Question

The point at which the line joining the points (2, -3, 1) and (3, -4, -5)
intersects the plane 2x + y + z = 7 is
[DSSE 1987; MP PET 1991]

• A. (1, 2, 7)
• B. (1, -2, 7)
• C. (-1, 2, 7)
• D. (1, -2, -7)

Answer 1

The correct option is B

(1, -2, 7)

Ratio $-\left[\frac{2\left(2\right)+(-3)\left(1\right)+\left(1\right)\left(1\right)-7}{2\left(3\right)+(-4)\left(1\right)+(-5\times 1)-7}\right]=-\left[\frac{-5}{-10}\right]=-\left(\frac{1}{2}\right)$
$\therefore x=\frac{2\left(2\right)-3\left(1\right)}{1}=1,y=\frac{-3\left(2\right)-(-4)}{1}=-2$
and $z=\frac{1\left(2\right)-(-5)}{1}=7$. Therefore , P(1, -2, 7).
Trick: As (1, -2, 7) and (-1, 2, 7) satisfy the equation 2x + y + z = 7, but the point (1, -2, 7) is collinear with (2, -3, 1) and (3, -4, -5).
Note : If a point dividing the join of two points in some particular ratio, then this point must be collinear with the given points.