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Calculus 3 : Vectors and Vector Operations

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Example Questions

Example Question #76 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} -6\\8 \\3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}-6 \\-2 \\7 \end{bmatrix}$ , 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.

$\overrightarrow{a}=\begin{bmatrix} a_1\\a_2 \\ ...\\ a_n \end{bmatrix}$

$\overrightarrow{b}=\begin{bmatrix}b_1 \\b_2 \\...\\b_n \end{bmatrix}$

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

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

$\overrightarrow{a}=\begin{bmatrix} -6\\8 \\3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}-6 \\-2 \\7 \end{bmatrix}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=36-16+21=41$

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Example Question #77 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} 4\\ -9\\-9 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -7\\ 7\\9 \end{bmatrix}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

108

214

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.

$\overrightarrow{a}=\begin{bmatrix} a_1\\a_2 \\ ...\\ a_n \end{bmatrix}$

$\overrightarrow{b}=\begin{bmatrix}b_1 \\b_2 \\...\\b_n \end{bmatrix}$

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

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

$\overrightarrow{a}=\begin{bmatrix} 4\\ -9\\-9 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -7\\ 7\\9 \end{bmatrix}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=-28-63-81=-172$

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Example Question #391 : Vectors And Vector Operations

Evaluate the dot product $<t,2t,t^3> \cdot <\sin(t),\cos(t),\ln(t)>$ .

Possible Answers:

$<t\cos(t),-2t\sin(t),t^2>$

None of the other answers

$<t\sin(t),2t\cos(t),t^3\ln(t)>$

<0,0,0>

Correct answer:

None of the other answers

Explanation:

The correct answer is $t\sin(t)+2t\cos(t)+t^3\ln(t)$ .

To compute the dot product, we take the corresponding components of each vector, and multiply them together. In this case, we have $(t)(\sin(t))+(2t)(\cos(t))+(t^3)(\ln(t))=t\sin(t)+2t\cos(t)+t^3\ln(t)$ .

Note that taking the dot product of any two vectors will always return a scalar-valued expression (or just a simple scalar). There should be no vector brackets in your answer.

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Example Question #392 : Vectors And Vector Operations

Find the dot product of the two vectors:

$\mathbf{u}=\left [ \frac{2\sqrt{2}}{3}, 7, 4\sqrt{3}\right ], \mathbf{v}=\left [ 6\sqrt{2}, 0, \frac{5}{2} \right ]$

Possible Answers:

22.14

25.32

19.07

23.58

20.72

Correct answer:

25.32

Explanation:

The dot product of two vectors is defined as:

$\mathbf{u} \cdot\mathbf{v}=u_xv_x+u_yv_y+u_zv_z$

For the given vectors, this is:

$\mathbf{u} \cdot\mathbf{v}=(\frac{2\sqrt{2}}{ {3}})(6\sqrt{2})+(7)(0)+(4\sqrt{3})(\frac{5}{2})$

$\mathbf{u} \cdot\mathbf{v}=8+0+10\sqrt{3}=25.32$

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Example Question #393 : Vectors And Vector Operations

Find the magnitude of the following vector:

$\mathbf{u}=6\mathbf{i}+3\mathbf{j}-4\mathbf{k}$

Possible Answers:

$\sqrt{61}$

$\sqrt{5}$

$\sqrt{70}$

$\sqrt{29}$

$\sqrt{13}$

Correct answer:

$\sqrt{61}$

Explanation:

The magnitude of a vector is given by:

$|\mathbf{u}|=\sqrt{u_x^2+u_y^2+u_z^2}=\sqrt{6^2+3^2+(-4)^2}=\sqrt{36+9+16}=\sqrt{61}$

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Example Question #2394 : Calculus 3

Find the dot product: $\left \langle -2,9\right \rangle\cdot \left \langle -3,7\right \rangle$

Possible Answers:

$\left \langle 6,63\right \rangle$

$\left \langle -5,16\right \rangle$

Correct answer:

Explanation:

Write the dot product formula.

$a\cdot b = a_1b_1+a_2b_2$

Substitute the values of the vectors and solve. The dot product will result in a number, not a vector.

(-2)(-3) + (9)(7) = 6 +63 =69

The dot product is:

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Example Question #391 : Vectors And Vector Operations

Find the length $\small ||v||$ of the vector $\small v=(1,\sqrt{3})$ .

Possible Answers:

$\small \small \small ||v||=\sqrt{1+\sqrt{3}}$

$\small \small ||v||=2$

$\small \small \small ||v||=1+\sqrt{3}$

$\small \small \small ||v||=4$

Correct answer:

$\small \small ||v||=2$

Explanation:

To find the length $\small ||v||$ of the vector $\small v=(1,\sqrt{3})$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small ||v||=\sqrt{v\cdot v}=\sqrt{1^2+\sqrt{3}^2}=\sqrt{4}=2$

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Example Question #2395 : Calculus 3

Find the length $\small ||v||$ of the vector $\small \small v=(15,12)$ .

Possible Answers:

$\small \small \small ||v||=27$

$\small \small \small ||v||=369$

$\small \small \small ||v||=\sqrt{370}$

$\small \small \small ||v||=\sqrt{369}$

Correct answer:

$\small \small \small ||v||=\sqrt{369}$

Explanation:

To find the length $\small ||v||$ of the vector $\small \small v=(15,12)$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small \small ||v||=\sqrt{v\cdot v}=\sqrt{15^2+12^2}=\sqrt{225+144}=\sqrt{369}$

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Example Question #2401 : Calculus 3

Find the length $\small ||v||$ of the vector $\small \small \small v=(5,5)$ .

Possible Answers:

$\small \small \small ||v||=50$

$\small \small ||v||=5\sqrt{2}$

$\small \small \small ||v||=2\sqrt{5}$

$\small \small \small ||v||=10$

Correct answer:

$\small \small ||v||=5\sqrt{2}$

Explanation:

To find the length $\small ||v||$ of the vector $\small \small \small v=(5,5)$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small \small \small \small ||v||=\sqrt{v\cdot v}=\sqrt{5^2+5^2}=\sqrt{25+25}=\sqrt{25\cdot 2}=5\sqrt{2}$

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Example Question #2402 : Calculus 3

Find the length $\small ||v||$ of the vector $\small \small \small \small v=(3,-1)$ .

Possible Answers:

$\small \small \small ||v||=2$

$\small \small \small ||v||=10$

$\small \small ||v||=\sqrt{10}$

$\small \small \small ||v||=4$

Correct answer:

$\small \small ||v||=\sqrt{10}$

Explanation:

To find the length $\small ||v||$ of the vector $\small \small \small \small v=(3,-1)$ , we take the square root of the dot product $\small v\cdot v$ :

$\small \small \small \small \small \small ||v||=\sqrt{v\cdot v}=\sqrt{3^2+(-1)^2}=\sqrt{9+1}=\sqrt{10}$

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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 #76 : Dot Product

Example Question #77 : Dot Product

Example Question #391 : Vectors And Vector Operations

Example Question #392 : Vectors And Vector Operations

Example Question #393 : Vectors And Vector Operations

Example Question #2394 : Calculus 3

Example Question #391 : Vectors And Vector Operations

Example Question #2395 : Calculus 3

Example Question #2401 : Calculus 3

Example Question #2402 : Calculus 3

All Calculus 3 Resources

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