# Planes and lines Q

1. Oct 26, 2013

### converting1

write the equation of the plane x - 2y - 2z = 27 in the form $$r.\hat{n} = d$$ done

write down the distance of the origin from the plane and show that the point which is the reflection of the origin is (6, -12, -12) done

A second point P has coordinates (-3,2,1). Find the direction cosines of the reflection of the line OP in the plane.

I need help with the last part. 1. How would I go about finding the line which is reflected? Also, how do I find the direction of cosines of a line - is it the just direction of cosines of the direction of the line?

2. Oct 26, 2013

### UltrafastPED

3. Oct 26, 2013

### converting1

ok here is what i have so far

I thought I would use the point P and reflect it in the plane, doing this I got (5,-14,-15) as P' (reflected P). Then I created a line through O and P and saw that it intersected the line at (9,-6,-3) (call it A) then I found the direction of P' and where OP and the plane intersected and got (-4,8,-12) and used this to get find the cosines

is this a correct method? for instance, I done: P' - A to find the direction of the line, if I had done A-P' I would get a different value for the cosines no?

4. Oct 26, 2013

### converting1

Let A be the point where $\vec{OP}$ meets the plane and P' be the point where P is reflected on the plane:

finding A:

$\vec{OP} = r = \lambda \begin{pmatrix} -3 \\ 2 \\ 1 \end{pmatrix}$ subbing this into the equation of the plane: $\lambda \begin{pmatrix} -3 \\ 2 \\ 1 \end{pmatrix}. \begin{pmatrix} 1 \\ -2 \\ -2 \end{pmatrix} = 27$ finding $\lambda = -3$and subbing this into the equation of OP I get A = $\begin{pmatrix} 9 \\ -6 \\ -3 \end{pmatrix}$

Finding P':

create a line which is normal to the plane and goes through p: $r = \begin{pmatrix} -3+\lambda \\2-2\lambda \\ 1-2\lambda \end{pmatrix}$ use this to see where it intersects with the plane:

$\begin{pmatrix} -3+\lambda \\2-2\lambda \\ 1-2\lambda \end{pmatrix}. \begin{pmatrix} 1 \\ -2 \\ -2 \end{pmatrix} = 27$ find $\lambda = 4$ double this and sub $\lambda = 8$ into the equation of the line with direction of the normal of the plane and find that $P' = \begin{pmatrix} 5 \\ -14 \\ -15 \end{pmatrix}$ now do P' - A and I get $P' - A = \begin{pmatrix} -4 \\ -8 \\ 18 \end{pmatrix}$ so the $cos(a,b,c) = \dfrac{-4}{2\sqrt{101}},\dfrac{-8}{2\sqrt{101}},\dfrac{18}{2\sqrt{101}}$