Suppose vectors \vec u and \vec v are given by \mathbb{R}^3, where \vec u = \langle ...

Question:

Suppose vectors {eq}\vec u {/eq} and {eq}\vec v {/eq} are given by {eq}\mathbb{R}^3 {/eq}, where {eq}\vec u = \langle 1,1,1\rangle, \vec v= \langle 4,-5,-3 \rangle {/eq}

a. Compute {eq}||\vec u- \vec v|| {/eq}

b. Find a unit vector parallel to {eq}\vec u - \vec v {/eq}

Unit Direction:

To determine a unit vector in the same direction as a given one:

we divide all its components by the norm of the vector, in this way, the resulting vector has the same direction and norm equal to 1.

a. Given the vectors {eq}\vec u = \langle 1,1,1\rangle, \vec v= \langle 4,-5,-3 \rangle {/eq}, first we need to compute the vector {eq}\vec u- \vec v {/eq}, that is: {eq}\vec u - \vec v = \left\langle { - 3,6,4} \right\rangle {/eq}. Now, we can calculate the norm of the vector:

{eq}\left\| {\vec u - \vec v} \right\| = \left\| {\left\langle { - 3,6,4} \right\rangle } \right\| = \sqrt {{{\left( { - 3} \right)}^2} + {6^2} + {4^2}} = \sqrt {61}. {/eq}

b. Given the vector, {eq}\vec u - \vec v = \left\langle { - 3,6,4} \right\rangle {/eq}, using the norm, we can write the unit vector as: {eq}\left\langle { - \frac{3}{{\sqrt {61} }},\frac{6}{{\sqrt {61} }},\frac{4}{{\sqrt {61} }}} \right\rangle. {/eq}