the options.