# vectors? a+b 100x greater than a-b

• Jan 18th 2007, 11:26 PM
thedoge
vectors? a+b 100x greater than a-b
Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A + B to be 100 times larger than the magnitude of A - B, what must be the angle between them?

I don't even know where to start. I attempt to solve this visually as any help I look up online includes formulas such as the "dot" formula which we have yet to discuss in class or even hint towards. Any help?
• Jan 19th 2007, 12:57 AM
earboth
Quote:

Originally Posted by thedoge
Two vectors A and B have precisely equal magnitudes. In order for the magnitude of A + B to be 100 times larger than the magnitude of A - B, what must be the angle between them?

I don't even know where to start. I attempt to solve this visually as any help I look up online includes formulas such as the "dot" formula which we have yet to discuss in class or even hint towards. Any help?

Hello,

the sum of two vectors can be described as the diagonal in a parallelogram with the vectors as its sides. The difference of two vectors gives the second diagonal in this parallelogram.

If $\vec {a}=\vec {b}$ then you are dealing with a rhombus and the diagonals are perpendicular that means you are dealing with a right triangle. (See attachment)

The angle between the vectors is $2 \alpha$. the ratio of the length of the diaogonals is 1 : 100. Thus

$\tan(\alpha)=\frac{\frac{1}{2} \cdot (\vec{a}-\vec{b})} {\frac{1}{2} \cdot (\vec{a}+\vec{b})}$. That means:

$\tan(\alpha)=\frac{\frac{1}{2} \cdot \left(\frac{1}{100}(\vec{a}+\vec{b})\right)} {\frac{1}{2} \cdot (\vec{a}+\vec{b})}=\frac{1}{100}$

Now you can calculate the angle between the vectors. I've got: $\approx 1.146^\circ$

EB
• Jan 19th 2007, 05:06 AM
thedoge
Thank you. I somehow ended up with 11.46 and couldnt figure out why.

Whoops I just realized this was in the wrong forum. Glad you found it anyway earboth.