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Vector components hard!

  • Thread starter dagg3r
  • Start date
  • #1
67
0
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
 

Answers and Replies

  • #2
StatusX
Homework Helper
2,564
1
1. You found the projection vector along w, but it sounds like they just wanted the length of this vector. That is, just [itex]\vec v \cdot \hat w[/itex]. 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).
 

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