# Find the equation of the plane through the points (2, 1, –1)

Question:

Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4) and perpendicular to the plane – 2+ 4= 10.

Solution:

We know that, equation of the plane passing through two points (x1, y1, z1) and (x2, y2, z2) with its normal’s direction ratios is

a(x – x1) + b(y – y1) + c(z – z1) = 0 …. (i)

Now, if the plane is passing through two points (2, 1, -1) and (-1, 3, 4) then

a(x2 -x1) + b(y2 – y1) + c(z2 – z1) = 0

a(-1 – 2) + b(3 – 1) + c(4 + 1) = 0

-3a + 2b + 5c = 0 …. (ii)

As the required plane is perpendicular to the given plane x – 2y + 4z = 10, then

1.a – 2.b + 4.c = 10 …. (iii)

On solving (ii) and (iii) we get,

$\frac{a}{8+10}=\frac{-b}{-12-5}=\frac{c}{6-2}=\lambda$

So, a = 18λ, b = 17λ and c = 4λ

Thus, the required plane is

18λ (x – 2) + 17λ (y – 1) + 4λ (z + 1) = 0

18x – 36 + 17y – 17 + 4z + 4 = 0

⇒ 18x + 17y + 4z – 49 = 0