Posted by j on Saturday, May 26, 2012 at 1:27am.
just take the cross-product of the normals to find the direction of the vector:
| i j k |
| 2 -5 3 |
| 3 4 -3 |
= 3i+15j+23k
Now find a point on the line:
if x=0, y=-18,z=-26
the line is thus
3ti + (-18+15t)j + (-26+23t)k
thanks steve.
do you know how to do these ones? :(
4. Write a vector equation of the line through the point (5, -2, 3) and parallel to the vector v=[4, -3, 1]
5. Determine an equation for the plane that is exactly between the points A(-1, 2, 4) and B(3, 1, -4).
6. Find out if the line r = (1, 3, 8) + t (-2, 5, 7) is parallel to the plane 3x + 4y - 2z = 1
7. Where does the line r = (1, 2, -5) + t (2, -3, 1) meet the plane 2x + 5y - 3z = 6?
for the #4 i got
r = (5, -2, 3) + t(4, -3, 1)
--- can you please help me on 5, 6, 7. instead since i got 4. thanks steve