Question

The length $\left(\text{in units}\right)$ 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. $\frac{2}{3}$
• B. $\frac{1}{3}$
• C. $\sqrt{\frac{2}{3}}$
• D. $\frac{2}{\sqrt{3}}$

Answer 1

The correct option is C $\sqrt{\frac{2}{3}}$

Let the given point be denoted by $B(5,-1,4)$ and $A(4,-1,3)$
Hence $\overrightarrow{AB}=(1,0,1)$
Let $\overrightarrow{AC}$ be the normal vector. $\overrightarrow{AC}=(1,1,1)$
Angle between $AB$ and $AC=\cos \theta =\frac{1+0+1}{\sqrt{2}\sqrt{3}}=\sqrt{\frac{2}{3}}$
We know $|\overrightarrow{AB}|=\sqrt{2}$
Required projection $=|\overrightarrow{AB}|\sin \theta$
Hence, required projection is $=\sqrt{2}\sqrt{1-\frac{2}{3}}=\sqrt{\frac{2}{3}}$