Discussion Forum

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

Answer:

Explanation:

Key Idea  length ofv projection of the line segment joining a₁ and a₂ on the plane r .n

= d is $\mid\frac{(a_{2}-a_{1})\times n}{\mid n\mid}\mid$

 Length of projection the line segment joining the points (5, -1, 4 ) and (4,-1,3)

   on the plane x+y+z = 7 is

$AC= \mid\frac{(a_{2}-a_{1})\times n}{\mid n\mid}\mid$

 $AC=  \frac{\mid (-\hat{i}-\hat{k})\times (\hat{i}+\hat{j}+\hat{k)} \mid }{ \mid i+j+k\mid}$

$AC= \frac{\mid \widetilde{i}-\widetilde{k}\mid}{\sqrt{3}}$

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

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

Alternative method

  Clearly , DR's of AB are 4-5, -1+1, 3-4, 

i.e. -1,0,-1

 and DR's of normal to plane are 1,1,1. Now, let θ be the angle between the line and plane , then θ is given by

  $\sin\theta=\frac{\mid -1+0-1\mid}{\sqrt{(-1)^{2}+(-1)^{2}}\sqrt{1^{2}+1^{2}+1^{2}}}$

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

$\Rightarrow\sin\theta=\sqrt{\frac{2}{3}}\Rightarrow \cos\theta=\sqrt{1-\sin^{2}\theta}$

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

Clearly, length of projection

= $AB \cos\theta=\sqrt{2}\frac{1}{\sqrt{3}}$     [ $\therefore AB=\sqrt{2}$]

= $=\sqrt{\frac{2}{3}}$