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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 #2353 : Calculus 3

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

Possible Answers:

$\displaystyle 0$

$\displaystyle 36$

$\displaystyle 10$

$\displaystyle 11$

$\displaystyle 49$

Correct answer:

$\displaystyle 10$

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

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(3)+(4)+(3)=10$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(3)+(4)+(3)=10$

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Example Question #2351 : Calculus 3

Given the following two vectors, $\overrightarrow{a}=\widehat{i}+4\widehat{j}+5\widehat{k}$ $\displaystyle \overrightarrow{a}=\widehat{i}+4\widehat{j}+5\widehat{k}$ and $\overrightarrow{b}=2\widehat{i}-3\widehat{k}$ $\displaystyle \overrightarrow{b}=2\widehat{i}-3\widehat{k}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

$\displaystyle 11$

$\displaystyle 15$

$\displaystyle -13$

$\displaystyle -9$

$\displaystyle -10$

Correct answer:

$\displaystyle -13$

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 $\overrightarrow{a}=\widehat{i}+4\widehat{j}+5\widehat{k}$ $\displaystyle \overrightarrow{a}=\widehat{i}+4\widehat{j}+5\widehat{k}$ and $\overrightarrow{b}=2\widehat{i}-3\widehat{k}$ $\displaystyle \overrightarrow{b}=2\widehat{i}-3\widehat{k}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(2)+(0)+(-15)=-13$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(2)+(0)+(-15)=-13$

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Example Question #31 : Dot Product

Find the dot product between the two vectors.

u=<1,2,1> $\displaystyle u=< 1,2,1>$

v=<0,-3,-7> $\displaystyle v=< 0,-3,-7>$

Possible Answers:

$\displaystyle 4$

$\displaystyle -6$

$\displaystyle 0$

$\displaystyle -13$

Correct answer:

$\displaystyle -13$

Explanation:

The dot product for two vectors <a_1,a_2,a_3> $\displaystyle < a_1,a_2,a_3>$ and <b_1,b_2,b_3> $\displaystyle < b_1,b_2,b_3>$

is defined as $\displaystyle a_1*b_1+a_2*b_2+a_3*b_3$

Fo the given vectors

u=<1,2,1> $\displaystyle u=< 1,2,1>$

v=<0,-3,-7> $\displaystyle v=< 0,-3,-7>$

$u\bullet v =<1,2,1>\bullet <0,-3,-7>$ $\displaystyle u\bullet v =< 1,2,1>\bullet < 0,-3,-7>$

$\displaystyle =1*0+2*(-3)+1*(-7)$

=0-6-7 $\displaystyle =0-6-7$

=-13 $\displaystyle =-13$

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Example Question #351 : Vectors And Vector Operations

Find the dot product of the two vectors.

u=<7,10> $\displaystyle u=< 7,10>$

v=<3,7> $\displaystyle v=< 3,7>$

Possible Answers:

700 $\displaystyle 700$

$\displaystyle 91$

1470 $\displaystyle 1470$

$\displaystyle 79$

Correct answer:

$\displaystyle 91$

Explanation:

The dot product for two vectors <a_1,a_2> $\displaystyle < a_1,a_2>$ and <b_1,b_2> $\displaystyle < b_1,b_2>$

is defined as $\displaystyle a_1*b_1+a_2*b_2$

Fo the given vectors

u=<7,10> $\displaystyle u=< 7,10>$

v=<3,7> $\displaystyle v=< 3,7>$

$u\bullet v = <7,10>\bullet <3,7>$ $\displaystyle u\bullet v = < 7,10>\bullet < 3,7>$

$\displaystyle =7*3+10*7$

=21+70 $\displaystyle =21+70$

=91 $\displaystyle =91$

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Example Question #351 : Vectors And Vector Operations

Find the dot product $v\cdot w$ $\displaystyle v\cdot w$ of the two vectors

$\displaystyle v=(1,-1,2)$

$\displaystyle w=(0,5,3)$

Possible Answers:

$\displaystyle 1$

$\displaystyle 5$

$\displaystyle 2$

$\displaystyle -1$

Correct answer:

$\displaystyle 1$

Explanation:

To find the dot product $v\cdot w$ $\displaystyle v\cdot w$ , we calculate

$v\cdot w=1\cdot 0+(-1)\cdot 5+2\cdot 3=-5+6=1$ $\displaystyle v\cdot w=1\cdot 0+(-1)\cdot 5+2\cdot 3=-5+6=1$

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Example Question #41 : Dot Product

Given the following two vectors, $\overrightarrow{a}=1\widehat{i}+11\widehat{j}$ $\displaystyle \overrightarrow{a}=1\widehat{i}+11\widehat{j}$ and $\overrightarrow{b}=15\widehat{i}-2\widehat{j}$ $\displaystyle \overrightarrow{b}=15\widehat{i}-2\widehat{j}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

$\displaystyle 37$

$\displaystyle -7$

$\displaystyle -156$

156 $\displaystyle 156$

$\displaystyle 52$

Correct answer:

$\displaystyle -7$

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 $\overrightarrow{a}=1\widehat{i}+11\widehat{j}$ $\displaystyle \overrightarrow{a}=1\widehat{i}+11\widehat{j}$ and $\overrightarrow{b}=15\widehat{i}-2\widehat{j}$ $\displaystyle \overrightarrow{b}=15\widehat{i}-2\widehat{j}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(15)+(-22)=-7$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(15)+(-22)=-7$

Report an Error

Example Question #2352 : Calculus 3

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

Possible Answers:

$\displaystyle 11$

$\displaystyle 23$

$\displaystyle 15$

$\displaystyle 17$

$\displaystyle -9$

Correct answer:

$\displaystyle 17$

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

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(14)+(3)=17$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(14)+(3)=17$

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Example Question #2353 : Calculus 3

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

Possible Answers:

$\displaystyle -72$

$\displaystyle -42$

$\displaystyle 56$

$\displaystyle 24$

$\displaystyle 48$

Correct answer:

$\displaystyle 24$

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

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(-24)+(48)=24$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(-24)+(48)=24$

Report an Error

Example Question #2361 : Calculus 3

Given the following two vectors, $\overrightarrow{a}=11\widehat{i}+17\widehat{j}$ $\displaystyle \overrightarrow{a}=11\widehat{i}+17\widehat{j}$ and $\overrightarrow{b}=3\widehat{i}-2\widehat{j}$ $\displaystyle \overrightarrow{b}=3\widehat{i}-2\widehat{j}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

$\displaystyle 13$

$\displaystyle 5$

$\displaystyle 12$

$\displaystyle -6$

$\displaystyle -1$

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 $\overrightarrow{a}=11\widehat{i}+17\widehat{j}$ $\displaystyle \overrightarrow{a}=11\widehat{i}+17\widehat{j}$ and $\overrightarrow{b}=3\widehat{i}-2\widehat{j}$ $\displaystyle \overrightarrow{b}=3\widehat{i}-2\widehat{j}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(33)+(-34)=-1$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(33)+(-34)=-1$

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Example Question #352 : Vectors And Vector Operations

Given the following two vectors, $\overrightarrow{a}=11\widehat{i}+3\widehat{j}$ $\displaystyle \overrightarrow{a}=11\widehat{i}+3\widehat{j}$ and $\overrightarrow{b}=10\widehat{i}+6\widehat{j}$ $\displaystyle \overrightarrow{b}=10\widehat{i}+6\widehat{j}$ , calculate the dot product between them, $\overrightarrow{a}\cdot\overrightarrow{b}$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}$ .

Possible Answers:

$\displaystyle 49$

154 $\displaystyle 154$

$\displaystyle 33$

$\displaystyle 96$

128 $\displaystyle 128$

Correct answer:

128 $\displaystyle 128$

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 $\overrightarrow{a}=11\widehat{i}+3\widehat{j}$ $\displaystyle \overrightarrow{a}=11\widehat{i}+3\widehat{j}$ and $\overrightarrow{b}=10\widehat{i}+6\widehat{j}$ $\displaystyle \overrightarrow{b}=10\widehat{i}+6\widehat{j}$

The dot product can be found following the example above:

$\overrightarrow{a}\cdot\overrightarrow{b}=(110)+(18)=128$ $\displaystyle \overrightarrow{a}\cdot\overrightarrow{b}=(110)+(18)=128$

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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 #2353 : Calculus 3

Example Question #2351 : Calculus 3

Example Question #31 : Dot Product

Example Question #351 : Vectors And Vector Operations

Example Question #351 : Vectors And Vector Operations

Example Question #41 : Dot Product

Example Question #2352 : Calculus 3

Example Question #2353 : Calculus 3

Example Question #2361 : Calculus 3

Example Question #352 : Vectors And Vector Operations

All Calculus 3 Resources

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