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

\(\displaystyle \theta = \frac{\pi}{6}\)