The vector equation of the line L in the plane passing through the points A and B with
position vectors a = (1,2) and b = (4,3) respectively is
r = a + t(b-a) = (1,2) + t(3,1)
(i.e. the set of vectors which are a plus some real multiple of the vector from a to b.)
(a) Use the dot product to find the point P on the line which is closest to the origin
(b) Give the vector equation of the line through the origin which is orthogonal to the
Anycnt know how to solve this question or can offer any advice?
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- Thread Starter
- 14-11-2008 19:28
- 14-11-2008 20:48
You want the dot-product of the gradient of r with the gradient of a vector running through (0,0) to be zero. This is because the closest point to (0,0) on r will be where the perpendicular vector of r, running through (0,0), intersects r. So < (3,1) , (x,y) > say. Solve for x,y, and this gives you the gradient of your perpendicular vector. So the perpendicular vector will be of the form n = s(x,y) for some parameter s, and your found values of x and y. Now solve r=n for t, insert that value of t back into the eqn for r to find the point, and you're done.