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Calculus 3 : Cross Product

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Example Questions

Example Question #11 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=2\widehat{i}+4\widehat{k}$ and $\overrightarrow{b}=\widehat{i}+3\widehat{j}-2\widehat{k}$

Possible Answers:

$-12\widehat{i}-8\widehat{j}+6\widehat{k}$

$12\widehat{i}+8\widehat{j}+6\widehat{k}$

$-12\widehat{i}+8\widehat{j}-6\widehat{k}$

$-12\widehat{i}+8\widehat{j}+6\widehat{k}$

$8\widehat{i}-12\widehat{j}+6\widehat{k}$

Correct answer:

$-12\widehat{i}+8\widehat{j}+6\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=2\widehat{i}+4\widehat{k}$ and $\overrightarrow{b}=\widehat{i}+3\widehat{j}-2\widehat{k}$

$\overrightarrow{a}\times \overrightarrow{b}=(-12)\widehat{i}+(4+4)\widehat{j}+(6)\widehat{k}=-12\widehat{i}+8\widehat{j}+6\widehat{k}$

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Example Question #12 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=5\widehat{i}-5\widehat{j}$ and $\overrightarrow{b}=\widehat{i}+\widehat{j}$

Possible Answers:

$15\widehat{k}$

$-20\widehat{k}$

$10\widehat{k}$

$25\widehat{k}$

Correct answer:

$10\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=5\widehat{i}-5\widehat{j}$ and $\overrightarrow{b}=\widehat{i}+\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+5+5+0)\widehat{k}=10\widehat{k}$

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Example Question #12 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=3\widehat{i}+9\widehat{j}$ and $\overrightarrow{b}=4\widehat{i}-16\widehat{j}$

Possible Answers:

$-84\widehat{k}$

$91\widehat{k}$

$-132\widehat{k}$

$-37\widehat{k}$

$156\widehat{k}$

Correct answer:

$-84\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=3\widehat{i}+9\widehat{j}$ and $\overrightarrow{b}=4\widehat{i}-16\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0-48-36+0)\widehat{k}=-84\widehat{k}$

Report an Error

Example Question #14 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=-10\widehat{j}$ and $\overrightarrow{b}=3\widehat{i}+6\widehat{j}$

Possible Answers:

$-90\widehat{k}$

$45\widehat{k}$

$90\widehat{k}$

$-30\widehat{k}$

$30\widehat{k}$

Correct answer:

$30\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=-10\widehat{j}$ and $\overrightarrow{b}=3\widehat{i}+6\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(30+0)\widehat{k}=30\widehat{k}$

Report an Error

Example Question #15 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=4\widehat{i}-4\widehat{j}+3\widehat{k}$ and $\overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}$

Possible Answers:

$18\widehat{i}+9\widehat{j}-12\widehat{k}$

$-18\widehat{i}-9\widehat{j}-12\widehat{k}$

$-18\widehat{i}-9\widehat{j}+12\widehat{k}$

$18\widehat{i}-9\widehat{j}+12\widehat{k}$

$18\widehat{i}+9\widehat{j}+12\widehat{k}$

Correct answer:

$-18\widehat{i}-9\widehat{j}+12\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=4\widehat{i}-4\widehat{j}+3\widehat{k}$ and $\overrightarrow{b}=\widehat{i}+2\widehat{j}+3\widehat{k}$

$\overrightarrow{a}\times \overrightarrow{b}=(-12-6)\widehat{i}+(-12+3)\widehat{j}+(8+4)\widehat{k}=-18\widehat{i}-9\widehat{j}+12\widehat{k}$

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Example Question #13 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=20\widehat{i}+2\widehat{j}$ and $\overrightarrow{b}=15\widehat{i}-8\widehat{j}$

Possible Answers:

$-190\widehat{k}$

$-130\widehat{k}$

$190\widehat{k}$

$130\widehat{k}$

Correct answer:

$-190\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=20\widehat{i}+2\widehat{j}$ and $\overrightarrow{b}=15\widehat{i}-8\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0-160-30+0)\widehat{k}=-190\widehat{k}$

Report an Error

Example Question #17 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=6\widehat{i}-\widehat{j}$ and $\overrightarrow{b}=11\widehat{i}+5\widehat{j}$

Possible Answers:

$41\widehat{k}$

$71\widehat{k}$

$19\widehat{k}$

$61\widehat{k}$

Correct answer:

$41\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=6\widehat{i}-\widehat{j}$ and $\overrightarrow{b}=11\widehat{i}+5\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+30+11+0)\widehat{k}=41\widehat{k}$

Report an Error

Example Question #121 : Vectors And Vector Operations

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=9\widehat{i}-9\widehat{j}$ and $\overrightarrow{b}=-2\widehat{i}+2\widehat{j}$

Possible Answers:

$36\widehat{k}$

$9\widehat{k}$

$-9\widehat{k}$

$-36\widehat{k}$

Correct answer:

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=9\widehat{i}-9\widehat{j}$ and $\overrightarrow{b}=-2\widehat{i}+2\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+18-18+0)\widehat{k}=0$

This zero result stems from the fact that these vectors are parallel, a fact which might be apparent from quick observation.

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Example Question #122 : Vectors And Vector Operations

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=2\widehat{i}+6\widehat{j}+3\widehat{k}$ and $\overrightarrow{b}=\widehat{i}-\widehat{j}-2\widehat{k}$

Possible Answers:

$-9\widehat{i}+7\widehat{j}-8\widehat{k}$

$9\widehat{i}-7\widehat{j}+8\widehat{k}$

$-9\widehat{i}-7\widehat{j}-8\widehat{k}$

$9\widehat{i}+7\widehat{j}-8\widehat{k}$

$9\widehat{i}+7\widehat{j}+8\widehat{k}$

Correct answer:

$-9\widehat{i}+7\widehat{j}-8\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=2\widehat{i}+6\widehat{j}+3\widehat{k}$ and $\overrightarrow{b}=\widehat{i}-\widehat{j}-2\widehat{k}$

$\overrightarrow{a}\times \overrightarrow{b}=(-12+3)\widehat{i}+(4+3)\widehat{j}+(-2-6)\widehat{k}=-9\widehat{i}+7\widehat{j}-8\widehat{k}$

Report an Error

Example Question #123 : Vectors And Vector Operations

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=14\widehat{i}+5\widehat{j}$ and $\overrightarrow{b}=4\widehat{i}+3\widehat{j}$

Possible Answers:

$31\widehat{k}$

$22\widehat{k}$

$62\widehat{k}$

$71\widehat{k}$

Correct answer:

$22\widehat{k}$

Explanation:

The cross product of two vectors is a new vector. In order to find the cross product, it is useful to know the following relationships between the directional vectors:

$\widehat{i}\times\widehat{j}=\widehat{k};\widehat{j}\times\widehat{i}=-\widehat{k}$

$\widehat{j}\times\widehat{k}=\widehat{i};\widehat{k}\times\widehat{j}=-\widehat{i}$

$\widehat{k}\times\widehat{i}=\widehat{j};\widehat{i}\times\widehat{k}=-\widehat{j}$

$\widehat{i}\times\widehat{i}=\widehat{j}\times\widehat{j}=\widehat{k}\times\widehat{k}=0$

Note how the sign changes when terms are reordered; order matters!

Scalar values (the numerical coefficients) multiply through, e.g:

$r\widehat{i}\times s\widehat{j}=(rs)\widehat{k}$

With these principles in mind, we can calculate the cross product of our vectors $\overrightarrow{a}=14\widehat{i}+5\widehat{j}$ and $\overrightarrow{b}=4\widehat{i}+3\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+42-20+0)\widehat{k}=22\widehat{k}$

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Calculus 3 : Cross Product

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #11 : Cross Product

Example Question #12 : Cross Product

Example Question #12 : Cross Product

Example Question #14 : Cross Product

Example Question #15 : Cross Product

Example Question #13 : Cross Product

Example Question #17 : Cross Product

Example Question #121 : Vectors And Vector Operations

Example Question #122 : Vectors And Vector Operations

Example Question #123 : Vectors And Vector Operations

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