Math problem with parametric equations and vectors in 3-space?

Question

Find parametric equations of the line that contains the point P and intersects the line L at a right angle.
P (0,2,1)
L: x = 2t , y = 1 - t , z = 2 + t

I just can't figure out how to go about solving this problem. Any help is appreciated

Answer 1

This is my best effort on this, and I hope this is right, or at least points you in a direction for a solution:

Consider a vector joining point P with a point on line L The coordinates of P are (0,2,1). The coordinates of the point on line L are (2t, 1-t, 2+t) The x, y, z components of the vector R joining the points are the differences in the point coordinates

R = [0-2t, 2-(1-t), 1-(2+t)]
R = (-2t, 1+t, -1-t)

This vector R must be perpendicular to L; the vector components of L are (dx/dt, dy/dt, dz/dt) = (2, -1, 1)

The condition for perpendicularity of two vectors is that the dot product is zero. Taking the dot product of the vector L and the vector R gives

R*L = -2t*2+(1+t)*(-1)+(-1-t)*1
R*L = -4t + (-1-t) + (-1-t) = -4*t - 2 - 2t = -6t-2 = 0
t=-1/3 This defines the point on L that the perpendicular intersects.

The vector R is then {from R = (-2t, 1+t, -1-t)}
R = (2/3, 2/3, -2/3)

The equation for the line starting at P = (0,2,1) and parallel to R is then x = 0+(2/3)t, y = 2+(2/3)t, z = 1-(2/3)t

Answer 2

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