# Vector equation of a line problem

• Mar 31st 2011, 03:58 PM
Mondy
Vector equation of a line problem
I need to find a line that passes through the point P (4,3,2) and is perpendicular to the line (1+2t, -1+t, 1-3t).

I only managed to use the dot product to get 2x +y-3Z =0 I have no idea how to proceed to get the exact values of x,y,z that passes through 4,3,2.

Any help will be appreciated.
• Apr 1st 2011, 10:53 AM
running-gag
Hi

The point P (4,3,2) and the line (1+2t, -1+t, 1-3t) determine a plane.
You can draw a sketch of this plane if you like.
Let \$\displaystyle H(x_H,y_H,z_H)\$ be the orthogonal projection of P on the line.

H being on the line there exists \$\displaystyle t_H\$ such that
\$\displaystyle x_H = 1+2t_H\$
\$\displaystyle y_H = -1+t_H\$
\$\displaystyle z_H = 1-3t_H\$

(PH) being perpendicular to the line the dot product of PH and (2,1,-3) is equal to 0
\$\displaystyle 2(x_H-4)+(y_H-3)-3(z_H-2)=0\$

Substituting \$\displaystyle x_H\$ with \$\displaystyle t_H\$ and so on ... you can fnd the value of \$\displaystyle t_H\$ then the coordinates of H, then the coordinates of PH, from which you can find an equation of (PH)