Find the equation of the plane passing through the point (3, 4, 1) and (0, 1, 0) and parallel to the line
We know that the equation of plane passing through (x1,y1,z1) is given by
a(x – x1) + b(y – y1) + c(z – z1) = 0 …… (1)
So, equation of plane passing through (3,4,1) is
a(x – 3) + b(y – 4) + c(z – 1) = 0 ……(2)
It also passes through (0,1,0)
So, equation (2) must satisfy the point (0,1,0)
∴ a(0 – 3) + b(1 – 4) + c(0 – 1) = 0
⇒ – 3a – 3b – c = 0
⇒ 3a + 3b + c = 0 ……(3)
We know that line is parallel to plane a2x + b2y + c2z + d2 = 0 if a1a2 + b1b2 + c1c2 = 0 …… (4)
So,
a×2 + b×7 + c×5 = 0
⇒ 2a + 7b + 5c = 0 ……(5)
Solving equation (3) and (5) by cross multiplication we have,
⇒
⇒
∴ a = 8k, b = – – 13k and c = 15k
Putting the value in equation (2)
8k(x – 3) – 13k(y – 4) + 15k(z – 1) = 0
8kx – 24k – 13ky + 52k + 15kz – 15k = 0
Dividing by k we have
8x – 13y + 15z + 13 = 0
Equation of required plane is 8x – 13y + 15z + 13 = 0