Question

1. what is the magnitude and direction of the $overrightarrow{pq}$ with tail and head points p(-6, 0) and q(2, 4)?

8.9 units, 26.6° north of east
13.4 units, 0° east
8.9 units, 63.4° north of east
13.4 units, 180° west

Explanation:

Step1: Find vector components

The vector $\overrightarrow{PQ}$ has $x$-component $x = 2-(-6)=8$ and $y$-component $y = 4 - 0=4$.

Step2: Calculate magnitude

The magnitude $|\overrightarrow{PQ}|$ of the vector is given by the formula $|\overrightarrow{PQ}|=\sqrt{x^{2}+y^{2}}=\sqrt{8^{2}+4^{2}}=\sqrt{64 + 16}=\sqrt{80}\approx8.9$ units.

Step3: Calculate direction

The direction $\theta$ of the vector is given by $\tan\theta=\frac{y}{x}=\frac{4}{8}=0.5$. So, $\theta=\arctan(0.5)\approx26.6^{\circ}$ north of east.

Answer:

8.9 units, 26.6° north of east