Question

Find a vector of magnitude $3$ in the direction opposite to the direction of $\vec v = \dfrac{1}{2}\hat i + \dfrac{1}{2}\hat j + \dfrac{1}{2}\hat k.$

Answer 1

Firstly, we will find mod of $\vec v$ then we will use the formula of unit vector to calculate $\hat v$.By using formula of unit vector $= \dfrac{{vector}}{{magnitude{\text{ of the vector}}}}$.

Complete step by step solution:
$\overrightarrow V = \dfrac{1}{2}\mathop i\limits^ \wedge - \dfrac{1}{2}\mathop f\limits^ \wedge - \dfrac{1}{2}\mathop K\limits^ \wedge$
Here,$\overrightarrow V = \dfrac{1}{2}i\dfrac{{ - 1}}{2}\mathop j\limits^ \wedge \dfrac{{ - 1}}{2}\mathop K\limits^ \wedge$
We will calculate:$\left| {\overrightarrow V } \right| = \left| {\dfrac{1}{2}\mathop i\limits^ \wedge - \dfrac{1}{2}\mathop j\limits^ \wedge - \dfrac{1}{2}\mathop K\limits^ \wedge } \right|$
$\left| {\overrightarrow V } \right| = \sqrt {{{\left( {\dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{ - 1}}{2}} \right)}^2} + {{\left( {\dfrac{{ - 1}}{2}} \right)}^2}}$

$$\left| {\overrightarrow V = \sqrt {\dfrac{1}{4} + \dfrac{1}{4} + \dfrac{1}{4}} } \right| \\\ \left| {\overrightarrow V } \right| = \sqrt {\dfrac{{1 + 1 + 1}}{4}} \\\ \left| {\overrightarrow V } \right| = \sqrt {\dfrac{3}{4}} \\\ \left| {\overrightarrow V } \right| = \dfrac{{\sqrt 3 }}{2} \\\$$

We will substitute the value of$\left| V \right|$in the formula as,
$\mathop V\limits^ \wedge = \dfrac{{ - \overrightarrow V }}{{\left| {\overrightarrow V } \right|}}$
$= \dfrac{{ - \left( { + \dfrac{1}{2}\mathop i\limits^ \wedge - \dfrac{1}{2}\mathop j\limits^ \wedge - \dfrac{1}{2}\mathop K\limits^ \wedge } \right)}}{{\dfrac{{\sqrt 3 }}{2}}} \\\ \dfrac{{ = + \left( { - \dfrac{1}{2}\mathop i\limits^ \wedge + \dfrac{1}{2}\mathop j\limits^ \wedge + \dfrac{1}{2}\mathop K\limits^ \wedge } \right)}}{{\dfrac{{\sqrt 3 }}{2}}} \\\$
Multiplying numerator and denominator by$2$, we have,
$\Rightarrow \mathop V\limits^ \wedge = \dfrac{{2\left( { - \dfrac{1}{2}\mathop j\limits^ \wedge + \dfrac{1}{2}\mathop j\limits^ \wedge + \dfrac{1}{2}R} \right)}}{{2\dfrac{{\sqrt 3 }}{2}}}$
$\Rightarrow \mathop V\limits^ \wedge = \dfrac{{\left( { - \mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop K\limits^ \wedge } \right)}}{{\sqrt 3 }}$
Multiply the unit vector by given magnitude$\left( 3 \right)$,
We have,
$\Rightarrow 32 = \dfrac{3}{{\sqrt 3 }}\left( { - \mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop K\limits^ \wedge } \right)$
$\Rightarrow 32 = \sqrt 3 \left( { - \mathop i\limits^ \wedge + \mathop j\limits^ \wedge + \mathop k\limits^ \wedge } \right)$

Additional information: The dot product between two vectors $a$ and $b$ is $a.b = ||a||||b||\cos \theta$, where $\theta$ is the angle between vectors $a\,\,and\,\,b$.

Note: Students should put the correct value in the formula for finding $\hat v$. In order to get the opposite direction $\vec v$ we need to get the direction of $\vec v$ first.