Vectors ll

[megan25]

Hello.. I nid help for a vector question urgently.. Its an old question from cambridge.. Nov 88.. It says 2 planes have equations r.(1 2 -3)=4 and r.(0 1 1)=1, denoted by p1 and p2 respectively. Am not being able to answer the 3rd part which says
' The line l passes through the point p with position vector (-2 0 -2) and is parallel to (4 1 2 ). Verify that the line lies in p1 and find the position vector of the point where l meets p2.
Also, find the length of 0P onto l. Pls help asap.. I've got exams tomorrow.. Thnk u!! :smile:

 

[pka]

Hello.. I nid help for a vector question urgently.. Its an old question from cambridge.. Nov 88.. It says 2 planes have equations r.(1 2 -3)=4 and r.(0 1 1)=1, denoted by p1 and p2 respectively. Am not being able to answer the 3rd part which says
' The line l passes through the point p with position vector (-2 0 -2) and is parallel to (4 1 2 ). Verify that the line lies in p1 and find the position vector of the point where l meets p2.
Also, find the length of 0P onto l.

I find this notation awkward.
Write the plane \(\displaystyle \pi_1:~x+2y-3z=4\) then it is clear that \(\displaystyle (-2,0,-2)\in\pi_1\) and the equation of the line is \(\displaystyle <4t-2,~t,~2t-2>\).

Because \(\displaystyle <4,1,2>\cdot<1,2,-3>=0\) the direction of the line is perpendicular to normal of the plane. So the line is a subset of the plane.

Substitute the ordinates of the line into \(\displaystyle \pi_2:~y+z=1\) and solve for \(\displaystyle t\) to get the point of intersection on \(\displaystyle \pi_2\).

I don't really understand what "find the length of 0P onto l" means. Is something like a vector projection?

 

[megan25]

Well yes, its written as the length of projection of OP onto the line... And thnk you by the way for your support.. That was truly helpful!! :smile: