Vectors, planes and niceness?

Question

Find the equation of a plane through the point (2,-1,6), perpendicular to the line of intersection of the planes 4x+4y-z-11=0 and 2x-y+7z+2=0.

Please show how for 10 points

Answer 1

10# - 42 sq. in. divided by 12, then add the square root of an ostlers triangle minus the girth of a container that is 2' x 2' x 3' x 5 '.  Draw a line perpendicular to a squirrel ls nest and add the length of the limb the nest is on.  Subtract your age and iq and add 2.  That's the equation.  

Answer 2

The directional vector v, of the line of intersection of the two given planes is orthogonal to the normal vectors n1 and n2, of both given planes.  Take the cross product.

v = n1 X n2 = <4, 4, -1> X <2, -1, 7> = <27, -30, -12>

Any non-zero multiple of v is also a directional vector of the line of intersection.  Divide by 3.

v = <9, -10, -4>

Since the desired plane is perpendicular to the line of intersection, the directional vector v, of the line of intersection is also the normal vector of the desired plane.

With a point P(2, -1, 6) in the plane and the normal vector v, of the plane we can write the equation of the plane.  Remember, the normal vector of the plane is orthgonal to any vector in the plane.  And the dot product of orthogonal vectors is zero.  Define R(x,y,z) to be an arbitrary point in the plane.  Then vector PR lies in the plane.

v • PR = 0

v • <R - P> = 0

<9, -10, -4> • <x - 2, y + 1, z - 6> = 0

9(x - 2) - 10(y + 1) - 4(z - 6) = 0

9x - 18 - 10y - 10 - 4z + 24 = 0

9x - 10y - 4z - 4 = 0