Chapter 12, Problem 19CLT

To determine

Find the coordinates of the point.

Answer to Problem 19CLT

  $u\times v$

  $=\boxed{\left[-18,-6,-14\right]}$

  $u\cdot v$ = $\boxed{-38}$

Explanation of Solution

Given information:

For the vectors $u=\left(2,-6,0\right)$ and $v=\left(-4,5,3\right)$ , find $u\cdot v$ and $u\times v$ .

Calculation:

Consider the vectors,

  $u=\left(2,-6,0\right)$

  $v=\left(-4,5,3\right)$

The standard form of vectors is.

  $\begin{array}{l}a=\left(x_1,y_1,z_1\right)\\ b=\left(x_2,y_2,z_2\right)\end{array}$

Then cross product of vector $a=\left(x_1,y_1,z_1\right)$ and $b=\left(x_2,y_2,z_2\right)$ will be,

  $\begin{array}{l}a\times b=\left|\begin{array}{ccc}i& j& k\\ x_1& y_1& z_1\\ x_2& y_2& z_2\end{array}\right|\\ =i\left[y_1\left(z_2\right)-y_2\left(z_1\right)\right]-j\left[x_1\left(z_2\right)-x_2\left(z_1\right)\right]+k\left[x_1\left(y_2\right)-x_2\left(y_1\right)\right]\end{array}$

  $\begin{array}{l}u\times v=\left|\begin{array}{ccc}i& j& k\\ 2& -6& 0\\ -4& 5& 3\end{array}\right|\\ =\left[\left(-6\right)\left(3\right)-\left(0\right)\left(5\right),\left(0\right)\left(-4\right)-\left(2\right)\left(3\right),\left(2\right)\left(5\right)-\left(-6\right)\left(-4\right)\right]\\ =\left[\left(-18-0\right),\left(0-6\right),\left(10-24\right)\right]\end{array}$

  $=\boxed{\left[-18,-6,-14\right]}$

Dot product of the vectors will be,

  $a\cdot b=x_1{x}_2+y_1{y}_2+z_1{z}_2$

  $u=\left(2,-6,0\right)$

  $v=\left(-4,5,3\right)$

Substitute the given values,

  $\begin{array}{l}u\cdot v=\left(2\right)\left(-4\right)+\left(-6\right)\left(5\right)+\left(0\right)\left(3\right)\\ =\left(-8\right)+\left(-30\right)+\left(0\right)\\ =\boxed{-38}\end{array}$

Hence $u\times v$ is $=\boxed{\left[-18,-6,-14\right]}$ and $u\cdot v$ = $\boxed{-38}$