1. Vectors and Lines

Hi guys! Need help with this question please,

The plane pi1 has vector EQN: r . (2, 2, 1) = 13. The points P and Q, which are not in pi1, have position vectors 4i + 5j + 7k and 10i + 8j + k respectively, and the line L1, which passes through P and Q, meets pi1 at point R. Find the position vector of R.
The line L2, is contained in pi1, passes through R and is perpendicular to L1. Find a vector which is parallel to L2.

I found OR = (-2, 2, 13)
Since L2 is perpendicular to L1,
Let F be the foot of perpendicular from L2 to L1.
OF lies on L1 => OF = (4, 5, 7) + s(2, 1, -2) for some s E R
OF = (4+2s, 5+s, 7-2s)
RF = (4+2s, 5+s, 7-2s) - (-2, 2, 13) = (6+2s, 3+s, -6-2s)
RF . (2, 1, -2) = 0 (since L1 and L2 are perpendicular)
Solving, s = -3
Then sub. s = -3 into RF which I ended up with a 0?

Any idea why this method doesnt work?

2. Re: Vectors and Lines

Originally Posted by Blizzardy
Hi guys! Need help with this question please,

The plane pi1 has vector EQN: r . (2, 2, 1) = 13. The points P and Q, which are not in pi1, have position vectors 4i + 5j + 7k and 10i + 8j + k respectively, and the line L1, which passes through P and Q, meets pi1 at point R. Find the position vector of R.
The line L2, is contained in pi1, passes through R and is perpendicular to L1. Find a vector which is parallel to L2.

I found OR = (-2, 2, 13)
Since L2 is perpendicular to L1,
Let F be the foot of perpendicular from L2 to L1.
OF lies on L1 => OF = (4, 5, 7) + s(2, 1, -2) for some s E R
OF = (4+2s, 5+s, 7-2s)
RF = (4+2s, 5+s, 7-2s) - (-2, 2, 13) = (6+2s, 3+s, -6-2s)
RF . (2, 1, -2) = 0 (since L1 and L2 are perpendicular)
Solving, s = -3
Then sub. s = -3 into RF which I ended up with a 0?

Any idea why this method doesnt work?
Your calculations are correct. You proved that point R and point F are equal.

1. The equation of the plane is

$2x+2y+z=13$ with the normal vector $\vec n = \langle 2,2,1 \rangle$

2. The line $L_1: \langle x,y,z \rangle = \langle 4,5,7 \rangle +s \cdot \langle 2,1,-2 \rangle$
that means your result is correct.

3. The line $L_2$ passes through R. The direction of this line must be perpendicular to $\vec n$ so $p_1$ contains this line and this line must be perpendicular to the direction of $L_1$. Let $\vec d$ denote the direction vector in question:

$\left|\begin{array}{l} \langle 2,1,-2 \rangle \cdot \vec d = 0 \\ \langle 2,2,1 \rangle \cdot \vec d = 0 \end{array}\right.$ ..... I've got $\vec d = \langle 5,-6,2 \rangle$