need help with finding the equation of the plane satisfying the conditions.?

Question

perpendicular to 3x-2y+z-5=0
contains (2,2,3)
help please!

wait...but how do you find the normal of a plane? oh...and the equation can be either vector or scalar

ok...so i only need an example of the equation of a plane that is perpendicular since there is no specific one...so yea basically i just need an explanation on the whole idea of it

Answer 1

I think your question may be incorrect. If you want the equation of the line satisfying those conditions, that makes sense:

The direction of the line is given by the coefficients of the plane: (3, -2, 1). A point on the line is given, so the equation of the line is
(x, y, z) = (2, 2, 3) + t(3, -2, 1)
or
x = 3t + 2, y = -2t + 2, z = t + 3

If you are looking for a plane oriented in a direction perpendicular to the original plane's orientation, any plane containing this line will do - there's an infinite number of possible answers.

Answer 2

If it is perpendicular to the given plane, then the normal of that plane (3,-2,1) is a direction vector of the plane you want. (2,2,3) is a point on that plane.

All you need is another direction vector!

You havent said what form you want the equation in...vector or scalar?

Answer 3

There is not a unique plane perpendicular to the given plane and passing thru (2,2,3). There is a unique line however. Please restate your question.

Answer 4

a million) the lines are parallel by way of way of reality the slopes are equivalent. 2) the lines are perpendicular by way of way of reality the slopes are detrimental reciprocals of one yet another 3) neither by way of way of reality the slopes are neither equivalent or detrimental reciprocals of one yet another.