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Calculus 3 : Cross Product

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Example Questions

Example Question #371 : Calculus 3

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

Possible Answers:

$27\widetilde{k}$

$-27\widetilde{k}$

$-12\widetilde{k}$

$12\widetilde{k}$

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 $\overrightarrow{a}=9\widehat{j}$ and $\overrightarrow{b}=3\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(27(0))\widehat{k}=0$

It may go without saying that these two vectors are parallel (afterall, both go strictly in the j-direction), and so the cross product is zero.

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Example Question #22 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=2\widehat{i}-6\widehat{j}$ and $\overrightarrow{b}=\widehat{i}-7\widehat{j}$

Possible Answers:

$-20\widehat{k}$

$20\widehat{k}$

$44\widehat{k}$

$-8\widehat{k}$

$-40\widehat{k}$

Correct answer:

$-8\widehat{k}$

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 $\overrightarrow{a}=2\widehat{i}-6\widehat{j}$ and $\overrightarrow{b}=\widehat{i}-7\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0-14+6+0)\widehat{k}=-8\widehat{k}$

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Example Question #21 : Cross Product

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

Possible Answers:

$5\widehat{k}$

$2\widehat{k}$

$3\widehat{k}$

$7\widehat{k}$

Correct answer:

$5\widehat{k}$

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

$\overrightarrow{a}\times \overrightarrow{b}=(6-1)\widehat{k}=5\widehat{k}$

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Example Question #125 : Vectors And Vector Operations

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

Possible Answers:

$-8\widehat{k}$

$6\widehat{k}$

$-6\widehat{k}$

$8\widehat{k}$

Correct answer:

$-6\widehat{k}$

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

$\overrightarrow{a}\times \overrightarrow{b}=(-6+0)\widehat{k}=-6\widehat{k}$

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Example Question #21 : Cross Product

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

Possible Answers:

$8\widehat{k}$

$-2\widehat{k}$

$14\widehat{k}$

$16\widehat{k}$

Correct answer:

$8\widehat{k}$

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

$\overrightarrow{a}\times \overrightarrow{b}=(0+3+5+0)\widehat{k}=8\widehat{k}$

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Example Question #22 : Cross Product

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

Possible Answers:

$8\widehat{k}$

$-8\widehat{k}$

$-6\widehat{k}$

$10\widehat{k}$

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 $\overrightarrow{a}=\widehat{i}+2\widehat{j}$ and $\overrightarrow{b}=2\widehat{i}+4\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+4-4+0)\widehat{k}=0$

Note that the zero answer means that these two vectors are parallel!

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Example Question #23 : Cross Product

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

Possible Answers:

$-18\widehat{k}$

$-82\widehat{k}$

$18\widehat{k}$

$80\widehat{k}$

Correct answer:

$-82\widehat{k}$

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 $\overrightarrow{a}=\widehat{i}+9\widehat{j}$ and $\overrightarrow{b}=9\widehat{i}-\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0-1-81+0)\widehat{k}=-82\widehat{k}$

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Example Question #24 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=7\widehat{i}-\widehat{j}$ and $\overrightarrow{b}=-21\widehat{i}+3\widehat{j}$ .

Possible Answers:

$-144\widehat{k}$

$72\widehat{k}$

$-150\widehat{k}$

$42\widehat{k}$

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 $\overrightarrow{a}=7\widehat{i}-\widehat{j}$ and $\overrightarrow{b}=-21\widehat{i}+3\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+21-21+0)\widehat{k}=0$

These zero results means that the two vectors are parallel.

Report an Error

Example Question #27 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=3\widehat{i}+4\widehat{j}$ and $\overrightarrow{b}=-8\widehat{i}+6\widehat{j}$ .

Possible Answers:

$50\widehat{k}$

$18\widehat{k}$

$32\widehat{k}$

$-14\widehat{k}$

Correct answer:

$50\widehat{k}$

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 $\overrightarrow{a}=3\widehat{i}+4\widehat{j}$ and $\overrightarrow{b}=-8\widehat{i}+6\widehat{j}$

$\overrightarrow{a}\times \overrightarrow{b}=(0+18+32+0)\widehat{k}=50\widehat{k}$

Take note that if you were to find the magnitude of each vector (5 and 10) respectively and found their product, it'd be the same as the absolute value of the cross product. This equivalence indicates that the two vectors are perpendicular!

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Example Question #25 : Cross Product

Determine the cross product $\overrightarrow{a}\times\overrightarrow{b}$ , if $\overrightarrow{a}=2\widehat{i}+11\widehat{j}$ and $\overrightarrow{b}=-\widehat{i}-\widehat{j}$ .

Possible Answers:

$13\widehat{k}$

$9\widehat{k}$

$20\widehat{k}$

$-9\widehat{k}$

$-13\widehat{k}$

Correct answer:

$9\widehat{k}$

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

$\overrightarrow{a}\times \overrightarrow{b}=(0-2+11+0)\widehat{k}=9\widehat{k}$

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Calculus 3 : Cross Product

Study concepts, example questions & explanations for Calculus 3

All Calculus 3 Resources

Example Questions

Example Question #371 : Calculus 3

Example Question #22 : Cross Product

Example Question #21 : Cross Product

Example Question #125 : Vectors And Vector Operations

Example Question #21 : Cross Product

Example Question #22 : Cross Product

Example Question #23 : Cross Product

Example Question #24 : Cross Product

Example Question #27 : Cross Product

Example Question #25 : Cross Product

All Calculus 3 Resources

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