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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 #53 : Dot Product

Given the following two vectors, $\overrightarrow{a}=14\widehat{i}-6\widehat{j}$ $\displaystyle \overrightarrow{a}=14\widehat{i}-6\widehat{j}$ and $\overrightarrow{b}=-5\widehat{i}-3\widehat{j}$ $\displaystyle \overrightarrow{b}=-5\widehat{i}-3\widehat{j}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

$\displaystyle -72$

$\displaystyle -88$

$\displaystyle 48$

$\displaystyle -12$

$\displaystyle -52$

Correct answer:

$\displaystyle -52$

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 $\overrightarrow{a}=14\widehat{i}-6\widehat{j}$ $\displaystyle \overrightarrow{a}=14\widehat{i}-6\widehat{j}$ and $\overrightarrow{b}=-5\widehat{i}-3\widehat{j}$ $\displaystyle \overrightarrow{b}=-5\widehat{i}-3\widehat{j}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(-70)+(18)=-52$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(-70)+(18)=-52$

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Example Question #54 : Dot Product

Given the following two vectors, $\overrightarrow{a}=16\widehat{i}+8\widehat{j}$ $\displaystyle \overrightarrow{a}=16\widehat{i}+8\widehat{j}$ and $\overrightarrow{b}=\widehat{i}+2\widehat{j}$ $\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

$\displaystyle 32$

$\displaystyle 16$

$\displaystyle 40$

$\displaystyle 24$

$\displaystyle 0$

Correct answer:

$\displaystyle 32$

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 $\overrightarrow{a}=16\widehat{i}+8\widehat{j}$ $\displaystyle \overrightarrow{a}=16\widehat{i}+8\widehat{j}$ and $\overrightarrow{b}=\widehat{i}+2\widehat{j}$ $\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(16)+(16)=32$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(16)+(16)=32$

Report an Error

Example Question #55 : Dot Product

Given the following two vectors, $\overrightarrow{a}=4\widehat{i}-24\widehat{j}$ $\displaystyle \overrightarrow{a}=4\widehat{i}-24\widehat{j}$ and $\overrightarrow{b}=7\widehat{i}+\widehat{j}$ $\displaystyle \overrightarrow{b}=7\widehat{i}+\widehat{j}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

$\displaystyle 4$

172 $\displaystyle 172$

$\displaystyle 26$

$\displaystyle 52$

$\displaystyle -162$

Correct answer:

$\displaystyle 4$

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 $\overrightarrow{a}=4\widehat{i}-24\widehat{j}$ $\displaystyle \overrightarrow{a}=4\widehat{i}-24\widehat{j}$ and $\overrightarrow{b}=7\widehat{i}+\widehat{j}$ $\displaystyle \overrightarrow{b}=7\widehat{i}+\widehat{j}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(28)+(-24)=4$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(28)+(-24)=4$

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Example Question #56 : Dot Product

Given the following two vectors, $\overrightarrow{a}=16\widehat{i}+17\widehat{j}$ $\displaystyle \overrightarrow{a}=16\widehat{i}+17\widehat{j}$ and $\overrightarrow{b}=\widehat{i}-\widehat{j}$ $\displaystyle \overrightarrow{b}=\widehat{i}-\widehat{j}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

$\displaystyle 33$

271 $\displaystyle 271$

$\displaystyle -1$

$\displaystyle -33$

$\displaystyle 1$

Correct answer:

$\displaystyle -1$

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 $\overrightarrow{a}=16\widehat{i}+17\widehat{j}$ $\displaystyle \overrightarrow{a}=16\widehat{i}+17\widehat{j}$ and $\overrightarrow{b}=\widehat{i}-\widehat{j}$ $\displaystyle \overrightarrow{b}=\widehat{i}-\widehat{j}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(16)+(-17)=-1$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(16)+(-17)=-1$

Report an Error

Example Question #2371 : Calculus 3

Given the following two vectors, $\overrightarrow{a}=46\widehat{i}+23\widehat{j}$ $\displaystyle \overrightarrow{a}=46\widehat{i}+23\widehat{j}$ and $\overrightarrow{b}=\widehat{i}+2\widehat{j}$ $\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

207 $\displaystyle 207$

$\displaystyle 0$

$\displaystyle 69$

$\displaystyle 92$

$\displaystyle 23$

Correct answer:

$\displaystyle 92$

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 $\overrightarrow{a}=46\widehat{i}+23\widehat{j}$ $\displaystyle \overrightarrow{a}=46\widehat{i}+23\widehat{j}$ and $\overrightarrow{b}=\widehat{i}+2\widehat{j}$ $\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(46)+(46)=92$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(46)+(46)=92$

Report an Error

Example Question #58 : Dot Product

Given the following two vectors, $\overrightarrow{a}=-3\widehat{j}$ $\displaystyle \overrightarrow{a}=-3\widehat{j}$ and $\overrightarrow{b}=15\widehat{i}+5\widehat{j}$ $\displaystyle \overrightarrow{b}=15\widehat{i}+5\widehat{j}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

$\displaystyle 15$

$\displaystyle -15$

$\displaystyle -60$

$\displaystyle -45$

$\displaystyle 45$

Correct answer:

$\displaystyle -15$

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 $\overrightarrow{a}=-3\widehat{j}$ $\displaystyle \overrightarrow{a}=-3\widehat{j}$ and $\overrightarrow{b}=15\widehat{i}+5\widehat{j}$ $\displaystyle \overrightarrow{b}=15\widehat{i}+5\widehat{j}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(0)+(-15)=-15$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(0)+(-15)=-15$

Report an Error

Example Question #59 : Dot Product

Given the following two vectors, $\overrightarrow{a}=3\widehat{i}+5\widehat{k}$ $\displaystyle \overrightarrow{a}=3\widehat{i}+5\widehat{k}$ and $\overrightarrow{b}=2\widehat{j}+6\widehat{k}$ $\displaystyle \overrightarrow{b}=2\widehat{j}+6\widehat{k}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

$\displaystyle 45$

$\displaystyle 15$

$\displaystyle 64$

$\displaystyle 36$

$\displaystyle 30$

Correct answer:

$\displaystyle 30$

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 $\overrightarrow{a}=3\widehat{i}+5\widehat{k}$ $\displaystyle \overrightarrow{a}=3\widehat{i}+5\widehat{k}$ and $\overrightarrow{b}=2\widehat{j}+6\widehat{k}$ $\displaystyle \overrightarrow{b}=2\widehat{j}+6\widehat{k}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(0)+(0)+(30)=30$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(0)+(0)+(30)=30$

Report an Error

Example Question #375 : Vectors And Vector Operations

Given the following two vectors, $\overrightarrow{a}=7\widehat{k}$ $\displaystyle \overrightarrow{a}=7\widehat{k}$ and $\overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}$ $\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

$\displaystyle 35$

$\displaystyle 21$

$\displaystyle 14$

$\displaystyle 42$

$\displaystyle 7$

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.

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

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

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

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 $\overrightarrow{a}=7\widehat{k}$ $\displaystyle \overrightarrow{a}=7\widehat{k}$ and $\overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}$ $\displaystyle \overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}$

The dot product can be found following the example above:

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

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Example Question #61 : Dot Product

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

Possible Answers:

$\displaystyle 18$

$\displaystyle -12$

$\displaystyle -9$

$\displaystyle 15$

$\displaystyle 6$

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.

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

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

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

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 $\overrightarrow{a}=\widehat{i}-6\widehat{j}-3\widehat{k}$ $\displaystyle \overrightarrow{a}=\widehat{i}-6\widehat{j}-3\widehat{k}$ and $\overrightarrow{b}=3\widehat{i}+\widehat{j}+2\widehat{k}$ $\displaystyle \overrightarrow{b}=3\widehat{i}+\widehat{j}+2\widehat{k}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(3)+(-6)+(-6)=-9$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(3)+(-6)+(-6)=-9$

Report an Error

Example Question #62 : Dot Product

Given the following two vectors, $\overrightarrow{a}=5\widehat{i}+\widehat{j}+13\widehat{k}$ $\displaystyle \overrightarrow{a}=5\widehat{i}+\widehat{j}+13\widehat{k}$ and $\overrightarrow{b}=-2\widehat{i}-\widehat{j}+\widehat{k}$ $\displaystyle \overrightarrow{b}=-2\widehat{i}-\widehat{j}+\widehat{k}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

$\displaystyle 1$

$\displaystyle 23$

$\displaystyle 24$

$\displaystyle 21$

$\displaystyle 2$

Correct answer:

$\displaystyle 2$

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 $\overrightarrow{a}=5\widehat{i}+\widehat{j}+13\widehat{k}$ $\displaystyle \overrightarrow{a}=5\widehat{i}+\widehat{j}+13\widehat{k}$ and $\overrightarrow{b}=-2\widehat{i}-\widehat{j}+\widehat{k}$ $\displaystyle \overrightarrow{b}=-2\widehat{i}-\widehat{j}+\widehat{k}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(-10)+(-1)+(13)=2$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(-10)+(-1)+(13)=2$

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Calculus 3 : Calculus 3

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #53 : Dot Product

Example Question #54 : Dot Product

Example Question #55 : Dot Product

Example Question #56 : Dot Product

Example Question #2371 : Calculus 3

Example Question #58 : Dot Product

Example Question #59 : Dot Product

Example Question #375 : Vectors And Vector Operations

Example Question #61 : Dot Product

Example Question #62 : Dot Product

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