# The vector \vec v starts at the point (-1,-6) and ends at the point (5,-5) a. What is the...

## Question:

The vector {eq}\vec v {/eq} starts at the point {eq}(-1,-6) {/eq} and ends at the point {eq}(5,-5) {/eq}

a. What is the component form of this vector?

b. Compute the magnitude,

c. The unit vector {eq}\vec u {/eq} parallel to {eq}\vec v, {/eq} but in the opposite direction has what component form?

## Magnitudes and Unit Vectors Over {eq}\mathbb{R}^2 {/eq}:

Suppose that {eq}\vec{v} {/eq} is a vector over {eq}\mathbb{R}^2 {/eq}. If {eq}\vec{v}=\left<a,b\right> {/eq} in component form, then the magnitude of {eq}\vec{v} {/eq} is the vector {eq}\sqrt{a^2+b^2} {/eq}.

Geometrically, the magnitude of a vector gives the length of the vector. A unit vector is a vector whose magnitude is 1.

a. If {eq}\vec{v} {/eq} starts at {eq}(-1,-6) {/eq} and ends at {eq}(5,-5) {/eq}, then the component form of {eq}\vec{v} {/eq} is:

{eq}\begin{align*} \vec{v}&=\left<5-(-1),(-5)-(-6)\right>\\ &=\left<6, 1\right>\, . \end{align*} {/eq}

That is, the component form of {eq}\vec{v} {/eq} is {eq}\boxed{\vec{v}=\left<6,1\right>}\, {/eq}.

b. The magnitude of {eq}\vec{v}=\left<6,1\right> {/eq} is:

{eq}\begin{align*} \|\vec{v}\|&=\sqrt{6^2+1^2}\\ &=\sqrt{37} \, . \end{align*} {/eq}

That is, {eq}\boxed{\|\vec{v}\|=\sqrt{37} \, .} {/eq}

c. To find the unit vector pointing in the same direction as {eq}\vec{v}=\left<6,1\right> {/eq}, we divide {eq}\vec{v} {/eq} by its magnitude. So this vector is:

{eq}\begin{align*} \frac{\|v\|}{\sqrt{37}}&=\frac{1}{\sqrt{37}}\left<6,1\right>\\ &=\left<\frac{6}{\sqrt{37}},\frac{1}{\sqrt{37}}\right> \, . \end{align*} {/eq}

The unit vector parallel to {eq}\vec{v} {/eq} but pointing in the opposite direction is the negation of this vector, which is {eq}\boxed{\left<-\frac{6}{\sqrt{37}},-\frac{1}{\sqrt{37}}\right>}\, {/eq}.