• :00Days
• :00Hours
• :00Mins
• 00Seconds
A new era for learning is coming soon Suggested languages for you:

Europe

Answers without the blur. Sign up and see all textbooks for free! Q. 23

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×w\mathrm{and}w×u$

The value of $u×w=23i+4j+14k$ and $w×u=-23i-4j-14k$.

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 $u×w\mathrm{and}w×u$

## Step 2. Firstly finding the value of u×w

The value of vectors $u=\left(2,1,-3\right),w=\left(-2,6,5\right)$

The cross product of $u×w=\mathrm{det}\left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 2& 1& -3\\ -2& 6& 5\end{array}\right]$

## Step 3. Now solving the matrix.

$u×w=\mathrm{det}\left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ 2& 1& -3\\ -2& 6& 5\end{array}\right]\phantom{\rule{0ex}{0ex}}u×w=\left(\left(1\right)\left(5\right)-\left(-3\right)\left(6\right)\right)i+\left(\left(2\right)\left(5\right)-\left(-2\right)\left(-3\right)\right)j+\left(\left(2\right)\left(6\right)-\left(1\right)\left(-2\right)\right)k\phantom{\rule{0ex}{0ex}}u×w=\left(5+18\right)i+\left(10-6\right)j+\left(12+2\right)k\phantom{\rule{0ex}{0ex}}u×w=23i+4j+14k$

## Step 4. Firstly finding the value of w×u

The value of vectors $w=\left(-2,6,5\right)\mathrm{and}\mathrm{u}=\left(2,1,-3\right)$

The cross product of $w×u=\mathrm{det}\left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ -2& 6& 5\\ 2& 1& -3\end{array}\right]$

## Step 5. Now solving the matrix.

$w×u=\mathrm{det}\left[\begin{array}{ccc}\mathrm{i}& \mathrm{j}& \mathrm{k}\\ -2& 6& 5\\ 2& 1& -3\end{array}\right]\phantom{\rule{0ex}{0ex}}w×u=\left(\left(6\right)\left(-3\right)-\left(5\right)\left(1\right)\right)i+\left(\left(-2\right)\left(-3\right)-\left(5\right)\left(2\right)\right)j+\left(\left(-2\right)\left(1\right)-\left(6\right)\left(2\right)\right)k\phantom{\rule{0ex}{0ex}}w×u=\left(-18-5\right)i+\left(6-10\right)j+\left(-2-12\right)k\phantom{\rule{0ex}{0ex}}w×u=-23i-4j-14k$ ### Want to see more solutions like these? 