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,
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.
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?
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