Question

a=2i+3j-k , b=6i-2j+5k find a unit vector parallel to the resultant of these vectors? here should i use (±) to find unit vector?
if needed then why? i can't figure out this. please show me in diagram, where it is +ve and where it is -ve

Answer 1

Answer on Question #51713 – Math – Vector Calculus

$a = 2i + 3j - k$, $b = 6i - 2j + 5k$ find a unit vector parallel to the resultant of these vectors? here should I use $(\pm)$ to find unit vector? If needed then why? I can't figure out this. please show me in diagram, where it is +ve and where it is -ve

Solution:

The resultant vector is given by

$$\vec{v} = \vec{a} + \vec{b} = 2\vec{i} + 3\vec{j} - \vec{k} + 6\vec{i} - 2\vec{j} + 5\vec{k} = 8\vec{i} + \vec{j} + 4\vec{k}$$

The unit vector parallel to it is

$$\overrightarrow{v_u} = \pm \frac{\vec{v}}{|\vec{v}|} = \pm \frac{8\vec{i} + \vec{j} + 4\vec{k}}{\sqrt{64 + 1 + 16}} = \pm \frac{8\vec{i} + \vec{j} + 4\vec{k}}{\sqrt{81}} = \pm \frac{8\vec{i} + \vec{j} + 4\vec{k}}{9}$$

In the diagram below two red vectors are unit vectors $\overrightarrow{v_u}$, parallel to $\vec{v}$ (the black one in the diagram). The big point in the diagram denotes the origin. These two red vectors are both parallel to $\vec{v}$, but they have opposite directions. So it's more properly to use $(\pm)$.

Answer: $\overrightarrow{v_u} = \pm \frac{8\vec{i} + \vec{j} + 4\vec{k}}{9}$.

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