Find two linearly independent vectors perpendicular to?

Find two linearly independent vectors perpendicular to the vector

This is not difficult. 1st make a vector that is LI of your given vector and also nonzero. Then just take the cross product of those two vectors and not only do you have 3 LI vectors, the last is perpendicular to the other two!!!

Linearly Independent Vectors

The dot product of perpendicular vectors is zero. So we want two vectors whose dot product with the given vector is zero. Let the vector to be found be:

Then

<-1,9,-3>• = 0
-a + 9b – 3c = 0

First vector.

Let c = 0. Then
-a + 9b = 0
a = 9b
Choose b = 1. Then a = 9.
= <9,1,0>

Second vector.

Let a = 0. Then
9b -3c = 0
3c = 9b
c = 3b
Choose b = 1. Then c = 3.
= <0,1,3>

The vectors <9,1,0> and <0,1,3> are both perpendicular to the given vector v and are linearly independent of one another since one is not a multiple of the other.

The vector (“cross”) product of this with any other vector is perpendicular to v. I expect if you do the cross product with two linearly independent vectors, the results are linearly independent? Check that.
Anyway, I choose
[1] ….. and ….. [0]
[0] ………………. [1]
[0] ………………. [0]

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and if you do v x each of those I think you get
[0] ……. [3]
[-3] ……. [0]
[-9] ……. [-1]

They’re certainly linearly independent. Just check that they’re each perpendicular to v.

i’m no longer a mathematician yet i am going to do my magnificent. you’re talking about a three-dimensional area. imagine unit vectors i, j, ok at proper angles to at least one yet another. A prevalent vector can then be written as ai + ҍʝ + ck. Collinear means interior a similar promptly line; both different vectors have a similar ratios a:b:c. Coplanar means interior a similar airplane; this isn’t my section and that i don’t be conscious of the thanks to convey the placement for this in words of the coefficients. a sequence of vectors are linearly autonomous even if it isn’t accessible to write down one in all them in words of each of the others. a million. in view that we in person-friendly words have 3 dimensions, we may have at maximum 3 at the same time autonomous vectors. 2. certain 3. once you’ve defined your airplane, you are able to in person-friendly words have 2 linearly autonomous vectors in it. So the answer to the first question is certain. i imagine the most suitable be conscious of this question should be “autonomous”. i wish someone more advantageous specialist than me will fill in the gaps, yet this may get you all started meantime.

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Answer 6

if vectors are perpendicular their dot product is 0.

if your vector v = <-1,9,-3> then
w = <9,1,0> and x = <0,1,3> will both be perpendicular to it, and they are clearly independent. <3,0,1> would also work.

see how I do it? 1 times one component, whatever-it-takes times another to get the opposite, and 0 times the third.

Answer 7

what you are looking for is 2 {r,t} vectors that are not scalar multiples(def. independence)
such that
r dot v=0
t dot v=0
I’m assuming your instructor does not want the zero vector as an answer
but if the question doesn’t state non-zero
I say it’s fair game and it is orthogonal to all
vectors that leaves 1 vector for you to find
<-1,9,-3>*=0

i feel like u want people to do ur homework

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