Tuesday

May 31, 2016
Posted by **j** on Saturday, May 26, 2012 at 1:27am.

- Calculus Grade 12 University -
**Steve**, Saturday, May 26, 2012 at 5:07amjust 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 - Calculus Grade 12 University -
**j**, Saturday, May 26, 2012 at 8:11amthanks 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? - Calculus Grade 12 University -
**j**, Saturday, May 26, 2012 at 8:34amfor 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