First slide

Planes in 3D

Question

The equation to the plane through the line of intersection of $2 x + y + 3 z - 2 = 0, x - y + z + 4 = 0$ such that each plane is at a distance of 2 unit from the origin is

Moderate

A

$x + y + 2 z + 13 = 0, x + y + z - 3 = 0$
B

$2 x + y - 2 z + 3 = 0, x - 2 y - 2 z - 3 = 0$
C

$15 x - 12 y + 16 z + 50 = 0$ , $x + 2 y + 2 z - 6 = 0$
D

x-y+2z-13=0, x+y-z-3=0

Solution

Equation of the plane is $(2 x + y + 3 z - 2) + k (x - y + z + 4) = 0$
$p = \frac{| - 2 + 4 k |}{\sqrt{{(2 + k)}^{2} + {(1 - k)}^{2} + {(3 + k)}^{2}}} = 2$
$\Rightarrow k = 13, - 1$
$∴$ Planes are $(2 x + y + 3 z - 2) + 13$
$(x - y + z + 4) = 0, (2 x + y + 3 z - 2) - 1$
$(x - y + z + 4) = 0$
$\Rightarrow 15 x - 12 y + 16 z + 50 = 0, x + 2 y + 2 z - 6 = 0$

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