Question

A vector $\overset{\to }{\mathop{R}}\,$ can be expressed as linear combination of a vector $\overset{\to }{\mathop{A}}\,$ and another vector perpendicular to $\overset{\to }{\mathop{A}}\,$ and coplanar with and $\overset{\to }{\mathop{R}}\,$and$\overset{\to }{\mathop{A}}\,$as, $\begin{aligned} & \overset{\to }{\mathop{R}}\,=\left( \dfrac{\overset{\to }{\mathop{R}}\,.\overset{\to }{\mathop{A}}\,}{\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{A}}\,} \right)\overset{\to }{\mathop{A}}\,-\left( \dfrac{1}{\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{A}}\,} \right)\left( \overset{\to }{\mathop{A}}\,\times \left( \overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{R}}\, \right) \right) \\\ & \\\ \end{aligned}$
Reason:
If$\overset{\to }{\mathop{a}}\,$,$\overset{\to }{\mathop{b}}\,$ , $\overset{\to }{\mathop{c}}\,$ are coplanar then $\overset{\to }{\mathop{a}}\,$+$\overset{\to }{\mathop{b}}\,$,$\overset{\to }{\mathop{b}}\,$+$\overset{\to }{\mathop{c}}\,$,$\overset{\to }{\mathop{c}}\,$+$\overset{\to }{\mathop{a}}\,$are also coplanar
(a) Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
(b) Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
(c) Assertion is correct but Reason is incorrect.
(d) Assertion is incorrect but Reason is correct.

Answer 1

First we will simplify the given vector expression then we will check if it is valid or not. For solving this you must know dot product and cross product of vectors. $\overset{\to }{\mathop{A}}\,$.$\overset{\to }{\mathop{B}}\,$=$|A|\,|B|\,\cos \theta$ and cross product is $\overset{\to }{\mathop{A}}\,\times$ $\overset{\to }{\mathop{B}}\,$=$|A|\,|B|\,\sin \theta$. Cross product of a vector with itself is zero

Complete step-by-step solution:
First simplify vector $\overset{\to }{\mathop{R}}\,$
$\begin{aligned} & \overset{\to }{\mathop{R}}\,=\left( \dfrac{\overset{\to }{\mathop{R}}\,.\overset{\to }{\mathop{A}}\,}{\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{A}}\,} \right)\overset{\to }{\mathop{A}}\,-\left( \dfrac{1}{\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{A}}\,} \right)\left( \overset{\to }{\mathop{A}}\,\times \left( \overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{R}}\, \right) \right) \\\ & \\\ \end{aligned}$
We know that, $\overset{\to }{\mathop{A}}\,$.$\overset{\to }{\mathop{B}}\,$=$|A|\,|B|\,\cos \theta$
Therefore, $\overset{\to }{\mathop{A}}\,$.$\overset{\to }{\mathop{A}}\,$becomes ${{A}^{2}}\cos (0)={{A}^{2}}$……(1)
Also, $\overset{\to }{\mathop{R}}\,$.$\overset{\to }{\mathop{A}}\,$will become $|A|\,|R|\,\cos \theta$
Now, when we take cross product, i.e., $\overset{\to }{\mathop{A}}\,\times$ $\overset{\to }{\mathop{B}}\,$=$|A|\,|B|\,\sin \theta$
So, when we find $\left( \overset{\to }{\mathop{A}}\,\times \left( \overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{R}}\, \right) \right)$, we will get a vector which will be perpendicular to $\left( \overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{R}}\, \right)$
i.e. perpendicular to the plane of $\overset{\to }{\mathop{A}}\,$and $\overset{\to }{\mathop{R}}\,$
Let the vector perpendicular to the plane of $\overset{\to }{\mathop{A}}\,$and $\overset{\to }{\mathop{R}}\,$be of $\overset{\to }{\mathop{B}}\,$
So, the equation will become,
$\begin{aligned} & \left( \overset{\to }{\mathop{A}}\,\times \left( \overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{R}}\, \right) \right)=\left( \overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{B}}\, \right)=|A||B|\sin \theta =|A||B|\sin (90) \\\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=|A||B| \\\ \end{aligned}$
Now substitute all the values in the equation.
$\begin{aligned} & \overset{\to }{\mathop{R}}\,=\left( \dfrac{\overset{\to }{\mathop{R}}\,.\overset{\to }{\mathop{A}}\,}{\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{A}}\,} \right)\overset{\to }{\mathop{A}}\,-\left( \dfrac{1}{\overset{\to }{\mathop{A}}\,.\overset{\to }{\mathop{A}}\,} \right)\left( \overset{\to }{\mathop{A}}\,\times \left( \overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{R}}\, \right) \right) \\\ & \overset{\to }{\mathop{R}}\,=\left( \dfrac{\overset{\to }{\mathop{R}}\,.\overset{\to }{\mathop{A}}\,}{{{A}^{2}}} \right)\overset{\to }{\mathop{A}}\,-\left( \dfrac{1}{{{A}^{2}}} \right)\left( \overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{B}}\, \right) \\\ \end{aligned}$
Now, the equation will become,
$\begin{aligned} & =\left( \dfrac{\overset{\to }{\mathop{R}}\,.\overset{\to }{\mathop{A}}\,+{{A}^{2}}}{{{A}^{2}}} \right)-\left( \dfrac{\overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{B}}\,}{{{A}^{2}}} \right) \\\ & =\left( \dfrac{\overset{\to }{\mathop{R}}\,.\overset{\to }{\mathop{A}}\,}{{{A}^{2}}} \right)+1-\left( \dfrac{\overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{B}}\,}{{{A}^{2}}} \right) \\\ & =\left( \dfrac{\overset{\to }{\mathop{R}}\,.\overset{\to }{\mathop{A-\overset{\to }{\mathop{A}}\,\times \overset{\to }{\mathop{B}}\,}}\,}{{{A}^{2}}} \right) \\\ & =\left( \dfrac{|R|.|\overset{{}}{\mathop{A|\cos \theta -|A|\,|B|\sin \theta }}\,}{{{A}^{2}}} \right) \\\ & =|A|\left( \dfrac{|R|\cos \theta -|B|\sin \theta }{{{A}^{2}}} \right) \\\ \end{aligned}$
And according to the Reason when $\overset{\to }{\mathop{a}}\,$,$\overset{\to }{\mathop{b}}\,$ , $\overset{\to }{\mathop{c}}\,$are coplanar i.e. they lie in same plane then $\overset{\to }{\mathop{a}}\,$+$\overset{\to }{\mathop{b}}\,$,$\overset{\to }{\mathop{b}}\,$+$\overset{\to }{\mathop{c}}\,$,$\overset{\to }{\mathop{c}}\,$+$\overset{\to }{\mathop{a}}\,$will also be coplanar.
As, two linear vectors are only being added therefore the resultant vector will also lie in same plane.
Therefore, Reason is correct.
Hence, the correct option will be (d).

Note: You can most probably make mistakes with the dot and cross vector multiplication.
Remember that dot product is $\overset{\to }{\mathop{A}}\,$.$\overset{\to }{\mathop{B}}\,$=$|A|\,|B|\,\cos \theta$ and cross product is $\overset{\to }{\mathop{A}}\,\times$ $\overset{\to }{\mathop{B}}\,$=$|A|\,|B|\,\sin \theta$
Also, remember that cross product of a vector with itself is zero.