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Calculus 3 : Calculus 3

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

Example Questions

Example Question #73 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} -7\\ -7\\5 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix} -7\\ -7\\5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 7\\-5 \\5 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix} 7\\-5 \\5 \end{bmatrix}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle -26$

$\displaystyle -14$

$\displaystyle 11$

$\displaystyle 13$

$\displaystyle 0$

Correct answer:

$\displaystyle 11$

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.

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

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

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

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

$\overrightarrow{a}=\begin{bmatrix} -7\\ -7\\5 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix} -7\\ -7\\5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 7\\-5 \\5 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix} 7\\-5 \\5 \end{bmatrix}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=-49+35+25=11$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=-49+35+25=11$

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Example Question #74 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} -3\\7 \\7 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix} -3\\7 \\7 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -4\\2 \\1 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix} -4\\2 \\1 \end{bmatrix}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle 33$

$\displaystyle 15$

$\displaystyle 45$

$\displaystyle -11$

$\displaystyle 9$

Correct answer:

$\displaystyle 33$

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.

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

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

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

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

$\overrightarrow{a}=\begin{bmatrix} -3\\7 \\7 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix} -3\\7 \\7 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -4\\2 \\1 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix} -4\\2 \\1 \end{bmatrix}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=12+14+7=33$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=12+14+7=33$

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Example Question #75 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} -6\\8 \\3 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix} -6\\8 \\3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix}-6 \\-2 \\7 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix}-6 \\-2 \\7 \end{bmatrix}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle 41$

$\displaystyle 15$

$\displaystyle 56$

$\displaystyle -5$

$\displaystyle 61$

Correct answer:

$\displaystyle 41$

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.

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

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

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

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

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

The dot product can be found following the example above:

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

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Example Question #76 : Dot Product

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

Possible Answers:

108 $\displaystyle 108$

214 $\displaystyle 214$

$\displaystyle -126$

$\displaystyle -172$

$\displaystyle -144$

Correct answer:

$\displaystyle -172$

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.

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

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

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

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

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

The dot product can be found following the example above:

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

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Example Question #2391 : Calculus 3

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

Possible Answers:

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

None of the other answers

<0,0,0> $\displaystyle < 0,0,0>$

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

Correct answer:

None of the other answers

Explanation:

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

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

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 #2392 : Calculus 3

Find the dot product of the two vectors:

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

Possible Answers:

22.14

19.07

20.72

23.58

25.32

Correct answer:

25.32

Explanation:

The dot product of two vectors is defined as:

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

For the given vectors, this is:

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

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

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Example Question #2393 : Calculus 3

Find the magnitude of the following vector:

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

Possible Answers:

$\sqrt{5}$ $\displaystyle \sqrt{5}$

$\sqrt{29}$ $\displaystyle \sqrt{29}$

$\sqrt{13}$ $\displaystyle \sqrt{13}$

$\sqrt{70}$ $\displaystyle \sqrt{70}$

$\sqrt{61}$ $\displaystyle \sqrt{61}$

Correct answer:

$\sqrt{61}$ $\displaystyle \sqrt{61}$

Explanation:

The magnitude of a vector is given by:

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

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Example Question #81 : Dot Product

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

Possible Answers:

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

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

$\displaystyle 10$

$\displaystyle 11$

$\displaystyle 69$

Correct answer:

$\displaystyle 69$

Explanation:

Write the dot product formula.

$a\cdot b = a_1b_1+a_2b_2$ $\displaystyle a\cdot b = a_1b_1+a_2b_2$

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

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

The dot product is: $\displaystyle 69$

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Example Question #82 : Dot Product

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Example Question #83 : Dot Product

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

Possible Answers:

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

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

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

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

Correct answer:

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

Explanation:

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

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

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #73 : Dot Product

Example Question #74 : Dot Product

Example Question #75 : Dot Product

Example Question #76 : Dot Product

Example Question #2391 : Calculus 3

Example Question #2392 : Calculus 3

Example Question #2393 : Calculus 3

Example Question #81 : Dot Product

Example Question #82 : Dot Product

Example Question #83 : Dot Product

All Calculus 3 Resources

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