Question

A vector of magnitude 2 along a bisector of the angle between the two vectors $2\overset{\hat{\ }}{\mathop{i}}\,-2\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,$ and $\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,-2\overset{\hat{\ }}{\mathop{k}}\,$ is
(A) $\dfrac{2}{\sqrt{10}}\left( 3\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{k}}\, \right)$
(B) $\dfrac{1}{\sqrt{26}}\left( \overset{\hat{\ }}{\mathop{i}}\,-4\overset{\hat{\ }}{\mathop{j}}\,+3\overset{\hat{\ }}{\mathop{k}}\, \right)$
(C) $\dfrac{2}{\sqrt{26}}\left( \overset{\hat{\ }}{\mathop{i}}\,-4\overset{\hat{\ }}{\mathop{j}}\,+3\overset{\hat{\ }}{\mathop{k}}\, \right)$
(D) None of these

Answer 1

We are given two vectors and we are asked to find the vector with a magnitude 2 which passes along the angle bisector of the angle of the given two vectors. So, we will have an internal vector and an external vector. The vector that passes through bisector of the angle is obtained by adding and subtracting the unit vectors of the given vectors and we get the two vectors as, $\dfrac{1}{3}\left( 3\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{k}}\, \right)and\dfrac{1}{3}\left( \overset{\hat{\ }}{\mathop{i}}\,-4\overset{\hat{\ }}{\mathop{j}}\,+3\overset{\hat{\ }}{\mathop{k}}\, \right)$. After deriving the vectors, we will first convert them to unit vectors and then we will multiply the obtained vectors by 2 so that we can get the required vectors with magnitude 2 and we will have a pair of vectors.

Complete step-by-step solution:
According to the given question, we are given two vectors and we have to find the vector of magnitude 2 which passes along the angle bisector of the given vectors.
Let us assume that the vectors be,
$\overset{\to }{\mathop{a}}\,=~2\overset{\hat{\ }}{\mathop{i}}\,-2\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,$ and $\overset{\to }{\mathop{b}}\,=\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,-2\overset{\hat{\ }}{\mathop{k}}\,$
And so we have to find the vectors passing through the angle bisectors of the given vectors. There are chances of two vectors passing through the angle bisector, one is the internal and the other is the external bisector.
Let us assume that the bisector vector be,
$\overset{\to }{\mathop{c}}\,=\overset{\hat{\ }}{\mathop{a}}\,\pm \overset{\hat{\ }}{\mathop{b}}\,$
That is, we can get the required angle bisectors by adding and subtracting the unit vectors of the given vectors. We have,
$\Rightarrow \overset{\to }{\mathop{c}}\,=\dfrac{\overset{\to }{\mathop{a}}\,}{|\overset{\to }{\mathop{a}}\,|}\pm \dfrac{\overset{\to }{\mathop{b}}\,}{|\overset{\to }{\mathop{b}}\,|}$
Substituting the values, we get,
$\Rightarrow \dfrac{2\overset{\hat{\ }}{\mathop{i}}\,-2\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,}{\sqrt{{{2}^{2}}+{{2}^{2}}+{{1}^{2}}}}\pm \dfrac{\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,-2\overset{\hat{\ }}{\mathop{k}}\,}{\sqrt{{{1}^{2}}+{{2}^{2}}+{{2}^{2}}}}$
$\Rightarrow \dfrac{2\overset{\hat{\ }}{\mathop{i}}\,-2\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,}{\sqrt{9}}\pm \dfrac{\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,-2\overset{\hat{\ }}{\mathop{k}}\,}{\sqrt{9}}$
$\Rightarrow \dfrac{2\overset{\hat{\ }}{\mathop{i}}\,-2\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,}{3}\pm \dfrac{\overset{\hat{\ }}{\mathop{i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,-2\overset{\hat{\ }}{\mathop{k}}\,}{3}$
Taking the common terms, we get,
$\Rightarrow \dfrac{1}{3}\left( (2\overset{\hat{\ }}{\mathop{i}}\,-2\overset{\hat{\ }}{\mathop{j}}\,+\overset{\hat{\ }}{\mathop{k}}\,)\pm \overset{\hat{\ }}{\mathop{(i}}\,+2\overset{\hat{\ }}{\mathop{j}}\,-2\overset{\hat{\ }}{\mathop{k}}\,) \right)$
$\Rightarrow \dfrac{1}{3}\left( 3\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{k}}\, \right)and\dfrac{1}{3}\left( \overset{\hat{\ }}{\mathop{i}}\,-4\overset{\hat{\ }}{\mathop{j}}\,+3\overset{\hat{\ }}{\mathop{k}}\, \right)$
So, we get the two vectors as,
$\overset{\to }{\mathop{c}}\,=\dfrac{1}{3}\left( 3\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{k}}\, \right)and\dfrac{1}{3}\left( \overset{\hat{\ }}{\mathop{i}}\,-4\overset{\hat{\ }}{\mathop{j}}\,+3\overset{\hat{\ }}{\mathop{k}}\, \right)$
Now, we will convert the obtained vectors into unit vectors, we get,
$\overset{\hat{\ }}{\mathop{c}}\,=\dfrac{\dfrac{1}{3}\left( 3\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{k}}\, \right)}{\sqrt{\dfrac{1}{9}\left( {{3}^{2}}+1 \right)}}and\dfrac{\dfrac{1}{3}\left( \overset{\hat{\ }}{\mathop{i}}\,-4\overset{\hat{\ }}{\mathop{j}}\,+3\overset{\hat{\ }}{\mathop{k}}\, \right)}{\sqrt{\dfrac{1}{9}\left( 1+{{(-4)}^{2}}+{{3}^{2}} \right)}}$
$\Rightarrow \overset{\hat{\ }}{\mathop{c}}\,=\dfrac{\dfrac{1}{3}\left( 3\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{k}}\, \right)}{\sqrt{\dfrac{1}{9}\left( 10 \right)}}and\dfrac{\dfrac{1}{3}\left( \overset{\hat{\ }}{\mathop{i}}\,-4\overset{\hat{\ }}{\mathop{j}}\,+3\overset{\hat{\ }}{\mathop{k}}\, \right)}{\sqrt{\dfrac{1}{9}\left( 26 \right)}}$
$\Rightarrow \overset{\hat{\ }}{\mathop{c}}\,=\dfrac{\dfrac{1}{3}\left( 3\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{k}}\, \right)}{\dfrac{1}{3}\sqrt{10}}and\dfrac{\dfrac{1}{3}\left( \overset{\hat{\ }}{\mathop{i}}\,-4\overset{\hat{\ }}{\mathop{j}}\,+3\overset{\hat{\ }}{\mathop{k}}\, \right)}{\dfrac{1}{3}\sqrt{26}}$
We will now cancel out the similar terms, we get,
$\Rightarrow \overset{\hat{\ }}{\mathop{c}}\,=\dfrac{\left( 3\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{k}}\, \right)}{\sqrt{10}}and\dfrac{\left( \overset{\hat{\ }}{\mathop{i}}\,-4\overset{\hat{\ }}{\mathop{j}}\,+3\overset{\hat{\ }}{\mathop{k}}\, \right)}{\sqrt{26}}$
Now, we will multiply the above unit vectors by 2 so that we get the magnitude 2, so we have,
$\overset{\hat{\ }}{\mathop{c}}\,=\dfrac{2}{\sqrt{10}}\left( 3\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{k}}\, \right)and\dfrac{2}{\sqrt{26}}\left( \overset{\hat{\ }}{\mathop{i}}\,-4\overset{\hat{\ }}{\mathop{j}}\,+3\overset{\hat{\ }}{\mathop{k}}\, \right)$
Therefore, the correct answers are (A) $\dfrac{2}{\sqrt{10}}\left( 3\overset{\hat{\ }}{\mathop{i}}\,-\overset{\hat{\ }}{\mathop{k}}\, \right)$ and (C) $\dfrac{2}{\sqrt{26}}\left( \overset{\hat{\ }}{\mathop{i}}\,-4\overset{\hat{\ }}{\mathop{j}}\,+3\overset{\hat{\ }}{\mathop{k}}\, \right)$.

Note: The conversion to unit vectors is important else the vectors will not be the standard one or the required one. And so that the obtained vectors have unit direction and unit magnitude. And do not forget to make the magnitude of the obtained vectors as 2 as it is asked in the question, else the entire solution will not be of any use.