The length of the projection of the line segment joining the points $(5,-1,4)$ and $(4,-1,3)$ on the plane $x+y+z=7$ is

Question

Aicharya · Accepted Answer

$\begin{aligned} & 	ext { Let } \mathrm{A}=(5,-1,4), \mathrm{B}=(4,-1,3) \ & \overline{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{k}} \Rightarrow|\overrightarrow{\mathrm{AB}}|=\sqrt{2}\end{aligned}$ Projection of $\overline{\mathrm{AB}}$ in the plane $x+y+z=7$ is $|\overline{\mathrm{AB}}| \cos 	heta=\left|\overline{\mathrm{A}^{\prime} \mathrm{B}^{\prime}}ight| \cos 	heta$ Direction ratios of normal to the given plane is $1,1,1$. $\begin{aligned} & \cos \left(90^{\circ}-	hetaight)=\left|\frac{1(-1)+1(0)+1(-1)}{\sqrt{1^2+1^2+1^2} \sqrt{1^2+0^2+1^2}}ight| \ & \begin{aligned} \Rightarrow \sin 	heta=\frac{2}{\sqrt{6}} \Rightarrow \cos 	heta=\sqrt{1-\frac{4}{6}}=\sqrt{\frac{1}{3}} \end{aligned} \ & 	ext { Required projection }=|\overline{\mathrm{AB}}| \cos 	heta \ & \qquad=\sqrt{2} 	imes \frac{1}{\sqrt{3}}=\sqrt{\frac{2}{3}} 	ext { units } \end{aligned}$

The length of the projection of the line segment joining the points $(5,-1,4)$ and $(4,-1,3)$ on the plane…

Solution