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.

 

[jedishrfu]

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

 

[Fredrik]

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)

 

[Mark44]

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|