find a vector a that has the same direction as <-8,9,8> but has a length of 3

Question

find a vector a that has the same direction as <-8,9,8> but has a length of 3

Answer 1

To maintain the same direction, we simply need to scale the vector by multiplying/dividing by a scalar. To obtain a length of 3, we need to divide by the scalar that is one-third of the current length. The reason for this is that an object of length L divided by L/3 will always yield 3.

We need to determine one-third of the current length, where the current length is simply the norm:

||<-8, 9, 8>|| = √((-8)² + 9² + 8²)

||<-8, 9, 8>|| = √209

The prime factorization of 209 is 11*19, so we cannot simplify the radical expression. We simply divide, or scale, the vector by √209 / 3 and we end up with the vector:

<-24/√209, 27/√209, 24/√209>

Since we simply scaled the vector, the direction remains the same. We can verify the length by computing the norm:

||<-24/√209, 27/√209, 24/√209>|| = √((-24/√209)² + (27/√209)² + (24/√209)²)

||<-24/√209, 27/√209, 24/√209>|| = √(576/209 + 729/209 + 576/209)

||<-24/√209, 27/√209, 24/√209>|| = √(1881/209)

||<-24/√209, 27/√209, 24/√209>|| = √9

||<-24/√209, 27/√209, 24/√209>|| = 3

Thus, <-24/√209, 27/√209, 24/√209> is the vector along the same direction of <-8, 9, 8> and of length 3.