# Problem with vector question

• Mar 6th 2010, 06:29 PM
Paymemoney
Problem with vector question
Hi
Can someone tell me where i have gone wrong in the following question?

Given that v=2i + 2j + k, w = 3i - j + k find a unit vector perpendicular to both v and w.

unit vector = {a x b}/{|a x b|}

| a x b |=$\displaystyle \sqrt{3^2+1^2+8^2}$

$\displaystyle =\sqrt{74}$

a x b = i(2 x 1 - 1 x -1)-j(2 x 1 - 1 x 3)+k(2 x -1 - 2 x 3)
=i(2+1)-j(2-3)+k(-2-6)
=3i+j-8k

therefore unit vector = $\displaystyle \frac{1}{\sqrt{74}}$(3i+j-8k)

answer says its$\displaystyle \frac{1}{\sqrt{74}}$(-3i-j+8k)

P.S
• Mar 6th 2010, 08:36 PM
Prove It
Quote:

Originally Posted by Paymemoney
Hi
Can someone tell me where i have gone wrong in the following question?

Given that v=2i + 2j + k, w = 3i - j + k find a unit vector perpendicular to both v and w.

unit vector = {a x b}/{|a x b|}

| a x b |=$\displaystyle \sqrt{3^2+1^2+8^2}$

$\displaystyle =\sqrt{74}$

a x b = i(2 x 1 - 1 x -1)-j(2 x 1 - 1 x 3)+k(2 x -1 - 2 x 3)
=i(2+1)-j(2-3)+k(-2-6)
=3i+j-8k

therefore unit vector = $\displaystyle \frac{1}{\sqrt{74}}$(3i+j-8k)

answer says its$\displaystyle \frac{1}{\sqrt{74}}$(-3i-j+8k)

P.S

$\displaystyle \mathbf{v}\times\mathbf{w} = \left|\begin{matrix}\mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & 2 & 1 \\ 3 & -1 & 1 \end{matrix}\right|$

$\displaystyle = \mathbf{i}[2 \cdot 1 - 1\cdot (-1)] - \mathbf{j}[2 \cdot 1 - 1 \cdot 3] + \mathbf{k}[2 \cdot (-1) - 2 \cdot 3]$

$\displaystyle = 3\mathbf{i} - \mathbf{j} - 8\mathbf{k}$.

I agree with the answer you have given for $\displaystyle \mathbf{v} \times \mathbf{w}$. Perhaps you copied the question down wrong or else the answer given is incorrect.

Anyway, to find the unit vector in the direction of $\displaystyle \mathbf{v} \times \mathbf{w}$, divide it by its length.

$\displaystyle |\mathbf{v}\times\mathbf{w}| = \sqrt{3^2 + (-1)^2 + (-8)^2}$

$\displaystyle = \sqrt{9 + 1 + 64}$

$\displaystyle = \sqrt{74}$.

Therefore:

$\displaystyle \frac{\mathbf{v}\times\mathbf{w}}{|\mathbf{v}\time s\mathbf{w}|} = \frac{3\mathbf{i} - \mathbf{j} - 8\mathbf{k}}{\sqrt{74}}$

$\displaystyle = \frac{3}{\sqrt{74}}\mathbf{i} - \frac{1}{\sqrt{74}}\mathbf{j} - \frac{8}{\sqrt{74}}\mathbf{k}$

$\displaystyle = \frac{3\sqrt{74}}{74}\mathbf{i} - \frac{\sqrt{74}}{74}\mathbf{j} - \frac{4\sqrt{74}}{37}\mathbf{k}$.
• Mar 6th 2010, 08:43 PM
Soroban
Hello, Paymemoney!

You've done nothing wrong . . .

Their answer is the negative of yours.

There are two vectors perpendicular to both $\displaystyle \vec v\text{ and }\vec w,$
. . one "up" and one "down".

They picked one, you picked the other.

• Mar 6th 2010, 09:22 PM
Paymemoney
ok oh ic, well to find the negative answer would you adjust the v and w values into negative? ie v=-2i + -2j - k, w = -3i + j - k