# Thread: vectors and dot product in 3-d

1. ## vectors and dot product in 3-d

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.

2. 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

3. 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 . . .

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

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

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

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

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

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

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# A man 2m tall stand on the same level ground as a vertical pole.He observed that the angle of elevasion of the top of the pole is 21degre and the angle of depression is 6degre Find correct to the nearest metre. (New concept Mathematic ss2)

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