# A triangle has vertices A (4,2,1), B (1,2,4), C (-1, 0, -4)

• Mar 14th 2011, 10:35 PM
brumby_3
A triangle has vertices A (4,2,1), B (1,2,4), C (-1, 0, -4)
Use dot products to find all three angles of the triangle.

I already found the lengths of the three sides of the triangle: AB = 3sqrt(2), AC = 3sqrt(6) and BC = 6sqrt(2) and I already know the answers because I found them using trigonometry, but not sure how to do it using dot products.
• Mar 14th 2011, 10:51 PM
Unknown008
To get the angles using dot product, you need the vectors, not the lengths.

Find:

the dot product of $\vec{AB}$ and $\vec{AC}$ to get the angle BAC.

the dot product of $\vec{BA}$ and $\vec{BC}$ to get the angle ABC.

the dot product of $\vec{CB}$ and $\vec{CA}$ to get the angle ACB.
• Mar 14th 2011, 10:54 PM
brumby_3
Could you please give the first one as an example? I already know AB is <-3, 0, 3> and AC is <-5,-2,-5> (that is AB and AC with the arrow on top)
• Mar 14th 2011, 11:04 PM
Unknown008
You will need the formula:

$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$

Applying this, we get:

$\left(\begin{array}{c} -3 \\ 0 \\ 3\end{array}\right) \cdot \left(\begin{array}{c} -5 \\ -2 \\ -5\end{array}\right) = (3\sqrt2)(3\sqrt6) \cos\theta_1$

There theta1 is the angle A.

Doing the dot product, we get:

$(-3)(-5) + (0)(-2) + (3)(-5) = 9\sqrt{12} \cos\theta_1$

$\cos\theta_1 = \dfrac{15-15}{9\sqrt{12}}$

theta1 is 90 degrees.
• Mar 14th 2011, 11:06 PM
brumby_3
Thanks a lot, I really appreciate your help :)