Question
Find the equation of the plane containing the line
and the point (0, 6, 0).
Solution
The equations of line are
...(1)
It passes through (-2, -3, 4) and has direction-ratios 2, 3, –2.
Since the plane contains line (1)
∴ equation of plane is
a(x + 2) + b(y + 3) + c(z – 4) = 0 ...(2)
where 2a + 3b – 2c = 0 ...(3)
Again plane (2) passes through (0, 6, 0)
∴ a (0 + 2) + b (6 + 3) + c (0 – 4) = 0
or 2 a + 9 b – 4c = 0 ...(4)
Solving (3) and (4), we get,
∴ a = 3 k, b = 2 k, c = 6 k
Putting these values of a, b. c in (2), we get,
3 k (x + 2) + 2 k (y + 3) + 6 k (z – 4) = 0
or 3 (x + 2) + 2 (y + 3) + 6 (z – 4) = 0
or 3x + 6 + 2y + 6 + 6 z –24 = 0
or 3 x + 2 y + 6 z – 12 = 0
...(1)It passes through (-2, -3, 4) and has direction-ratios 2, 3, –2.
Since the plane contains line (1)
∴ equation of plane is
a(x + 2) + b(y + 3) + c(z – 4) = 0 ...(2)
where 2a + 3b – 2c = 0 ...(3)
Again plane (2) passes through (0, 6, 0)
∴ a (0 + 2) + b (6 + 3) + c (0 – 4) = 0
or 2 a + 9 b – 4c = 0 ...(4)
Solving (3) and (4), we get,
∴ a = 3 k, b = 2 k, c = 6 k
Putting these values of a, b. c in (2), we get,
3 k (x + 2) + 2 k (y + 3) + 6 k (z – 4) = 0
or 3 (x + 2) + 2 (y + 3) + 6 (z – 4) = 0
or 3x + 6 + 2y + 6 + 6 z –24 = 0
or 3 x + 2 y + 6 z – 12 = 0
