| 
June 21st, 2009, 04:01 PM
| | Junior Member | | Join Date: Feb 2009
Posts: 36
Country:  Thanks: 11
Thanked 0 Times in 0 Posts
| |  Cartesian eqn of a plane
A plane passes through the P, with position vector i + 2j - k, and is perpendicular to the line L with eqn r = 3i -2k + w(-i + 2j + 3k)
Show that the Cartesian eqn of the plane is x - 5y -3z = -6
I'm not sure how to work out the normal of the plane? |
June 21st, 2009, 04:54 PM
| | MHF Contributor | | Join Date: Aug 2006
Posts: 6,717
Thanks: 69
Thanked 2,485 Times in 2,279 Posts
| |
Quote:
Originally Posted by Erghhh  A plane passes through the P, with position vector i + 2j - k, and is perpendicular to the line L with eqn r = 3i -2k + w(-i + 2j + 3k)
Show that the Cartesian eqn of the plane is x - 5y -3z = -6 | Please check the details of your post.
That given answer is answer is not perpendicular to the line L.
The answer I get is  . |
June 21st, 2009, 04:57 PM
| | Junior Member | | Join Date: Feb 2009
Posts: 36
Country: Thanks: 11
Thanked 0 Times in 0 Posts
| |
Yeah, I was careful. It was from a past paper and I thought there was a typo. And on the mark scheme it says that the normal was 1i + 5j + 3k. I was skeptical about there being two typos on two separate documents. But I do agree with your answer. | | Thread Tools | | | | Display Modes |  Linear Mode | 
Posting Rules
| You may not post new threads You may not post replies You may not post attachments You may not edit your posts
HTML code is Off | | | All times are GMT -7. The time now is 04:47 PM. | | |
 | |  |
