I'm doing a bunch of practice finals and I don't know how to approach this problem.
Find a vector a such that a is orthogonal to < 1, 5, 2 > and has length equal to 6.
If I want to find a vector that is orthogonal to <1,5,2>, I must take the cross product?
that is one way. Pick any other vector, say <1,2,3> then the cross product is orthogonal to both vectors.
Then you wind up with some vector u. Take u/|u|*6 to get magnitude 6.
Wow I never thought of approaching it that way thanks it makes sense now!
You can take the cross-product if you have two vectors, but you only have one
so how about this ...
let the vector be (a,b,c)
then we know that (a,b,c)dot(1,5,2) = 0
a + 5b + 2c = 0
pick any value for any variable,
e.g.
let b = -1 , c = 2
a - 5 + 4 = 0
a = 1
so (1 , -1 , 2) is orthogonal to our given vector,
now we have to adjust its length to be 6
length of our new vector
= √(1 + 1 +4) = √6
so a unit vector in the direction of (1,-1,2) is (1/√6, -1/√6, 2/√6)
= (√6/6, -√6/6, 2√6/6)
so 6 times that unit vector
= (√6, -√6, 2√6)
This answer is not unique, and there is an infinite number of solutions,
We could let two of the a,b,c be anything we wanted ,except all or two of them equal to zero
e.g. let b=0
then a + 2c = 0
a = -2c
let c = 1
then a = -2, b=0 ,c=1
our vector would be (-2,0,1)
and (-2,0,1)dot(1,5,2) = -2+0+2 = 0 , so it is orthogonal
length of (-2,0,1) = √5
unit vector = (-2√5 , 0 , 1/√5 ) = (-2√5/5 , 0 , √5/5)
make it 6 units long ...
= (-12√5/5 , 0 , 6√5/5) , another answer.
Thanks Reiny!!
I understand everything except for the part making the length equal to 6.
So I find a vector <1, -1, 2> orthogonal to <1, 5, 2>
Then find the unit vector u/lul and get (1/√6, -1/√6, 2/√6)
but how did you get (√6/6, -√6/6, 2√6/6) as the final answer?
Yes, to find a vector that is orthogonal (or perpendicular) to a given vector, you can use the cross product.
To find the cross product of two vectors, let's denote the given vector as u = <1, 5, 2>. The cross product of a vector u = <u1, u2, u3> and a vector v = <v1, v2, v3> is given by the formula:
u x v = <(u2*v3) - (u3*v2), (u3*v1) - (u1*v3), (u1*v2) - (u2*v1)>
In this case, we want to find a vector a that is orthogonal to u = <1, 5, 2>. So, we can take any vector v and calculate the cross product u x v. The resulting vector will be orthogonal to u.
Next, we want the resulting vector to have a length of 6, which means its magnitude or length should be equal to 6. To calculate the magnitude of a vector, we use the formula:
|a| = sqrt(a1^2 + a2^2 + a3^2)
where a = <a1, a2, a3> is the vector.
To summarize the steps to find vector a:
1. Take the given vector u = <1, 5, 2>.
2. Choose any vector v.
3. Calculate the cross product of u x v to obtain a new vector orthogonal to u.
4. Determine the magnitude of the resulting vector.
5. Adjust the vector to have a magnitude of 6 by multiplying each component by a scaling factor.
By following these steps, you can find a vector a that is orthogonal to <1, 5, 2> and has a length of 6.