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

Hint: 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.