Find the equation of the plane passing through intersection of the plane 4x-y+z=10 and x+y-z=4 and parallel to the line - Maths - Relations and Functions

Question

Find the equation of the plane passing through intersection of
the plane 4x-y+z=10 and x+y-z=4 and parallel to the line with
direction ratios proportional to (2,1,1) Find also perpendicular
Distance of (1,1,1) From this plane

Ajanta Trivedi · Accepted Answer

equation of the plane passing through the intersection of the plane 4x-y+z=10 and the plane x+y-z=4 is given by:

$(4 x - y + z - 10) + λ (x + y - z - 4) = 0 (4 + λ) x + (- 1 + λ) y + (1 - λ) z + (- 10 - 4 λ) = 0 ........ (1)$

since the plane (1) is parallel to the line with the direction ratios proportional to (2,1,1)

therefore

$2 (4 + λ) + 1. (- 1 + λ) + 1. (1 - λ) = 0 8 + 2 λ - 1 + λ + 1 - λ = 0 2 λ = - 8 λ = - 4$

substituting the value of λ in (1)

-5y+5z+6=0 i.e. 5y-5z-6=0...(1a)

(ii) let perpendicular distance of (1,1,1) from plane 1(a) be l.

$l = \frac{| 0 + 5 \times 1 - 5 \times 1 - 6 |}{\sqrt{5^{2} + 5^{2}}} = \frac{| - 6 |}{5 \sqrt{2}} = \frac{3}{5} \sqrt{2} u n i t$

Find the equation of the plane passing through intersection of the plane 4x-y+z=10 and x+y-z=4 and parallel to the line with direction ratios proportional to (2,1,1). Find also perpendicular Distance of (1,1,1) From this plane.