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

• Correct answer
  A

  $\sqrt{\frac{2}{3}}$
• B

  $\frac{2}{\sqrt{3}}$
• C

  $\frac{2}{3}$
• D

  $\frac{1}{3}$

Solution

DR's of $A{A}^{\text{'}}$ $(1, 1, 1)$, as this will be normal to the plane.

DR's of $AB$ is $(1, 0, 1)$

$\Rightarrow \cos \theta =\frac{\left|\left(1\times 1\right)+\left(1\times 0\right)+\left(1\times 1\right)\right|}{\sqrt{1^2+1^2+1^2}\sqrt{1^2+0^2+1^2}}=\frac{2}{\sqrt{6}}=\sqrt{\frac{2}{3}} ,$ also $AB=\sqrt{2}$

Projection $A_1{B}_1=\sqrt{2}.\sin \theta$

$=\sqrt{2}.\sqrt{1-\frac{2}{3}}$

$=\sqrt{\frac{2}{3}}$
 

Hence, the correct answer is option: 'A'.