# Check my work regarding vector magnitude and direction?

• Jun 30th 2010, 04:04 PM
Mattpd
Check my work regarding vector magnitude and direction?
Sorry if this is the wrong section, but it came from my Calc book.

The question is : find a vector of norm $10\sqrt{2}$ in the direction opposite of $3i + 4j - 5k$.

As I understand it, the direction vector is the $\mathbf{V}/\left \| \mathbf{V} \right \|$

So the direction vector of the given vector is:

$3i/\sqrt{50} + 4j/\sqrt{50} - 5k/\sqrt{50}$ right?

New vector needs to be opposite so switch the signs:

$-3i/\sqrt{50} - 4j/\sqrt{50} + 5k/\sqrt{50}$

Now do I multiply the required norm into this direction vector to get the final answer:?

$-30\sqrt{2}i/\sqrt{50} - 40\sqrt{2}j/\sqrt{50} + 50\sqrt{2}k/\sqrt{50}$

I just took a stab at it, so I likely messed up...
• Jun 30th 2010, 04:20 PM
jakncoke
if u take the vector $\begin{bmatrix} 3 \\ 4 \\ -5 \end{bmatrix}$ and multiply it by -2. u get a vector $\begin{bmatrix} -6 \\ -8 \\ 10 \end{bmatrix}$ with a norm of 10*root(2) in the oppsite direction. The direction of the vector is simply 3 in the x direction, 4 in the y direction, -5 in the z direction. the norm is $\sqrt{x^2 + y^2 + z^2}$ . $\mathbf{V}/\left \| \mathbf{V} \right \|$ is the formula for the unit vectors in the given direction, meaning its the same direction as the original vector but just scaled to have unit magnitude.
• Jun 30th 2010, 08:00 PM
Mattpd
Can I ask how you knew to multiply by -2?
• Jun 30th 2010, 10:11 PM
Prove It
Jakncoke had measured the length of your original vector and found it to be $5\sqrt{2}$. So you had to magnify it by $2$ to get the length of $10\sqrt{2}$, and then by $-1$ to reverse its direction.
• Jul 1st 2010, 08:38 AM
HallsofIvy
Quote:

Originally Posted by Mattpd
Sorry if this is the wrong section, but it came from my Calc book.

The question is : find a vector of norm $10\sqrt{2}$ in the direction opposite of $3i + 4j - 5k$.

As I understand it, the direction vector is the $\mathbf{V}/\left \| \mathbf{V} \right \|$

So the direction vector of the given vector is:

$3i/\sqrt{50} + 4j/\sqrt{50} - 5k/\sqrt{50}$ right?

New vector needs to be opposite so switch the signs:

$-3i/\sqrt{50} - 4j/\sqrt{50} + 5k/\sqrt{50}$

Now do I multiply the required norm into this direction vector to get the final answer:?

$-30\sqrt{2}i/\sqrt{50} - 40\sqrt{2}j/\sqrt{50} + 50\sqrt{2}k/\sqrt{50}$

I just took a stab at it, so I likely messed up...

It helps to know that 50= 2*25 so that $\frac{\sqrt{2}}{\sqrt{50}}= \frac{\sqrt{2}}{\sqrt{2}\sqrt{25}}$ $= \frac{1}{\sqrt{25}}= \frac{1}{5}$