# Dot product Question

• May 24th 2009, 08:03 AM
carnage
Dot product Question
The vectors x = (-4, p, -2) and y = (-2, 3, 6) are such that $\displaystyle \arccos(\frac{4}{21}) = \theta$, where $\displaystyle \theta$ is the angle between x and y. Determine the value(s) of p.

I know that x • y must equal to 4 in this case, and the product between the magnitudes of x and y must equal to 21,
but how do you set it up so that both conditions will be true?
• May 24th 2009, 08:27 AM
e^(i*pi)
Quote:

Originally Posted by carnage
The vectors x = (-4, p, -2) and y = (-2, 3, 6) are such that $\displaystyle \arccos(\frac{4}{21}) = \theta$, where $\displaystyle \theta$ is the angle between x and y. Determine the value(s) of p.

I know that x • y must equal to 4 in this case, and the product between the magnitudes of x and y must equal to 21,
but how do you set it up so that both conditions will be true?

$\displaystyle cos\theta = \frac{4}{21}$

The dot product is $\displaystyle a.b = |a||b|cos\theta$

$\displaystyle (-4\cdot -2) + (p \cdot 3) + (-2 \cdot 6) = \sqrt{(-4)^2 + p^2+(-2)^2} \times \sqrt{(-2)^2+3^2+6^2} \times \frac{4}{21}$

$\displaystyle \rightarrow 8 + 3p - 12 = \sqrt{20+p^2} \times \sqrt{49} \times \frac{4}{21}$

$\displaystyle \rightarrow 3p - 4 = 7\sqrt{20+p^2} \times \frac{4}{21} = \frac{4\sqrt{20+p^2}}{3}$

Solve for p