Find an equation of the plane that contains the point (-1,0,0) , and is perpendicular to the plane 4x-8y+3z=-14.
What about this question? How should I solve it? Thank you very much.
One way of going about this (not necessarily the best but one that
comes to mind) is the following.
All the points p(lambda)=(0,7/4,0)+lambda(4,-8,3) are on a line perpendicular
to the given plane. This is because the vector (4, -8, 3) is the gradient
of 4x-8y+3z+14, and hence is normal to the given plane.
Then a plane perpendicular to the given plane through the given point is:
Ax+By+Cz=K,
where:
1.) -A = K (force the given point (-1,0,0) onto the plane))
2.) (7/4) B = K (force the point p(0) onto the plane)
3.) 4A + (7/4-8)B +3C=K (force the point p(1) onto the plane)
Which defines all three coefficients in terms of K. So pick any
non-zero value for K, and the equation of the plane through
the three points follows.
RonL