Question

Find the ratio in which the line segment joining points (2 , -1, 3) and (-1, 2, 1 ) is divided by the plane x + y + z = 5.​

Answer 1

$\bigstar{\fbox{\fbox{\orange{\large{Answer:-}}}}}$

Let the plane x + y + z = 5 divides the line joining points ( 2, -1, 3) and (-1, 2, 1) ratio k : 1.

$\sf {So \:the \:required \:coordinates\: are}$

$\sf{\blue{ (\frac{ - k + 2}{k + 1} , \frac{2k - 1}{k + 1} ,\frac{k + 3}{k + 1} )}}$

Since this point lies on the plane x + y + z = 5

$\sf{\green{ (\frac{ - k + 2}{k + 1} +\frac{2k - 1}{k + 1} +\frac{k + 3}{k + 1} )= 5}}$

$\sf{\green{ => - k + 2 + 2k - 1 + k + 3 = 5(k + 1)}}$

$\sf{\green{ => 2k + 4 = 5k + 5}}$

$\sf{\green{=> 3k = -1}}$

$\sf{\green{ therefore, k = - \frac{1}{3} }}$

So the required ratio is $\sf{-\frac{1}{3}: or 1: 3 }$ externally divides the plane.

The coordinates of point of division are

$\sf{\purple{\frac{5}{2} ,\frac{ - 5}{2}, 4}}$