Question

A vector r is equally inclined with the co-ordinate axes. If the tip of r is in the positive octant and |r| = 6, then $\mathbf{r}$ is

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

Answer 1

Answer: (D)

$2\sqrt{3}(\mathbf{i} + \mathbf{j} + \mathbf{k})$

Answer 2

Let $l,m,n$ be the d.c's of $\mathbf{r}.$ Then $l = m = n$, (given)

∴ $l^{2} + m^{2} + n^{2} = 1 \Rightarrow 3l^{2} = 1 \Rightarrow l = \frac{1}{\sqrt{3}} = m = n$

Now, $\mathbf{r} = |\mathbf{r}|(l\mathbf{i} + m\mathbf{j} + n\mathbf{k}) = 6\left( \frac{1}{\sqrt{3}}\mathbf{i} + \frac{1}{\sqrt{3}}\mathbf{j} + \frac{1}{\sqrt{3}}\mathbf{k} \right)$

Hence,$\mathbf{r} = 2\sqrt{3}(\mathbf{i} + \mathbf{j} + \mathbf{k})$.