Question

The projections of the line segment joining points P(-1, 2, 3) and Q(1, 0, 4) on a line inclined equal acute angles with the coordinates axis is

• A. $-\frac{1}{\sqrt{3}}$
• B. $\sqrt{3}$
• C. $\frac{1}{3}$
• D. $\frac{1}{\sqrt{2}}$

Answer 1

The correct option is A $-\frac{1}{\sqrt{3}}$

Consider the points be $P(-1,2,3)$ and $Q(1,0,4)$

Now,${\cos }^2\alpha +{\cos }^2\beta +{\cos }^2\gamma =1$

$3{\cos }^2\alpha =1$ since $\alpha =\beta =\gamma$

$\Rightarrow \cos \alpha =\frac{1}{\sqrt{3}}$

Then,$\overrightarrow{v}=L(\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}},\frac{1}{\sqrt{3}})$

$\overrightarrow{u}=\overrightarrow{PQ}=\overrightarrow{OQ}-\overrightarrow{OP}$

$=(\hat{i}+4\hat{k})-(-\hat{i}+2\hat{j}+3\hat{k})=-2\hat{j}+\hat{k}$

$\overrightarrow{u}=(0,-2,1)$

Now,$\left|\overrightarrow{u}\right|\cos \theta =\left|\overrightarrow{u}\right|\frac{(\overrightarrow{u}.\overrightarrow{v})}{\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|}$

$=\frac{0\times \frac{1}{\sqrt{3}}-2\times \frac{1}{\sqrt{3}}+1\times \frac{1}{\sqrt{3}}}{\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}}$

$=\frac{-\frac{2}{\sqrt{3}}+\frac{1}{\sqrt{3}}}{\sqrt{\frac{3}{3}}}$

$=-\frac{1}{\sqrt{3}}$

Hence the projection is $-\frac{1}{\sqrt{3}}$