# Vector Product

• Feb 15th 2010, 02:27 PM
axa121
Vector Product
Hi there I was attempting a previous exam question on vector products but I was not sure it I had done it correctly. Could someone please look over my solution and help me out to part B.

Show that the line L: $\displaystyle \frac{x-2}{1} = \frac{y+3}{-2}=\frac{z-1}{1}$

lies on the Plane P, $\displaystyle 3x + y - z = 2$ and find:

(a) the equation of the plane that contains L and is perpendicular to P.
(b) equations for the line P that is perpendicular to L and intersects it in (2,-3,1)

Solution:

To show that the line lies on the plane:

I got the parametric form of L, giving x = t + 2, y = -2t -3 and z = t + 1

I then put this in the equation of the plane which gave me 2, so I said since LHS = RHS the line lies on the plane.

for part (a):

I did the vector product of the normal vector x direction vector of L:

which gave me (-1,-4,-3) and then I got the resulting plane equation as
x + 4y +3z = 17. Is this correct?

Thanks.
• Feb 15th 2010, 02:36 PM
icemanfan
1. I calculated the cross product of $\displaystyle \vec{i} - 2\vec{j} + \vec{k}$ (the vector along the given line) and $\displaystyle 3\vec{i} + \vec{j} - \vec{k}$ (the normal vector to the given plane) and got $\displaystyle \vec{i} +4\vec{j} + 7\vec{k}$. Then, knowing that the point $\displaystyle (2, -3, 1)$ is on the line, I determined the equation of the perpendicular plane to be $\displaystyle x + 4y + 7z = -3$.
• Feb 15th 2010, 02:45 PM
axa121
Oh ok I had done the normal x the direction vector and I think I made an arithmetic mistake when I the point back into the plane. Anyway I've worked it out again using what you said and I've now got the same plane equation.

Any ideas on how to do part (b)

Thanks.
• Feb 15th 2010, 02:54 PM
Plato
Quote:

Originally Posted by axa121
Any ideas on how to do part (b)

Write the equation of a line containing the point $\displaystyle (2,-3,1)$
having direction vector $\displaystyle 3i+j-k$.
• Feb 15th 2010, 02:56 PM
icemanfan
Never mind, Plato is correct.
• Feb 15th 2010, 03:06 PM
axa121
Thanks Pato and Iceman:

For B i got:

(2,-3,1) + t(3,1,-1)
=(3t +2, t -3, -t)

giving the equations:

x = 3t + 2
y = t -3
z = -t
• Feb 15th 2010, 03:09 PM
icemanfan
A slight correction: $\displaystyle z = -t + 1$

Alternatively, if you want to write this line in the same form as the line you were given in the problem, it would be:

$\displaystyle \frac{x-2}{3} = y+3 = -(z-1)$
• Feb 15th 2010, 03:14 PM
axa121
K, will put it back in the same form. I'm starting to make silly mistakes now.

Time to go to bed, Thanks once again.