# Vectors and 3D Geometry

• May 10th 2011, 12:43 AM
Blizzardy
Vectors and 3D Geometry
Hi guys, I am new to this forum, hope I posted this thread in the correct place...:)
Anyway,I got a question on vectors which I need help in:

The vector q is inclined at angles 135, 60 and \alpha where \alpha is an obtuse angle, to the x-,y- and z-axes respectively. The line L passes through the point A(2,-2,3) and is parallel to q. The plane 'pie' passes through B(5,-3,2) and contains the y-axis.

(i) Find a vector EQN of line L.
(ii) Find the cartesian EQN of plane 'pie'
(iii) Find the acute angle b/w L and 'pie'
(iv) Find the length of projection of AB onto 'pie'

I cant solve (ii) and (iv). :( By the way, the ans to (i) is r = (2,-2,3) + \lambda (-2^1/2,1,-1) if it is of any help.

From what i understand from 'contains y-axis', I think it means the plane is parallel to y-axis and this would mean the direction vectors of x- and z-axes is 0 right? But what does it mean? I am bad with planes + axes... I think (ii) and (iv) are related thats why I can't do (iv) as well but I do know about length of projection.

Sorry if my post is messy. Thank you!
• May 10th 2011, 04:25 AM
earboth
Quote:

Originally Posted by Blizzardy
Hi guys, I am new to this forum, hope I posted this thread in the correct place...:)
Anyway,I got a question on vectors which I need help in:

The vector q is inclined at angles 135, 60 and \alpha where \alpha is an obtuse angle, to the x-,y- and z-axes respectively. The line L passes through the point A(2,-2,3) and is parallel to q. The plane 'pie' passes through B(5,-3,2) and contains the y-axis.

...
(ii) Find the cartesian EQN of plane 'pie'
...

From what i understand from 'contains y-axis', I think it means the plane is parallel to y-axis and this would mean the direction vectors of x- and z-axes is 0 right? But what does it mean? I am bad with planes + axes... I think (ii) and (iv) are related thats why I can't do (iv) as well but I do know about length of projection.

Sorry if my post is messy. Thank you!

1. You know about the plane pi (I assume that you don't refer to a kind of pastry(?)) that
• the point B(5, -3, 2) belongs to pi
• the y-axis belong to pi, so the origin belongs to pi too

2. Use the origin as a fixed point of pi, then the vectors
$\vec u = \langle 0,1,0 \rangle , \vec v = \langle 5,-3,2 \rangle$
span the plane pi.

3. A parametric equation of pi could be:

$\pi:\langle x,y,z \rangle = s \cdot \vec u + t \cdot \vec v$

Plug in the appropriate value.