Question

Find the projection of the line segment joining $(2,-1,3)$ and $(4,2,5)$ on a line which makes equal acute angle with co-ordinate axes.

Answer 1

Consider the points be

$P(2,-1,3)$ and $Q(4,2,5)$

Now,

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

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

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

Then,

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

$\overrightarrow{u}=PQ(2,3,2)$

Now,

$\begin{array}{c}\left|\overrightarrow{u}\right|\cos \theta =\left|\overrightarrow{u}\right|\frac{\left(\overrightarrow{u}\cdot \overrightarrow{v}\right)}{\left|\overrightarrow{u}\right|\left|\overrightarrow{v}\right|}\\ \\ =\frac{2\times \frac{1}{\sqrt{3}}+\frac{2}{\sqrt{3}}+\frac{2}{\sqrt{3}}}{\sqrt{\frac{1}{3}+\frac{1}{3}+\frac{1}{3}}}\\ =\frac{7}{\sqrt{3}}\end{array}$

Hence, the projection is $\frac{7}{\sqrt{3}}$