Find the vector and Cartesian equations of the plane passing through t

1.	Find the vector and Cartesian equations 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 the d.r.’s of the normal to the plane passing through the point (–1, 3, 2) be A, B, C. So, eq. of plane is : A(x + 1) + B(y - 3) + C(z - 2) = 0 ...(i) As (i) is perpendicular to the planes x + 2y + 3z = 5 and 3x + 3y + z = 0 So, A + 2B + 3C = 0 ...(ii) and, 3A + 3B + 3C = 0 ...(iii) Solving (ii) and (iii), we get : A/(2 - 9) = B/(9 - 1) = C/(3 - 6) i.e., A/-7 = B/8 = C/-3 i.e., the d.r.’s are –7, 8, –3. By (i), -7(x + 1) + 8(y - 3) - 3(z - 2) = 0 i.e., 7x - 8y + 3z + 25 = 0, which is Cartesian equation. Also the vector equation of plane is, vector r.(7i - 8j + 3k) + 25 = 0

Find the vector and Cartesian equations 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 the d.r.’s of the normal to the plane passing through the point (–1, 3, 2) be A, B, C.

So, eq. of plane is : A(x + 1) + B(y - 3) + C(z - 2) = 0 ...(i)

As (i) is perpendicular to the planes x + 2y + 3z = 5 and 3x + 3y + z = 0

So, A + 2B + 3C = 0 ...(ii) and, 3A + 3B + 3C = 0 ...(iii)

Solving (ii) and (iii), we get :

A/(2 - 9) = B/(9 - 1) = C/(3 - 6) i.e.,

A/-7 = B/8 = C/-3 i.e., the d.r.’s are –7, 8, –3.

By (i), -7(x + 1) + 8(y - 3) - 3(z - 2) = 0 i.e., 7x - 8y + 3z + 25 = 0,

which is Cartesian equation. Also the vector equation of plane is, vector r.(7i - 8j + 3k) + 25 = 0

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