# Vector with the same direction but a different magnitude

Calpalned

## Homework Statement

Find a vector that has the same direction as <-2, 4, 2> but has length 6.

## Homework Equations

Length of vector in 3D = |v| = (x2 + y2 + z2)0.5

## The Attempt at a Solution

I can see that the length of the given vector is (-2)2 + 42 + 22 = 24
36 is (1.5)24, so I tried multiplying each component of the vector by the scalar 1.5. However, this doesn't give me the desired length of six (rather, it gives me the square root of 54). I had a feeling this method would be faulty, but nothing else comes to mind.

Mentor
Compare lengths not the squares of lengths to get the right factor.

Staff Emeritus
Gold Member
I can see that the length of the given vector is (-2)2 + 42 + 22 = 24
Not according to the formula you included under "relevant equations".

Calpalned

## Homework Statement

Find a vector that has the same direction as <-2, 4, 2> but has length 6.

## Homework Equations

Length of vector in 3D = |v| = (x2 + y2 + z2)0.5

## The Attempt at a Solution

I can see that the length of the given vector is (-2)2 + 42 + 22 = 24
36 is (1.5)24, so I tried multiplying each component of the vector by the scalar 1.5. However, this doesn't give me the desired length of six (rather, it gives me the square root of 54). I had a feeling this method would be faulty, but nothing else comes to mind.

I tried it again with the length of the desired vector = (F(-2)2 + F42 + F22)0.5 Where F = 1.5, and the answer seems to be correct...

Calpalned
I just noticed my error. In my original attempt, (F * IJK)2 for each component I did when it should have been F(IJK2)

Mentor
I just noticed my error. In my original attempt, (F * IJK)2 for each component I did when it should have been F(IJK2)
Perhaps you understand what you mean, but this is very weird notation. If u = <a, b, c> and v = 2<a, b, c>, then |v| = 2|u|