Question 1011206: Find a vector perpendicular to the given vector <5,-7,-8>
Show all your work step by step.
Found 3 solutions by Alan3354, rothauserc, Edwin McCravy:
Answer by Alan3354(69443)   (Show Source):
You can put this solution on YOUR website! Do it like this one, done by tutor Mathmate.
=========================================================
Question:
Find a vector perpendicular to the given vector 2i+9j-6k
Please Show all your work step by step.
Solution:
A vector will be denoted where x,y,z are the respective components.
We're looking for a vector perpendicular to P=<2,9,-6>.
In fact, there is an infinite number of vectors perpendicular to P.
We know that the cross product of two non-parallel vectors P, Q is perpendicular to both P and Q.
So by finding the cross product P and an arbitrary vector Q=, we can obtain the vector R=PxQ such that R is perpendicular to both P and Q, with the restriction that Q is not parallel to P, or Q does not equal kQ where k is a real number.
The cross product can be obtained by evaluation of the determinant
|i j k |
|2 9 -6|
|a b c |
which gives R=<9c+6b, -2c-6a, 2b-9a>
Thus
R=<9c+6b, -2c-6a, 2b-9a> is a vector perpendicular to P=<2,9,-6> for any vector Q= such that Q≠kP and k is a real number.
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! Find a vector perpendicular to the given vector <5,-7,-8>
there are an infinite number of vectors that are perpendicular to <5,-7,-8>, namely all c1, c2, c3 that satisfy the dot product = 0
5c1 -7c2 -8c3 = 0
for example,
suppose c2=1 and c3=1 then c1 = 3, therefore
<3,1,1> is perpendicular to <5,-7,-8>
Answer by Edwin McCravy(20064)  (Show Source):
You can put this solution on YOUR website!
Two non-zero vectors are perpendicular when
their dot product is zero.
Let < x,y,z > be a vector perpendicular to <5,-7,-8>.
Then,
<5,-7,-8> ∙ < x,y,z > = 5x-7y-8z = 0
We can pick any values of x,y, and z that
will make the dot product above be 0.
For instance, since -8-7 =-15 and we can
make the 5 become a +15 to cancel that by
multiplying it by 3, so one solution would
be to choose y and z to be 1 each and x to
be 3.
So
<5,-7,-8> ∙ <3,1,1> = (5)(3)+(-7)(1)+(-8)(1)
= 15-7-8 = 0
So <3,1,1> is perpendicular to <5,-7,-8>
There are many other possibilities.
Edwin
|
|
|
