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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 #64 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} 1\\-5 \\3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -2\\ 5\\6 \end{bmatrix}$ , 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

$\overrightarrow{a}=\begin{bmatrix} 1\\-5 \\3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -2\\ 5\\6 \end{bmatrix}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=-2-25+18=-9$

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Example Question #65 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} -8\\-6 \\-6 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 4\\-2 \\6 \end{bmatrix}$ , 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

$\overrightarrow{a}=\begin{bmatrix} -8\\-6 \\-6 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 4\\-2 \\6 \end{bmatrix}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=-32+12-36=-56$

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Example Question #66 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} 0\\ -3\\0 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 5\\7 \\8 \end{bmatrix}$ , 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

$\overrightarrow{a}=\begin{bmatrix} 0\\ -3\\0 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 5\\7 \\8 \end{bmatrix}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=0-21+0=-21$

Report an Error

Example Question #67 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} -1\\7 \\5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 0\\0 \\5 \end{bmatrix}$ , 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

$\overrightarrow{a}=\begin{bmatrix} -1\\7 \\5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 0\\0 \\5 \end{bmatrix}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=0+0+25=25$

Report an Error

Example Question #68 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix}3 \\-7 \\-3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -9\\-4 \\3 \end{bmatrix}$ , 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

$\overrightarrow{a}=\begin{bmatrix}3 \\-7 \\-3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -9\\-4 \\3 \end{bmatrix}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=-27+28-9=-8$

Report an Error

Example Question #71 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} 6\\ 4\\3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 7\\6 \\1 \end{bmatrix}$ , 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

$\overrightarrow{a}=\begin{bmatrix} 6\\ 4\\3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 7\\6 \\1 \end{bmatrix}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=42+24+3=69$

Report an Error

Example Question #71 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} 9\\2 \\1 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -1\\4 \\-7 \end{bmatrix}$ , 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

$\overrightarrow{a}=\begin{bmatrix} 9\\2 \\1 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -1\\4 \\-7 \end{bmatrix}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=-9+8-7=-8$

Report an Error

Example Question #72 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} 0\\2 \\-5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\\-6 \\6 \end{bmatrix}$ , 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

$\overrightarrow{a}=\begin{bmatrix} 0\\2 \\-5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\\-6 \\6 \end{bmatrix}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=0-12-30=-42$

Report an Error

Example Question #71 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} -7\\ -7\\5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 7\\-5 \\5 \end{bmatrix}$ , 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

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

The dot product can be found following the example above:

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

Report an Error

Example Question #73 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} -3\\7 \\7 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -4\\2 \\1 \end{bmatrix}$ , 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

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

The dot product can be found following the example above:

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

Report an Error

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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 #64 : Dot Product

Example Question #65 : Dot Product

Example Question #66 : Dot Product

Example Question #67 : Dot Product

Example Question #68 : Dot Product

Example Question #71 : Dot Product

Example Question #71 : Dot Product

Example Question #72 : Dot Product

Example Question #71 : Dot Product

Example Question #73 : Dot Product

All Calculus 3 Resources

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