How to Find Vector Components and Parametric Equations for Plane Intersections?

[dagg3r]

hi all got some problems with vectors so waswondering if anyone can check what I've done
is correct thanks
question1.

find the vector component of 2i+j-k in the direction of i-3j+2k
basically i use the rule
a=(v*w^)*w^
w^= i-3j+2k / sqrt(14)

thus v=2i+j-k
therefore a=(2i+j-k)*( i-3j+2k / sqrt(14) ) * (i-3j+2k / sqrt(14))
a= [2(1)-3(1)-1(2) /sqrt(14)] * [i-3j+2k / sqrt(14)]
a= -3/sqrt(14)* i-3j+2k / sqrt(14)

thus the vector componentis a=-3/14( i-3j+2k) ?

2. find the parametric equationsof the straightline of intersection of the planes
x-2y+3z=5
3x+y-2z=1

i used gaussian elimination and got the tableu
1 -2 3 | 5
3 1 -2 | 1 R2-3R1

1 -2 3 | 5
0 7 -11| -14

x-2y+3z=5
7y-11z=-14
z=t

sub z=t into 7y-11z=-14
y=11t/7 - 2

sub y=11t/7 - 2 into x-2y+3z=5
x= 5 + 1/7t

therefore are the parametric equations
x= 5 + 1/7t
y=11t/7 - 2
z=t

??
thanks all

 

[StatusX]

1. You found the projection vector along w, but it sounds like they just wanted the length of this vector. That is, just $\vec v \cdot \hat w$. Think of it like this: if you rearranged your axes so that w pointed along the new x axis, what would be the new x component of v?

2. The easiest way to check this is to plug x, y, and z back into the two plane equations and see if they solve them for all t (ie, t drops out and you get something like 1=1).

 

[quicknote]

Hello! It looks like you have correctly found the vector component in the first question. Your steps and calculations are correct.

For the second question, your parametric equations are correct as well. Great job using Gaussian elimination to solve for the intersection of the two planes. Keep up the good work!