1. ## problem with vectors

I need some help with the the following problem:

If vectors a and b are orthogonal, |a|=3, |b|=4.
Calculate:
a) p=|(a+b)x(a-b)|
b) q=|(3a-b)x(a-2b)|

2. Originally Posted by sillyme
I need some help with the the following problem:
If vectors a and b are orthogonal, |a|=3, |b|=4.
Calculate:
a) p=|(a+b)x(a-b)|
b) q=|(3a-b)x(a-2b)|
Is that a cross product $u\times v$?
Or is it a dot product: $u \cdot v$?
I suspect it is a dot product.

3. Originally Posted by Plato
Is that a cross product $u\times v$?
Or is it a dot product: $u \cdot v$?
I suspect it is a dot product.
It is a cross product.
p is the norm of the cross product between the vector sum (a+b) and the vector difference (a-b)

$p = |(a+b)\times (a-b)|$

4. Some useful properties:
• $(u + v) \times w = (u \times w) + (v \times w)$
• $w \times (u + v) = (w \times u) + (w \times v)$
• $u \times v = -(v \times u)$
• $|u \times v | = |u||v|\sin \theta$

\begin{aligned} (a+b) \times (a-b) & = ( a \times (a-b)) + (b \times (a-b)) \\ & = (a \times a - (a \times b)) \ + \ ( b \times a - b \times b) \\ & = (0 - a \times b) + (-(a \times b) - 0) \\ & = -2 (a \times b) \end{aligned}

So: $p = | -2 (a \times b) | = 2 |a \times b | = 2 |a||b| \sin \theta$

But since they're orthogonal ....

5. Thank you o_O.

6. A very simple way to do this is to set up a coordinate system so the $\vec{a}= 3\vec{i}$ and $\vec{b}= 4\vec{j}$.
Then $\vec{a}+ \vec{b}= 3\vec{i}+ 4\vec{j}$, $\vec{a}- \vec{b}= 3\vec{i}- 4\vec{j}$ , $3\vec{a}- \vec{b}= 9\vec{i}- 4\vec{j}$, and $\vec{a}- 2\vec{b}= 3\vec{i}-8\vec{j}$.