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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 #63 : Dot Product

Given the following two vectors, $\overrightarrow{a}=6\widehat{i}-8\widehat{j}+11\widehat{k}$ $\displaystyle \overrightarrow{a}=6\widehat{i}-8\widehat{j}+11\widehat{k}$ and $\overrightarrow{b}=4\widehat{i}+\widehat{j}-2\widehat{k}$ $\displaystyle \overrightarrow{b}=4\widehat{i}+\widehat{j}-2\widehat{k}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle 14$

$\displaystyle -6$

$\displaystyle 54$

$\displaystyle -11$

$\displaystyle 8$

Correct answer:

$\displaystyle -6$

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.

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

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

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

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 $\overrightarrow{a}=6\widehat{i}-8\widehat{j}+11\widehat{k}$ $\displaystyle \overrightarrow{a}=6\widehat{i}-8\widehat{j}+11\widehat{k}$ and $\overrightarrow{b}=4\widehat{i}+\widehat{j}-2\widehat{k}$ $\displaystyle \overrightarrow{b}=4\widehat{i}+\widehat{j}-2\widehat{k}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(24)+(-8)+(-22)=-6$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(24)+(-8)+(-22)=-6$

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Example Question #379 : Vectors And Vector Operations

Given the following two vectors, $\overrightarrow{a}=\widehat{i}+5\widehat{j}+8\widehat{k}$ $\displaystyle \overrightarrow{a}=\widehat{i}+5\widehat{j}+8\widehat{k}$ and $\overrightarrow{b}=-2\widehat{i}+2\widehat{j}-\widehat{k}$ $\displaystyle \overrightarrow{b}=-2\widehat{i}+2\widehat{j}-\widehat{k}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle 4$

$\displaystyle 20$

$\displaystyle 0$

$\displaystyle -2$

$\displaystyle 16$

Correct answer:

$\displaystyle 0$

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.

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

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 $\overrightarrow{a}=\widehat{i}+5\widehat{j}+8\widehat{k}$ $\displaystyle \overrightarrow{a}=\widehat{i}+5\widehat{j}+8\widehat{k}$ and $\overrightarrow{b}=-2\widehat{i}+2\widehat{j}-\widehat{k}$ $\displaystyle \overrightarrow{b}=-2\widehat{i}+2\widehat{j}-\widehat{k}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(-2)+(10)+(-8)=0$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(-2)+(10)+(-8)=0$

Observe from this zero result that the two vectors must be perpendicular!

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Example Question #64 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} 1\\-5 \\3 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix} 1\\-5 \\3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -2\\ 5\\6 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix} -2\\ 5\\6 \end{bmatrix}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle -7$

$\displaystyle 17$

$\displaystyle 13$

$\displaystyle -9$

$\displaystyle 8$

Correct answer:

$\displaystyle -9$

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

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

The dot product can be found following the example above:

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

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Example Question #65 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} -8\\-6 \\-6 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix} -8\\-6 \\-6 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 4\\-2 \\6 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix} 4\\-2 \\6 \end{bmatrix}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle -80$

$\displaystyle 64$

$\displaystyle -56$

$\displaystyle 32$

$\displaystyle -28$

Correct answer:

$\displaystyle -56$

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

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

The dot product can be found following the example above:

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

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Example Question #66 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} 0\\ -3\\0 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix} 0\\ -3\\0 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 5\\7 \\8 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix} 5\\7 \\8 \end{bmatrix}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle 25$

$\displaystyle -15$

$\displaystyle 17$

$\displaystyle -24$

$\displaystyle -21$

Correct answer:

$\displaystyle -21$

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

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

The dot product can be found following the example above:

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

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Example Question #67 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} -1\\7 \\5 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix} -1\\7 \\5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 0\\0 \\5 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix} 0\\0 \\5 \end{bmatrix}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle 65$

$\displaystyle 35$

$\displaystyle -5$

$\displaystyle 55$

$\displaystyle 25$

Correct answer:

$\displaystyle 25$

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

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

The dot product can be found following the example above:

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

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Example Question #68 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix}3 \\-7 \\-3 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix}3 \\-7 \\-3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -9\\-4 \\3 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix} -9\\-4 \\3 \end{bmatrix}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle -64$

$\displaystyle 32$

$\displaystyle -8$

$\displaystyle 18$

$\displaystyle -36$

Correct answer:

$\displaystyle -8$

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

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

The dot product can be found following the example above:

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

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Example Question #71 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} 6\\ 4\\3 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix} 6\\ 4\\3 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 7\\6 \\1 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix} 7\\6 \\1 \end{bmatrix}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle 66$

$\displaystyle 45$

$\displaystyle 41$

$\displaystyle 27$

$\displaystyle 69$

Correct answer:

$\displaystyle 69$

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

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

The dot product can be found following the example above:

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

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Example Question #71 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} 9\\2 \\1 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix} 9\\2 \\1 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} -1\\4 \\-7 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix} -1\\4 \\-7 \end{bmatrix}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle -8$

$\displaystyle 16$

$\displaystyle -24$

$\displaystyle 2$

$\displaystyle -14$

Correct answer:

$\displaystyle -8$

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

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

The dot product can be found following the example above:

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

Report an Error

Example Question #72 : Dot Product

Given the following two vectors, $\overrightarrow{a}=\begin{bmatrix} 0\\2 \\-5 \end{bmatrix}$ $\displaystyle \overrightarrow{a}=\begin{bmatrix} 0\\2 \\-5 \end{bmatrix}$ and $\overrightarrow{b}=\begin{bmatrix} 2\\-6 \\6 \end{bmatrix}$ $\displaystyle \overrightarrow{b}=\begin{bmatrix} 2\\-6 \\6 \end{bmatrix}$, calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$.

Possible Answers:

$\displaystyle -42$

$\displaystyle 44$

$\displaystyle -56$

$\displaystyle -28$

$\displaystyle 28$

Correct answer:

$\displaystyle -42$

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

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

The dot product can be found following the example above:

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

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #63 : Dot Product

Example Question #379 : Vectors And Vector Operations

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

All Calculus 3 Resources

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