# Thread: Check my work regarding vector magnitude and direction?

1. ## 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 $\displaystyle 10\sqrt{2}$ in the direction opposite of $\displaystyle 3i + 4j - 5k$.

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

So the direction vector of the given vector is:

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

New vector needs to be opposite so switch the signs:

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

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

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

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

2. if u take the vector $\displaystyle \begin{bmatrix} 3 \\ 4 \\ -5 \end{bmatrix}$ and multiply it by -2. u get a vector $\displaystyle \begin{bmatrix} -6 \\ -8 \\ 10 \end{bmatrix}$ 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 $\displaystyle \sqrt{x^2 + y^2 + z^2}$ . $\displaystyle \mathbf{V}/\left \| \mathbf{V} \right \|$ 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.

3. Can I ask how you knew to multiply by -2?

4. Jakncoke had measured the length of your original vector and found it to be $\displaystyle 5\sqrt{2}$. So you had to magnify it by $\displaystyle 2$ to get the length of $\displaystyle 10\sqrt{2}$, and then by $\displaystyle -1$ to reverse its direction.

5. 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 $\displaystyle 10\sqrt{2}$ in the direction opposite of $\displaystyle 3i + 4j - 5k$.

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

So the direction vector of the given vector is:

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

New vector needs to be opposite so switch the signs:

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

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

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

I just took a stab at it, so I likely messed up...
It helps to know that 50= 2*25 so that $\displaystyle \frac{\sqrt{2}}{\sqrt{50}}= \frac{\sqrt{2}}{\sqrt{2}\sqrt{25}}$$\displaystyle = \frac{1}{\sqrt{25}}= \frac{1}{5}$