# Vector with the same direction but a different magnitude

## 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.

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

Fredrik
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".

## 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...

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

Mark44
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|