Lets remember the formula for finding the angle between two vectors.

$\displaystyle \cos(\theta)=\frac{(u\cdot v)}{\left \| u\right \|\left \| v\right \|}$

$\displaystyle \cos(\theta)=\frac{((3\cdot 9)+(4\cdot (-2)))}{\sqrt{3^2+4^2}\sqrt{9^2+(-2)^2}}$

$\displaystyle \cos(\theta)=\frac{19}{\sqrt{25}\sqrt{85}}$

$\displaystyle \cos(\theta)=\frac{19}{5\sqrt{85}}$

$\displaystyle \theta=\cos^{-1}\Big(\frac{19}{5\sqrt{85}}\Big)\approx65.66^\circ$

Lets recall the equation for finding the angle between vectors.

$\displaystyle \cos(\theta)=\frac{(u\cdot v)}{\left \| u\right \|\left \| v\right \|}$

$\displaystyle \cos(\theta)=\frac{((1\cdot 3)+(2\cdot 4))}{\sqrt{1^2+2^2}\sqrt{3^2+4^2}}$

$\displaystyle \cos(\theta)=\frac{11}{\sqrt{5}\sqrt{25}}$

$\displaystyle \cos(\theta)=\frac{11}{5\sqrt{5}}$

$\displaystyle \theta=\cos^{-1}\Big(\frac{11}{5\sqrt{5}}\Big)\approx 10.3^\circ$

To find the angle between vectors, we must use the dot product formula

$\displaystyle a\cdot b=\left | a\right |\left | b\right |cos\theta$

where $\displaystyle a\cdot b$ is the dot product of the vectors  $\displaystyle a$ and $\displaystyle b$, respectively.

  $\displaystyle \left | a\right |$ and $\displaystyle \left | b\right |$ are the magnitudes of vectors $\displaystyle a$ and $\displaystyle b$, respectively.

$\displaystyle cos\theta$ is the angle between the two vectors.

Let vector $\displaystyle a$ be represented as  $\displaystyle \vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vector $\displaystyle b$  be represented as  $\displaystyle \vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$.

The dot product of the vectors  $\displaystyle a$ and $\displaystyle b$ is $\displaystyle a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$.

The magnitude of vector $\displaystyle a$ is $\displaystyle \left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vector $\displaystyle a$ is $\displaystyle \left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$.

Rearranging the dot product formula to solve for $\displaystyle \theta$ gives us$\displaystyle \theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$\displaystyle a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=1(-2)+(-3)(1)+(7)(4)$

$\displaystyle a\cdot b=-2-3+28=23$

$\displaystyle \left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(1)^{2}+(-3)^{2}+(7)^{2}}$

$\displaystyle \left | a\right |=\sqrt{1+9+49}=\sqrt{59}$

$\displaystyle \left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(-2)^{2}+(1)^{2}+(4)^{2}}$

$\displaystyle \left | b\right |=\sqrt{4+1+16}=\sqrt{21}$

$\displaystyle \theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{23}{\sqrt{59}\sqrt{21}}\right )=cos^{-1}\left ( \frac{23}{\sqrt{1239}}\right )$

$\displaystyle \theta\approx49.2^{\circ}$

To find the angle between vectors, we must use the dot product formula

$\displaystyle a\cdot b=\left | a\right |\left | b\right |cos\theta$

where $\displaystyle a\cdot b$ is the dot product of the vectors  $\displaystyle a$ and $\displaystyle b$, respectively.

  $\displaystyle \left | a\right |$ and $\displaystyle \left | b\right |$ are the magnitudes of vectors $\displaystyle a$ and $\displaystyle b$, respectively.

$\displaystyle cos\theta$ is the angle between the two vectors.

Let vector $\displaystyle a$ be represented as  $\displaystyle \vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vector $\displaystyle b$  be represented as  $\displaystyle \vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$.

The dot product of the vectors  $\displaystyle a$ and $\displaystyle b$ is $\displaystyle a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$.

The magnitude of vector $\displaystyle a$ is $\displaystyle \left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vector $\displaystyle a$ is $\displaystyle \left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$.

Rearranging the dot product formula to solve for $\displaystyle \theta$ gives us$\displaystyle \theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$\displaystyle a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=2(1)+(-1)(2)+(7)(0)$

$\displaystyle a\cdot b=2-2+0=0$

$\displaystyle \left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(2)^{2}+(1)^{2}+(7)^{2}}$

$\displaystyle \left | a\right |=\sqrt{4+1+49}=\sqrt{54}$

$\displaystyle \left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(1)^{2}+(2)^{2}+(0)^{2}}$

$\displaystyle \left | b\right |=\sqrt{1+4}=\sqrt{5}$

$\displaystyle \theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{0}{\sqrt{54}\sqrt{5}}\right )=cos^{-1}\left (0\right )$

$\displaystyle \theta\approx90^{\circ}$

The vectors are perpendicular

To find the angle between vectors, we must use the dot product formula

$\displaystyle a\cdot b=\left | a\right |\left | b\right |cos\theta$

where $\displaystyle a\cdot b$ is the dot product of the vectors  $\displaystyle a$ and $\displaystyle b$, respectively.

  $\displaystyle \left | a\right |$ and $\displaystyle \left | b\right |$ are the magnitudes of vectors $\displaystyle a$ and $\displaystyle b$, respectively.

$\displaystyle cos\theta$ is the angle between the two vectors.

Let vector $\displaystyle a$ be represented as  $\displaystyle \vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vector $\displaystyle b$  be represented as  $\displaystyle \vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$.

The dot product of the vectors  $\displaystyle a$ and $\displaystyle b$ is $\displaystyle a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$.

The magnitude of vector $\displaystyle a$ is $\displaystyle \left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vector $\displaystyle a$ is $\displaystyle \left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$.

Rearranging the dot product formula to solve for $\displaystyle \theta$ gives us$\displaystyle \theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$\displaystyle a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=-2(1)+(-3)(2)+(1)(5)$

$\displaystyle a\cdot b=-2-6+5=-3$

$\displaystyle \left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(-2)^{2}+(-3)^{2}+(1)^{2}}$

$\displaystyle \left | a\right |=\sqrt{4+9+1}=\sqrt{14}$

$\displaystyle \left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(1)^{2}+(2)^{2}+(5)^{2}}$

$\displaystyle \left | b\right |=\sqrt{1+4+25}=\sqrt{30}$

$\displaystyle \theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{-3}{\sqrt{14}\sqrt{30}}\right )=cos^{-1}\left (\frac{-3}{\sqrt{420}}\right )$

$\displaystyle \theta\approx98.42^{\circ}$

To find the angle between vectors, we must use the dot product formula

$\displaystyle a\cdot b=\left | a\right |\left | b\right |cos\theta$

where $\displaystyle a\cdot b$ is the dot product of the vectors  $\displaystyle a$ and $\displaystyle b$, respectively.

  $\displaystyle \left | a\right |$ and $\displaystyle \left | b\right |$ are the magnitudes of vectors $\displaystyle a$ and $\displaystyle b$, respectively.

$\displaystyle cos\theta$ is the angle between the two vectors.

Let vector $\displaystyle a$ be represented as  $\displaystyle \vec{a}= a_{1}\hat{i}+a_{2}\hat{j} +a_{3}\hat{k}$ and vector $\displaystyle b$  be represented as  $\displaystyle \vec{b}= b_{1}\hat{i}+b_{2}\hat{j} +b_{3}\hat{k}$.

The dot product of the vectors  $\displaystyle a$ and $\displaystyle b$ is $\displaystyle a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})$.

The magnitude of vector $\displaystyle a$ is $\displaystyle \left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}$ and vector $\displaystyle a$ is $\displaystyle \left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}$.

Rearranging the dot product formula to solve for $\displaystyle \theta$ gives us$\displaystyle \theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )$

For this problem,

$\displaystyle a\cdot b=a_{1}( b_{1})+a_{2}( b_{2})+a_{3}( b_{3})=1(2)+(3)(6)+(5)(10)$

$\displaystyle a\cdot b=2+18+50=70$

$\displaystyle \left | a\right |=\sqrt{(a_{1})^{2}+(a_{2})^{2}+(a_{3})^{2}}=\sqrt{(1)^{2}+(3)^{2}+(5)^{2}}$

$\displaystyle \left | a\right |=\sqrt{1+9+25}=\sqrt{35}$

$\displaystyle \left | b\right |=\sqrt{(b_{1})^{2}+(b_{2})^{2}+(b_{3})^{2}}=\sqrt{(2)^{2}+(6)^{2}+(10)^{2}}$

$\displaystyle \left | b\right |=\sqrt{4+36+100}=\sqrt{140}$$\displaystyle \theta=cos^{-1}\left ( \frac{a\cdot b}{\left | a\right |\left | b\right |}\right )=cos^{-1}\left ( \frac{70}{\sqrt{35}\sqrt{140}}\right )=cos^{-1}\left (\frac{70}{\sqrt{4900}}\right )$

$\displaystyle \theta=cos^{-1}\left (\frac{70}{70}\right )=cos^{-1}\left (1\right )$

$\displaystyle \theta\approx0^{\circ}$

The two vectors are parallel.

To find the angle between two vectors, use the formula

$\displaystyle \theta = \cos^{-1}(\frac{\mathbf{a}\cdot\mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|})$.

$\displaystyle \theta = \cos^{-1}(\frac{< 2,2,5>\cdot< 0,5,-1>}{\sqrt{4+4+25}\sqrt{0+25+1}})$

$\displaystyle \theta = \cos^{-1}(\frac{5}{\sqrt{858}})$

$\displaystyle \theta \approx 80.22$

In order to find the angle between two vectors, we need to take the quotient of their dot product and their magnitudes:

$\displaystyle \frac{\vec{A}\cdot \vec{B}}{\left|\vec{A} \right| \left| \vec{B} \right| } = \cos{\theta}\Rightarrow \theta = \cos^{-1} \left( \frac{\vec{A}\cdot \vec{B}}{\left|\vec{A} \right| \left| \vec{B} \right| }\right )$

Therefore, we find that

$\displaystyle \vec{A}\cdot\vec{B}=3(6)+5(-2)-1(5)=3,$

$\displaystyle \left| \vec{A}\right| = \sqrt{3^2+5^2+1^2}=\sqrt{35}, \left| \vec{B}\right| = \sqrt{6^2+2^2+5^2}=\sqrt{65}$

$\displaystyle \theta = \cos^{-1} \left( \frac{3}{\sqrt{35}\sqrt{65} }\right)=86.39 ^{\circ}$.

To find the angle between vectors, we use the formula

$\displaystyle \theta = \cos^{-1}(\frac{\mathbf{a} \cdot \mathbf{b}}{\|\mathbf{a}\|\|\mathbf{b}\|})$.

Substituting in our values, we get

$\displaystyle \theta = \cos^{-1}(\frac{< 1,0,\frac{1}{2}> \cdot < \frac{3}{2},0,12>}{\sqrt{1^2+0^2+\frac{1}{2}^2}\sqrt{\frac{3}{2}^2+0^2+12^2}}) \approx \cos^{-1}(\frac{7.5}{13.521}) = 56.310$

To find the angle between two vector we use the following formula

$\displaystyle cos \, \theta = \frac{u\bullet v}{\left \| u\right \|*\left \| v\right \|}$

and solve for $\displaystyle \theta$.

Given

$\displaystyle u=< 2,0>$
$\displaystyle v=< \sqrt{3}, 1>$ we find

$\displaystyle u\bullet v= 2*\sqrt3+0*1=2\sqrt3$

$\displaystyle \left \| u\right \|=\sqrt{2^2+0^2}=\sqrt4=2$

$\displaystyle \left \| v\right \|=\sqrt{\sqrt{3}^2+1^2}=\sqrt4=2$

Plugging these values in we get

$\displaystyle cos\, \theta = \frac{2\sqrt3}{2*2}=\frac{\sqrt3}{2}$

To find $\displaystyle \theta$ we calculate the $\displaystyle cos^{-1}$ of both sides

$\displaystyle cos^{-1}[cos(\theta)]=cos^{-1}(\frac{\sqrt3}{2})$

and find that

$\displaystyle \theta = \frac{\pi}{6}$