Vector Calculus: Find Line Perp. to L Through P in 6x-4y+2z=1 Plane

yoyomath
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Homework Statement



1. Let L (line) be the line given by x = 6 - t, y = 4 + t , z= 4 + t. L intersects the plane 6x - 4y + 2z = 1 at the point P = (6, 4, - 8). Find parametric equations for the line through P which lies in the plane and is perpendicular to L




I don't even know where to start...
I was stuck to this problem for HOURS yesterday
 
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What is the direction vector of the line and normal vector of the plane?

Once you have those two vectors, you need to determine what condition the asked line must meet to be in the plane and be perpendicular to the given line.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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