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Q. 22

Expert-verified
Found in: Page 824

Calculus

Book edition 1st
Author(s) Peter Kohn, Laura Taalman
Pages 1155 pages
ISBN 9781429241861

In Exercises 22–29 compute the indicated quantities when $u=\left(2,1,-3\right),v=\left(4,0,1\right),\mathrm{and}w=\left(-2,6,5\right)$$u×v\mathrm{and}v×u$

The value of $u×v=1i+14j-4k$ and localid="1649397999598" $v×u=-1i-14j+4k$

See the step by step solution

Step 1. Given Information

In Exercises 22–29 compute the indicated quantities when$u=\left(2,1,-3\right),v=\left(4,0,1\right),\mathrm{and}w=\left(-2,6,5\right)$

We have to find the value of $u×v\mathrm{and}v×u$

Step 2. Firstly finding the value of u×v

The value of vectors $u=\left(2,1,-3\right),v=\left(4,0,1\right)$

The cross product of $u×v$

localid="1649398046712" $u×v=\mathrm{det}\left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 2& 1& -3\\ 4& 0& 1\end{array}\right]$

Step 3. Now solving the matrix.

$u×v=\mathrm{det}\left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 2& 1& -3\\ 4& 0& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}u×v=\left(\left(1\right)\left(1\right)-\left(-3\right)\left(0\right)\right)i+\left(\left(2\right)\left(1\right)-\left(-3\right)\left(4\right)\right)j+\left(\left(2\right)\left(0\right)-\left(1\right)\left(4\right)\right)k\phantom{\rule{0ex}{0ex}}u×v=\left(1+0\right)i+\left(2+12\right)j+\left(0-4\right)k\phantom{\rule{0ex}{0ex}}u×v=1i+14j-4k$

Step 4. Now finding the value of v×u

The value of vectors $u=\left(2,1,-3\right),v=\left(4,0,1\right)$

The cross product of $v×u=\mathrm{det}\left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 4& 0& 1\\ 2& 1& -3\end{array}\right]$

Step 5. Now solving the matrix.

$v×u=\mathrm{det}\left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 4& 0& 1\\ 2& 1& -3\end{array}\right]\phantom{\rule{0ex}{0ex}}v×u=\left(\left(0\right)\left(-3\right)-\left(1\right)\left(1\right)\right)i+\left(\left(4\right)\left(-3\right)-\left(1\right)\left(2\right)\right)j+\left(\left(4\right)\left(1\right)-\left(0\right)\left(2\right)\right)k\phantom{\rule{0ex}{0ex}}v×u=\left(0-1\right)i+\left(-12-2\right)j+\left(4-0\right)k\phantom{\rule{0ex}{0ex}}v×u=-1i-14j+4k$