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Calculus 3 : Vectors and Vector Operations

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6 Diagnostic Tests 373 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #4 : Dot Product

What is the length of the vector

$\small \small \bold{v}=(1,1,\sqrt{2})$ ?

Possible Answers:

$\small 4$

$\small 2$

$\small \sqrt{2}$

Correct answer:

$\small 2$

Explanation:

We can compute the length of a vector by taking the square root of the dot product of $\small \bold{v}$ and $\small \bold{v}$ , so the length of $\small \small \bold{v}=(1,1,\sqrt{2})$ is:

$\small \small |\bold{v}|=\sqrt{\bold{v}\cdot \bold{v}}=\sqrt{1^2+1^2+\sqrt{2}^2}=\sqrt{1+1+2}$

$\small \small =\sqrt{4}=2$

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Example Question #5 : Dot Product

Which of the following cannot be used as a definition of the dot product of two real-valued vectors?

Possible Answers:

$\mathbf{a} \cdot \mathbf{b} = \sum_{i = 1}^{n}a_ib_i$

$\mathbf{a} \cdot \mathbf{b} = \sum_{i = 1}^{n}a_i+b_i$

$\mathbf{a} \cdot \mathbf{b} = \|\mathbf{a}\|\|\mathbf{b}\|\cos{\theta}$ , where $\theta$ is the angle between $\mathbf{a}, \mathbf{b}$ .

They may all be used

Correct answer:

$\mathbf{a} \cdot \mathbf{b} = \sum_{i = 1}^{n}a_i+b_i$

Explanation:

$\mathbf{a} \cdot \mathbf{b} = \sum_{i = 1}^{n}a_i+b_i$ is not correct. This is saying effectively to add all the components of the two vectors together. The other two definitions are commonly used in computing angles between vectors and other objects, and can also be derived from each other.

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Example Question #5 : Dot Product

Which of the following is true concerning the dot product of two vectors?

Possible Answers:

$\mathbf{a}\cdot\mathbf{b}\cdot\mathbf{c}$ is well-defined as long as each vector is the same dimension

The dot product of two vectors is never negative.

None of the other statements are true.

$\mathbf{a} \cdot \mathbf{b} =0$ if and only if $\mathbf{a}, \mathbf{b}$ are orthogonal.

The dot product of two vectors is never a scalar.

Correct answer:

$\mathbf{a} \cdot \mathbf{b} =0$ if and only if $\mathbf{a}, \mathbf{b}$ are orthogonal.

Explanation:

This statement is true; it can be derived from the definition $\mathbf{a}\cdot\mathbf{b} = \|\mathbf{a}\|\|\mathbf{b}\|\cos\theta$ by setting the acute angle between the vectors to be $\theta = \frac{\pi}{2}$ ; the requirement for orthogonality. Additionally, if either vector has length , the vectors are still said to be orthogonal.

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Example Question #9 : Dot Product

What is the dot product of vectors $\vec{a}= -\hat{i}-\hat{j} -3\hat{k}$ and $\vec{b}= -3\hat{i}-4\hat{j} -5\hat{k}$ ?

Possible Answers:

Correct answer:

Explanation:

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

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

In this problem

$a\cdot b=-1(-3)-1(-4)-3(-5)=22$

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Example Question #10 : Dot Product

What is the dot product of vectors $\vec{a}= 2\hat{i}+5\hat{j} +7\hat{k}$ and $\vec{b}= -2\hat{i}-3\hat{j} +\hat{k}$ ?

Possible Answers:

Does not exist

Correct answer:

Explanation:

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

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

In this problem

$a\cdot b=2(-2)+5(-3)+7(1)=-12$

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Example Question #321 : Vectors And Vector Operations

For what angle(s) is the dot product of two vectors $\vec{A} \cdot \vec{B}=0$ ?

Possible Answers:

$0^{\circ}$

$45^{\circ}$

$90 ^{\circ}, 270 ^{\circ}$

$360^{\circ}$

$180^{\circ}$

Correct answer:

$90 ^{\circ}, 270 ^{\circ}$

Explanation:

We have the following equation that shows the relation between the dot product of two vectors, $\vec{A} \cdot \vec{B}$ , to the relative angle between them $\theta$ ,

$\frac{\vec{A} \cdot \vec{B}}{\left| \vec{A}\right| \left| \vec{B} \right|} = \cos \theta$ .

From this, we can see that the numerator $\vec{A} \cdot \vec{B}$ will be whenever $\cos \theta=0$ .

$\cos \theta = 0$ for all odd-multiples of $\frac{\pi}{2}$ , which in one rotation, includes $\frac{\pi}{2}=90 ^{\circ},3\frac{\pi}{2}=270^{\circ}$ .

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Example Question #321 : Vectors And Vector Operations

Compute $<0,1,x,t> \cdot <2,4>$

Possible Answers:

2+2x+4t

None of the other answers

<0,4,2x,4t>

4+4x+4t

Correct answer:

None of the other answers

Explanation:

There is no correct way to compute the above. In order to take the dot product, the two vectors must have the same number of components. These vectors have and components respectively.

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Example Question #322 : Vectors And Vector Operations

Compute $<t_1,0,5> \cdot <t_8 ,8, 4>$

Possible Answers:

t_1t_8-20

None of the other answers

<t_1t_8, 0, 20>

t_1t_8 +20

Correct answer:

t_1t_8 +20

Explanation:

To computer the dot product, we multiply the values of common components together and sum their totals. The outcome is a scalar value, not a vector.

So we have

$<t_1,0,5> \cdot <t_8 ,8, 4> = (t_1)(t_8)+(0)(8)+(5)(4) = t_1t_8+20$

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Example Question #2331 : Calculus 3

Given the following two vectors, $\overrightarrow{a}=3\widehat{i}-1\widehat{j}+4\widehat{k}$ and $\overrightarrow{b}=2\widehat{i}+6\widehat{j}-2\widehat{k}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

Correct answer:

Explanation:

The dot product of a paired set of vectors can be found by summing up the individual products of the multiplications between matched directional vectors.

$\overrightarrow{a}=a_1\widehat{x_1}+a_2\widehat{x_2}+a_3\widehat{x_3}+...+a_n\widehat{x_n}$

$\overrightarrow{b}=b_1\widehat{x_1}+b_2\widehat{x_2}+b_3\widehat{x_3}+...+b_n\widehat{x_n}$

$\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3+...+a_nb_n$

Note that the dot product is a scalar value rather than a vector; there's no directional term.

Now considering our problem, we're given the vectors $\overrightarrow{a}=3\widehat{i}-1\widehat{j}+4\widehat{k}$ and $\overrightarrow{b}=2\widehat{i}+6\widehat{j}-2\widehat{k}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(3*2)+(-1*6)+(4*-2)=-8$

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Example Question #2332 : Calculus 3

Given the following two vectors, $\overrightarrow{a}=4\widehat{i}+5\widehat{j}+2\widehat{k}$ and $\overrightarrow{b}=3\widehat{i}-3\widehat{j}+3\widehat{k}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

Correct answer:

Explanation:

The dot product of a paired set of vectors can be found by summing up the individual products of the multiplications between matched directional vectors.

$\overrightarrow{a}=a_1\widehat{x_1}+a_2\widehat{x_2}+a_3\widehat{x_3}+...+a_n\widehat{x_n}$

$\overrightarrow{b}=b_1\widehat{x_1}+b_2\widehat{x_2}+b_3\widehat{x_3}+...+b_n\widehat{x_n}$

$\overrightarrow{a}\cdot\overrightarrow{b}=a_1b_1+a_2b_2+a_3b_3+...+a_nb_n$

Note that the dot product is a scalar value rather than a vector; there's no directional term.

Now considering our problem, we're given the vectors $\overrightarrow{a}=4\widehat{i}+5\widehat{j}+2\widehat{k}$ and $\overrightarrow{b}=3\widehat{i}-3\widehat{j}+3\widehat{k}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(4*3)+(5*-3)+(2*3)=3$

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Calculus 3 : Vectors and Vector Operations

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #4 : Dot Product

Example Question #5 : Dot Product

Example Question #5 : Dot Product

Example Question #9 : Dot Product

Example Question #10 : Dot Product

Example Question #321 : Vectors And Vector Operations

Example Question #321 : Vectors And Vector Operations

Example Question #322 : Vectors And Vector Operations

Example Question #2331 : Calculus 3

Example Question #2332 : Calculus 3

All Calculus 3 Resources

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