Question

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:

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

Answer 1

Answer: (D)

Normal to the plane $x+y+z=7$ is $\vec{n}=\hat{i}+\hat{j}+\hat{k}$
$\overrightarrow{AB}=−\hat{i}−\hat{k}⇒âˆ\pounds \overrightarrow{AB}âˆ\pounds =AB=\sqrt{2}$
$BC=\text{ Length of projection of }\overrightarrow{AB}\text{ on }\vec{n}=âˆ\pounds \overrightarrow{AB}â‹\dots \hat{n}âˆ\pounds$
$=â(−\hat{i}−\hat{k})â‹\dots \frac{(\hat{i}+\hat{j}+\hat{k})}{\sqrt{3}}â=\frac{2}{\sqrt{3}}$
Length of projection of the line segment on the plane is $AC$
$A{C}^2=A{B}^2−B{C}^2=2−\frac{4}{3}=\frac{2}{3}$
$A{C}^2=\sqrt{\frac{2}{3}}$