Question

If the scalar product of the vector $\hat{i}+\hat{j}+2\hat{k}$ with the unit vector along $m\hat{i}+2\hat{j}+3\hat{k}$ is equal to $2$, then one of the values of $m$ is

• A. 3
• B. 4
• C. 5
• D. 6

Answer 1

Answer: (D)

6

Answer 2

The unit vector along $m\hat{i}+2\hat{j}+3\hat{k}$ is $\frac{m\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{{{m}^{2}}+4+9}}=\frac{m\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{{{m}^{2}}+13}}$ Now, $(\hat{i}+\hat{j}+2\hat{k}).\left( \frac{m\hat{i}+2\hat{j}+3\hat{k}}{\sqrt{{{m}^{2}}+13}} \right)=2$
$\Rightarrow$ $\frac{m}{\sqrt{{{m}^{2}}+13}}+\frac{2}{\sqrt{{{m}^{2}}+13}}+\frac{6}{\sqrt{{{m}^{2}}+13}}=2$
$\Rightarrow$ $\frac{m+8}{\sqrt{{{m}^{2}}+13}}=2$
Squaring on both sides, we get
$\Rightarrow$ ${{(m+8)}^{2}}=4({{m}^{2}}+13)$
$\Rightarrow$ ${{m}^{2}}+16m+64=4{{m}^{2}}+52$
$\Rightarrow$ $3{{m}^{2}}-16m-12=0$
$\Rightarrow$ $m=\frac{16\pm \sqrt{256+144}}{6}=\frac{16\pm 20}{6}$
$\Rightarrow$ $m=6,-\frac{2}{3}$
$\Rightarrow$ $m=6,$ as $-\frac{2}{3}$ is not possible.