Can someone please help me find the vector equation for the line through point R(3,0,-3) normal to the plane-4x-12y+4z=-24 ?
I have the normal as n = -4i -12j +4k, but am at lost to the next step.
Any help will be greatly appreciated.
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Can someone please help me find the vector equation for the line through point R(3,0,-3) normal to the plane-4x-12y+4z=-24 ?
I have the normal as n = -4i -12j +4k, but am at lost to the next step.
Any help will be greatly appreciated.
We can use the normalQuote:
Originally Posted by tammmyl
so the the equation of the line is
.
Recall that a line is perpendicular to a plane if it is parallel to the normal of the plane.
The obvious point (which I just have to mention) is that when t= 0, this is (3, 0, -3), the given point, and its direction is given by the "slope", <1, 3, -1>, as given by the normal to the plane.