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
• E. 7

Answer 1

Answer: (D)

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.