Question

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

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

Answer 1

The correct option is B

$\sqrt{\frac{2}{3}}$

Find the length of the line segment joining the given points

Points given, $\left(5,-1,4\right)$ and $\left(4,-1,3\right)$

The direction ratios of line $AB$ are $\left(5-4,-1+1,4-3\right)$

Hence the direction ratios of $AB$ are $(1,0,1)$

Let the angle between $AB$ and normal to the plane be $\theta$

$\cos \theta =\frac{1\times 1+1\times 0+1\times 1}{\sqrt{1^2+0^2+1^2}\sqrt{1^2+1^2+1^2}}$

$\therefore$ $\cos \theta =\frac{\sqrt{2}}{\sqrt{3}}$

$\therefore$ $\sin \theta =\sqrt{1-{\cos }^2\theta }$

$\therefore$ $\sin \theta =\sqrt{1-\frac{2}{3}}=\frac{1}{\sqrt{3}}$

$\left|AB\right|=\sqrt{{\left(5-4\right)}^2+{(-1+1)}^2+{\left(4-3\right)}^2}$

$\therefore$$\left|AB\right|=\sqrt{2}$

Projection of $AB$ on the plane is given by $\left|AB\right|\sin \theta$

$=\sqrt{2}\times \frac{1}{\sqrt{3}}$

$=\sqrt{\frac{2}{3}}$

So, the projection of $AB$ on the plane $x+y+z=7$ is $\sqrt{\frac{2}{3}}$

Hence, option $\left(B\right)$ $\sqrt{\frac{2}{3}}$, is the correct answer.