Chapter 6.1, Problem 11E

Solution

To find: The component form and magnitude of the vector.

The component form of vector $3\overrightarrow{QR}+\overrightarrow{PS}$ is $\langle -11,-7\rangle$ .

The magnitude form of vector is $\sqrt{170}$ .

Given information:

Let $P=\left(-2,2\right),Q\left(3,4\right),R\left(-2,5\right)$ and $S\left(2,-8\right)$ . Find the component form and magnitude of the vector $3\overrightarrow{QR}+\overrightarrow{PS}$ .

Concept used:

Let the initial side of the vector is $\left(x_1,y_1\right)$ and the terminal side of the vector be $\left(x_2,y_2\right)$ .

Then, the component form of the vector will be:

  $\langle \left(x_2-x_1,y_2-y_1\right)\rangle \text{or }\langle x,y\rangle$

The magnitude form of the vector will be:

  $\sqrt{x^2+y^2}$

Explanation:

The component form of vector $\overrightarrow{QR}$ will be:

  $\begin{array}{l}\langle \left(-2-3,5-4\right)\rangle \\ =\langle \left(-5,1\right)\rangle \end{array}$

The component form of $3\overrightarrow{QR}$ will be:

  $\begin{array}{l}3\langle \left(-5,1\right)\rangle \\ =\langle \left(-15,3\right)\rangle \end{array}$

The component form of vector $\overrightarrow{PS}$ will be:

  $\begin{array}{l}\langle 2-\left(-2\right),-8-2\rangle \\ =\langle 2+2,-10\rangle \\ =\langle 4.-10\rangle \end{array}$

The component form of vector $3\overrightarrow{QR}+\overrightarrow{PS}$ will be:

  $\begin{array}{l}\langle \left(-15,3\right)\rangle +\langle \left(4,-10\right)\rangle \\ =\langle -15+4,3-10\rangle \\ =\langle -11,-7\rangle \end{array}$

The magnitude form of vector $\parallel 3\overrightarrow{QR}+\overrightarrow{PS}\parallel$ will be:

  $\begin{array}{l}\sqrt{{\left(-11\right)}^2+{\left(-7\right)}^2}\\ =\sqrt{121+49}\\ =\sqrt{170}\end{array}$