Calculus 3 : Cross Product

Study concepts, example questions & explanations for Calculus 3

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Example Questions

Example Question #377 : Calculus 3

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

Possible Answers:

\displaystyle 0

\displaystyle 12\widetilde{k}

\displaystyle 27\widetilde{k}

\displaystyle -12\widetilde{k}

\displaystyle -27\widetilde{k}

Correct answer:

\displaystyle 0

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:

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

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

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

\displaystyle \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:

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

With these principles in mind, we can calculate the cross product of our vectors \displaystyle \overrightarrow{a}=9\widehat{j} and \displaystyle \overrightarrow{b}=3\widehat{j}

\displaystyle \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.

Example Question #378 : Calculus 3

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

Possible Answers:

\displaystyle -40\widehat{k}

\displaystyle 20\widehat{k}

\displaystyle -8\widehat{k}

\displaystyle -20\widehat{k}

\displaystyle 44\widehat{k}

Correct answer:

\displaystyle -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:

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

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

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

\displaystyle \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:

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

With these principles in mind, we can calculate the cross product of our vectors \displaystyle \overrightarrow{a}=2\widehat{i}-6\widehat{j} and \displaystyle \overrightarrow{b}=\widehat{i}-7\widehat{j}

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

Example Question #121 : Vectors And Vector Operations

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

Possible Answers:

\displaystyle 3\widehat{k}

\displaystyle 5\widehat{k}

\displaystyle 2\widehat{k}

\displaystyle 7\widehat{k}

\displaystyle 0

Correct answer:

\displaystyle 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:

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

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

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

\displaystyle \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:

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

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

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

Example Question #380 : Calculus 3

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

Possible Answers:

\displaystyle -6\widehat{k}

\displaystyle -8\widehat{k}

\displaystyle 0

\displaystyle 8\widehat{k}

\displaystyle 6\widehat{k}

Correct answer:

\displaystyle -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:

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

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

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

\displaystyle \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:

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

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

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

Example Question #131 : Vectors And Vector Operations

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

Possible Answers:

\displaystyle 8\widehat{k}

\displaystyle 14\widehat{k}

\displaystyle -2\widehat{k}

\displaystyle 16\widehat{k}

\displaystyle 0

Correct answer:

\displaystyle 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:

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

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

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

\displaystyle \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:

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

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

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

Example Question #132 : Vectors And Vector Operations

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

Possible Answers:

\displaystyle 8\widehat{k}

\displaystyle 10\widehat{k}

\displaystyle -8\widehat{k}

\displaystyle -6\widehat{k}

\displaystyle 0

Correct answer:

\displaystyle 0

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:

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

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

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

\displaystyle \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:

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

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

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

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

Example Question #133 : Vectors And Vector Operations

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

Possible Answers:

\displaystyle 80\widehat{k}

\displaystyle 18\widehat{k}

\displaystyle -18\widehat{k}

\displaystyle -82\widehat{k}

\displaystyle 0

Correct answer:

\displaystyle -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:

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

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

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

\displaystyle \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:

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

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

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

Example Question #134 : Vectors And Vector Operations

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

Possible Answers:

\displaystyle 0

\displaystyle -150\widehat{k}

\displaystyle 42\widehat{k}

\displaystyle 72\widehat{k}

\displaystyle -144\widehat{k}

Correct answer:

\displaystyle 0

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:

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

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

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

\displaystyle \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:

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

With these principles in mind, we can calculate the cross product of our vectors \displaystyle \overrightarrow{a}=7\widehat{i}-\widehat{j} and \displaystyle \overrightarrow{b}=-21\widehat{i}+3\widehat{j}

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

These zero results means that the two vectors are parallel.

Example Question #135 : Vectors And Vector Operations

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

Possible Answers:

\displaystyle 32\widehat{k}

\displaystyle 50\widehat{k}

\displaystyle 18\widehat{k}

\displaystyle -14\widehat{k}

\displaystyle 0

Correct answer:

\displaystyle 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:

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

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

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

\displaystyle \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:

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

With these principles in mind, we can calculate the cross product of our vectors \displaystyle \overrightarrow{a}=3\widehat{i}+4\widehat{j} and \displaystyle \overrightarrow{b}=-8\widehat{i}+6\widehat{j}

\displaystyle \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!

Example Question #136 : Vectors And Vector Operations

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

Possible Answers:

\displaystyle -9\widehat{k}

\displaystyle 9\widehat{k}

\displaystyle -13\widehat{k}

\displaystyle 20\widehat{k}

\displaystyle 13\widehat{k}

Correct answer:

\displaystyle 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:

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

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

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

\displaystyle \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:

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

With these principles in mind, we can calculate the cross product of our vectors \displaystyle \overrightarrow{a}=2\widehat{i}+11\widehat{j} and \displaystyle \overrightarrow{b}=-\widehat{i}-\widehat{j}

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

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