Find the equation of the plane passing through the intersection of the planes x-2y+z=1 and 2x+y+z=8 and parallel to th - Maths - Three Dimensional Geometry

Question

Find the equation of the plane passing through the intersection
of the planes x-2y+z=1 and 2x+y+z=8 and parallel to th eline with
direction ratios proportional to 1,2,1 Find also the perpendicular
distance of (1,1,1) from this plane

Mayank Bhardwaj · Accepted Answer

The equation of the plane passing through the point of intersection of the planes x – 2y + z = 1 and 2x + y + z = 8 is:

(x – 2y + z – 1) + λ (2x + y + z – 8) = 0

⇒ (1 + 2λ) x + (–2 + λ) y + (1 + λ) z – (1 + 8λ) = 0

Direction ratios of normal to this plane are 1 + 2λ, –2 + λ and (1 + λ).

Normal to the plane will perpendicular to the line whose direction ratios are 1, 2 and 1.

Therefore, 1 × (1 + 2λ) + 2 (–2 + λ) + 1 (1 + λ) = 0

⇒ 1 + 2λ – 4 + 2λ + 1 + λ = 0

⇒ 5λ – 2 = 0

$\Rightarrow λ = \frac{2}{5}$

On putting the value of 5λ in the equation of the plane, we obtain

$(1 + 2 \times \frac{2}{5}) x + - (2 + \frac{2}{5}) y + (1 + \frac{2}{5}) z - (1 + 8 \times \frac{2}{5}) = 0 \frac{1}{5} (9 x - 12 y + 7 z - 21) = 0 \Rightarrow 9 x - 12 y - 7 z - 21 = 0$

Let the perpendicular distance of the point (1, 1, 1) from the obtained plane be d.

$d = | \frac{9 \times 1 + (- 12) \times 1 + (- 7) \times 1 - 21}{\sqrt{9^{2} + (- 12)^{2} + (- 7)^{2}}} | \Rightarrow d = | \frac{9 - 12 - 7 - 21}{\sqrt{81 + 144 + 49}} | \Rightarrow d = | \frac{- 31}{\sqrt{274}} | \Rightarrow d = | \frac{- 31}{16.5} | (approximately) \Rightarrow d = \frac{31}{16.5} \Rightarrow d = \frac{310}{165} \Rightarrow d = 1.87 (a p p r o x i m a t e l y)$

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Find the equation of the plane passing through the intersection of the planes x-2y+z=1 and 2x+y+z=8 and parallel to th eline with direction ratios proportional to 1,2,1. Find also the perpendicular distance of (1,1,1) from this plane.