vectors and dot product in 3-d

**ahawk1** · April 25th 2009, 02:53 PM

can someone tell me how to do this descriptively?
Find, correct to the nearest degree, the three angles of the triangle with the given vertices.
A(0, 1, 1), B(-2, 4, 2), C(1, 1, -2),

CAB = 115

ABC = 29

BCA = 3 36

those are the correct answers....i dont think im picking the correct vertices when im doing them or im just retarded and completely missing how to do this.

**TheEmptySet** · April 25th 2009, 03:13 PM

Originally Posted by ahawk1

can someone tell me how to do this descriptively?
Find, correct to the nearest degree, the three angles of the triangle with the given vertices.
A(0, 1, 1), B(-2, 4, 2), C(1, 1, -2),

CAB = 115

ABC = 29

BCA = 3 36

those are the correct answers....i dont think im picking the correct vertices when im doing them or im just retarded and completely missing how to do this.

I will show you how to do one
angle CAB

First we need to find the vectors $\vec{AC} \mbox{ and } \vec{AB}$

$\vec{AC}=\vec i + -3 \vec k$

$\vec{AB}=-2\vec i + 3\vec j+1 \vec k$

Now we can use the dot production to find the angle between two vectors

$\vec a \cdot \vec b =||a||||b||\cos(\theta)$

$(-2-3)=(\sqrt{1^2+(-3)^2})(\sqrt{(-2)^2+(3)^2+(1)^2})\cos(\theta)$
$\cos(\theta)=\frac{-5}{(\sqrt{10})(\sqrt{14})}$

$\theta = \cos^{-1} \left( \frac{-5}{\sqrt{140}} \right) \approx 140^\circ$

This same method will work for the other two angles. Good luck

**Soroban** · April 25th 2009, 03:52 PM

Hello, ahawk1!

Find, correct to the nearest degree, the three angles of the triangle
. . with the given vertices: . $A(0, 1, 1),\quad B(-2, 4, 2),\quad C(1, 1, -2)$

Answers: . $A = 115^o,\;\;B = 29^o,\;\;C = 36^o$

I would list all the vectors first . . .

$\begin{array}{ccc}\overrightarrow{AB} &=&\langle\text{-}2,3,1\rangle \\<br /> \overrightarrow{BA} &=& \langle 2,\text{-}3,\text{-}1\rangle \\<br /> |\overrightarrow{AB}| &=& \sqrt{14} \end{array}$ . . . $\begin{array}{ccc}\overrightarrow{BC} &=& \langle 3,\text{-}3,\text{-}4\rangle \\<br /> \overrightarrow{CB} &=& \langle \text{-}3,3,4\rangle \\ |\overrightarrow{BC}| &=& \sqrt{34}\end{array}$ . . . $\begin{array}{ccc}\overrightarrow{AC} &=&\langle 1,0,\text{-}3\rangle \\ \overrightarrow{CA} &=& \langle \text{-}1,0,3\rangle \\ |\overrightarrow{AC}| &=& \sqrt{10} \end{array}$

$\cos A \;=\;\frac{AB\bullet AC}{|AB||AC|} \;=\;\frac{\langle\text{-}2,3,1\rangle\bullet\langle1,0,\text{-}3\rangle}{\sqrt{14}\sqrt{10}} \;=\;\frac{\text{-}2+0-3}{\sqrt{140}} \;=$ . $\frac{\text{-}5}{2\sqrt{35}}\;=\;-0.422577127$

. . $A \;=\;114.9973996^o \;\approx\;\boxed{115^o}$

$\cos B \;=\;\frac{BA\bullet BC}{|BA||BC|} \;=\;\frac{\langle2,\text{-}3,\text{-}1\rangle\bullet\langle3,\text{-}3,\text{-}4\rangle}{\sqrt{14}\sqrt{34}} \;=\;\frac{6+9+4}{\sqrt{476}} \;=\;\frac{19}{2\sqrt{119}} \;=$ . $0.870863512$

. . $B \;=\;29.44085226^o \;\approx\;\boxed{29^o}$

And: . $C \;=\;180^o - 115^o - 29^o \;=\;\boxed{36^o}$

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