First, find the position vector.

[latex]\begin{align}\boldsymbol{u}&=\langle -2,-\left(-8\right),-5 - 1\rangle \\ &=\langle 6,-6\rangle \end{align}[/latex]

We use the Pythagorean Theorem to find the magnitude.

[latex]\begin{align}|\boldsymbol{u}|&=\sqrt{{\left(6\right)}^{2}+{\left(-6\right)}^{2}} \\ &=\sqrt{72} \\ &=6\sqrt{2} \end{align}[/latex]

The direction is given as

[latex]\begin{align}&\tan \theta =\frac{-6}{6}=-1\\ &\theta ={\tan }^{-1}\left(-1\right) =-45^\circ \end{align}[/latex]

However, the angle terminates in the fourth quadrant, so we add 360° to obtain a positive angle. Thus, [latex]-45^\circ +360^\circ =315^\circ[/latex].

Draw the vector [latex]\boldsymbol{v}[/latex] starting at initial [latex]\left(5,-3\right)[/latex] and terminal point [latex]\left(-1,2\right)[/latex]. Draw the vector [latex]\boldsymbol{u}[/latex] with initial point [latex]\left(-1,-3\right)[/latex] and terminal point [latex]\left(-7,2\right)[/latex]. Find the standard position for each.

Next, find and sketch the position vector for v and u. We have

[latex]\begin{align}\boldsymbol{v}&=\langle -1 - 5,2-\left(-3\right)\rangle \\ &=\langle -6,5\rangle \\ \text{ } \\ \boldsymbol{u}&=\langle -7-\left(-1\right),2-\left(-3\right)\rangle \\ &=\langle -6,5\rangle \end{align}[/latex]

Since the position vectors are the same, v and u are the same.

An alternative way to check for vector equality is to show that the magnitude and direction are the same for both vectors. To show that the magnitudes are equal, use the Pythagorean Theorem.

[latex]\begin{align}|\boldsymbol{v}|&=\sqrt{{\left(-1 - 5\right)}^{2}+{\left(2-\left(-3\right)\right)}^{2}} \\ &=\sqrt{{\left(-6\right)}^{2}+{\left(5\right)}^{2}} \\ &=\sqrt{36+25} \\ &=\sqrt{61} \\ |\boldsymbol{u}|&=\sqrt{{\left(-7-\left(-1\right)\right)}^{2}+{\left(2-\left(-3\right)\right)}^{2}} \\ &=\sqrt{{\left(-6\right)}^{2}+{\left(5\right)}^{2}} \\ &=\sqrt{36+25} \\ &=\sqrt{61} \end{align}[/latex]

As the magnitudes are equal, we now need to verify the direction. Using the tangent function with the position vector gives

[latex]\begin{align}&\tan \theta =-\frac{5}{6}\\ &\theta ={\tan }^{-1}\left(-\frac{5}{6}\right) =-39.8^\circ \end{align}[/latex]

However, we can see that the position vector terminates in the second quadrant, so we add [latex]180^\circ[/latex]. Thus, the direction is [latex]-39.8^\circ +180^\circ =140.2^\circ[/latex].