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High School Math : Understanding Vector Calculations

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Example Question #2171 : High School Math

Find the magnitude of v=6i-6j .

Possible Answers:

$6\sqrt{2}$

$\sqrt{36}$

$\sqrt{24}$

Correct answer:

$6\sqrt{2}$

Explanation:

v=6i-6j therefore the vector is <6,-6>

To solve for the magnitude:

$\sqrt{(6)^2+(-6)^2}$

$\sqrt{(36)+(36)}$

$\sqrt{72}$

$6\sqrt{2}$

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Example Question #22 : Calculus Ii — Integrals

Let $\vec{}$ $\vec{a}$ and $\vec{b}$ be the following vectors: $\vec{a}=\left \langle 1,2,1\right \rangle$ and $\vec{b}=\left \langle 2,3,-1\right \rangle$ . If $\theta$ is the acute angle between the vectors, then which of the following is equal to $\theta$ ?

Possible Answers:

$\tan^{-1}\left(\frac{\sqrt{21}}{14}\right)$

$\cos^{-1}\left(\frac{\sqrt{21}}{14}\right)$

$\cos^{-1}\left(\frac{\sqrt{21}}{7}\right)$

$\sin^{-1}\left(\frac{\sqrt{21}}{7}\right)$

Correct answer:

$\cos^{-1}\left(\frac{\sqrt{21}}{14}\right)$

Explanation:

The cosine of the acute angle between two vectors is given by the following formula:

$\cos\theta=\frac{\vec{a}\cdot \vec{b}}{\left \| \vec{a}\right \| \left \| \vec{b} \right \|}$ , where $\vec{a} \cdot \vec{b}$ represents the dot product of the two vectors, $\left \| \vec{a} \right \|$ is the magnitude of vector a, and $\left \| \vec{b} \right \|$ is the magnitude of vector b.

First, we will need to compute the dot product of the two vectors. Let's say we have two general vectors in space (three dimensions), $\vec{a}$ and $\vec{b}$ . Let the components of $\vec{a}$ be $<a_{1}, a_{2}, a_{3}>$ and the components of $\vec{b}$ be $<b_{1}, b_{2}, b_{3}>$ . Then the dot product $\vec{a} \cdot \vec{b}$ is defined as follows:

$\vec{a} \cdot \vec{b}=a_{1}(b_{1})+a_{2}(b_{2})+a_{3}(b_{3})$ .

Going back to the original problem, we can use this definition to find the dot product of $\vec{a}=<1,2,1>$ and $\vec{b}=<-2,3,-1>$ .

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

=1(-2)+2(3)+1(-1)

=-2+6-1=3

The next two things we will need to compute are $\left \| \vec{a} \right \|$ and $\left \| \vec{b} \right \|$ .

Let the components of a general vector $\vec{a}$ be $<a_{1}, a_{2}, a_{3}>$ . Then $\left \| \vec{a} \right \|$ is defined as $\left \| \vec{a}\right \|=\sqrt{(a_{1})^2+(a_{2})^2+(a_3)^2}$ .

Thus, if $\vec{a}=<1,2,1>$ and $\vec{b}=<-2,3,-1>$ , then

$\left \| \vec{a}\right \|=\sqrt{(a_{1})^2+(a_{2})^2+(a_3)^2}$

$=\sqrt{(1)^2+(2)^2+(1)^2}=\sqrt{1+4+1}=\sqrt{6}$

and $\left \| \vec{b}\right \|=\sqrt{(b_{1})^2+(b_{2})^2+(b_3)^2}=\sqrt{(-2)^2+(3)^2+(-1)^2}$

$=\sqrt{4+9+1}=\sqrt{14}$ .

Now, we put all of this information together to find the cosine of the angle between the two vectors.

$\cos\theta=\frac{\vec{a}\cdot \vec{b}}{\left \| \vec{a}\right \| \left \| \vec{b} \right \|}=\frac{3}{\sqrt{6}\cdot\sqrt{14}}$

We just need to simplify this.

$\frac{3}{\sqrt{6}\cdot\sqrt{14}}=\frac{3}{\sqrt{6\cdot14}}=\frac{3}{\sqrt{2\cdot3\cdot2\cdot7}}$

$=\frac{3}{2\sqrt{21}}$ .

In order to get it completely simplified, we have to rationalize the denominator by multiplying the numerator and denominator by the sqare root of 21.

$\frac{3}{2\sqrt{21}}=\frac{3\cdot\sqrt{21}}{2\sqrt{21}\cdot\sqrt{21}}=\frac{3\sqrt{21}}{2(21)}$

$=\frac{\sqrt{21}}{2\cdot7}=\frac{\sqrt{21}}{14}$ .

We just have one more step. We need to solve for the value of the angle. In order to do this, we can take the inverse cosine of both sides of the equation.

$\cos\theta=\frac{\sqrt{21}}{14}$

$\theta=\cos^{-1}(\frac{\sqrt{21}}{14})$ .

The answer is $\cos^{-1}(\frac{\sqrt{21}}{14})$ .

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High School Math : Understanding Vector Calculations

Study concepts, example questions & explanations for High School Math

All High School Math Resources

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Example Question #2171 : High School Math

Example Question #22 : Calculus Ii — Integrals

All High School Math Resources

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