# Alternative solution for problem on Vectors

• Feb 15th 2009, 11:17 AM
scherz0
Alternative solution for problem on Vectors
Hello everyone,

Could anyone please tell me how to solve the following problem without knowledge of dot product or cross product?

Thank you very much.

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1. If $\displaystyle |\vec x|= 11, |\vec y| = 23,$ and $\displaystyle |\vec x - \vec y | = 30$, find $\displaystyle |\vec x + \vec y|$.

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My current solution:

$\displaystyle 30 = \sqrt{11^2 + 23^2 - 2(11)(23)\cos \theta}$ $\displaystyle \Rightarrow \cos \theta = -0.49407$

Since $\displaystyle |\vec x + \vec y| = \sqrt{x^2 + y^2 + 2xy\cos \theta}$, $\displaystyle |\vec x + \vec y|$ = 20.

As seen, my current solution involves the fact that $\displaystyle \vec x \cdot \vec y = |\vec x||\vec y|\cos \theta$.
• Feb 16th 2009, 06:50 AM
Soroban
Hello, scherz0!

Quote:

Could anyone please tell me how to solve the following
problem without knowledge of dot product or cross product?

1. If $\displaystyle |\vec x|= 11,\;|\vec y| = 23,$ and $\displaystyle |\vec x - \vec y | = 30$, find $\displaystyle |\vec x + \vec y|$.

It can be solved with just the Law of Cosines.

Code:

    D * * * * * * * * * * C       *  *              *         *    *  30      *       23 *        *        *           *          *    *           * 119.6°      *  *           A * * * * * * * * * * B                     11

I used this variation to find angle A.

. . $\displaystyle \cos A \:=\:\frac{11^2 + 23^2 - 30^2}{2(11)(23)} \:=\:-0.494071146 \quad\Rightarrow\quad A \:\approx\:119.6^o$

Then in parallelogram $\displaystyle ABCD,\;\angle B \,=\,60.4^o$

Code:

    D * * * * * * * * * * C       *              *  *         *          *    *         *        *        * 23           *    *          *           *  *        60.4° *           A * * * * * * * * * * B                     11

In $\displaystyle \Delta ABC\!:\;\;AC^2 \;=\;11^2 + 24^2 - 2(11)(23)\cos60.4^o \;=\;400.0654155$

Therefore: .$\displaystyle |\vec x + \vec y| \;=\;AC \;\approx\;20$