# Chapter 6 - Section 6.6 - Vectors - Exercise Set - Page 782: 64

The magnitude is 3 and the direction angle is $180{}^\circ$.

#### Work Step by Step

Simplify the provided vector as: \begin{align} & \mathbf{v=}\left( 7\mathbf{i}-3\mathbf{j} \right)-\left( 10\mathbf{i}-3\mathbf{j} \right) \\ & =\left( 7-10 \right)\mathbf{i}+\left( -3+3 \right)\mathbf{j} \\ & =-3\mathbf{i}+0\mathbf{j} \end{align} If $\mathbf{v}=a\mathbf{i}+b\mathbf{j}$ is a vector, then its magnitude is $\left\| \mathbf{v} \right\|=\sqrt{{{a}^{2}}+{{b}^{2}}}$. Use this definition of norm then, \begin{align} & \left\| \mathbf{v} \right\|=\sqrt{{{\left( -3 \right)}^{2}}+{{\left( 0 \right)}^{2}}} \\ & =\sqrt{9+0} \\ & =\sqrt{9} \\ & =3 \end{align} Hence, the magnitude of the vector $\mathbf{v}$ is $3$. If $\mathbf{v}=a\mathbf{i}+b\mathbf{j}$ is a vector then its direction angle $\theta$ is $\theta ={{\tan }^{-1}}\left( \frac{b}{a} \right)$.so, \begin{align} & \theta ={{\tan }^{-1}}\left( \frac{0}{-3} \right) \\ & ={{\tan }^{-1}}\left( -0 \right) \\ & =180{}^\circ \end{align}. Hence, the angle is $180{}^\circ$.

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