question_answer 1)

Find the equation of the plane passing through the point (–1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0. 

Answer:

Let a, b, c are direction ratios of require plane. Therefore, required equation of plane passing through(-1, 3, 2) is given by       a(x + 1) + b (y ? 3) + c(z ? 2) = 0      ? (1)       Also direction ratios of plane x + 2y + 3z = 5 are 1, 2, 3 and direction ratios of plane 3x + 3y + z = 0 are 3, 3, 1       Now it is given that required plane (1) is perependicular to planes x + 2y + 3z = 5 and 3x + 3y + z = 0        a × 1 + b × 2 + c × 3 = 0       and a × 3 + b × 3 + c × 1 = 0       i.e. a + 2b + 3c = 0                          ?. (2)       and 3a + 3b + c = 0                         ?.. (3)       On cross-multiplying (2) and (3), we get                         Putting these values in (1), we get       ?7K (x + 1) + 8K (y ? 3)P ? 3K (z ? 2) = 0        ?7x ? 7 + 8y ? 24 ? 3z + 6 = 0       7x ? 8y + 3z + 25 = 0       which is the required equation of the plane.