Question

Find the equation of the plane through the point (2,5,7) that is parallel to the line r = (3i + 2j - 2k) + t(i + 2j + 6 k) and perpendicular to the plane 4x + 5y + 6z= 14. Write the equation in the form indicated.
Equation: ? (x - 2) + ? (y - ? ) -3(z - ? ) = 0

Answer 1

Find the equation of the plane through the point P(2,5,7) that is parallel to the
line r = (3i + 2j - 2k) + t(i + 2j + 6k) and perpendicular to the plane 4x + 5y + 6z = 14.

If the desired plane is parallel to the given line, the directional vector v, of the line is also a directional vector of the desired plane. Also,since the given plane is perpendicular to the desired plane, the normal vector n1, of the given plane is a directional vector of the desiredplane.

The normal vector n, of the desired plane is orthogonal to both its directional vectors. Take the cross product.

n = v X n1 = <1, 2, 6> X <4, 5, 6> = <-3, 6, -3>

Any non-zero multiple of n is also a normal vector of the desired plane. Divide by-3.

n = <1, -2, 1>

With a point P(2, 5, 7) in the plane and the normal vector n, of the plane we can write the equation of the plane. Remember, the normal vector of theplane is orthogonal any vector that lies in the plane. And the dot product of orthogonalvectors is zero. Define R(x,y,z) to be an arbitrary point in the plane. Then vector PR lies in the plane.

n • PR = 0
n • <R - P> = 0
<1, -2, 1> • <x - 2, y - 5, z - 7> = 0
1(x - 2) - 2(y - 5) + 1(z - 7) = 0
x - 2 - 2y + 10 + z - 7 = 0
x - 2y + z + 1 = 0