Question

If the vectors $\vec{b}=\left( tan\alpha ,1,\sqrt{4\sin \dfrac{\alpha }{2}} \right)$ and $\vec{c}=\left( tan\alpha ,tan\alpha ,\dfrac{-3}{\sqrt{\sin \dfrac{\alpha }{2}}} \right)$ are orthogonal and a vector $\vec{a}=\left( 1,3,sin2\alpha \right)$ makes an obtuse angle with z – axis, then the value of $\alpha $ is
(a)$(4n+1)\pi -{{\tan }^{-1}}(2)$
(b)$(4n+2)\pi -{{\tan }^{-1}}(2)$
(c)$(4n-1)\pi +{{\tan }^{-1}}(2)$
(d)$(4n-2)\pi -{{\tan }^{-1}}(2)$

Answer 1

Hint: To solve this question, first we will write vector $\vec{b}=\left( tan\alpha ,1,\sqrt{4\sin \dfrac{\alpha }{2}} \right)$ and $\vec{c}=\left( tan\alpha ,tan\alpha ,\dfrac{-3}{\sqrt{\sin \dfrac{\alpha }{2}}} \right)$ in terms of vector and then using condition of orthogonality which is dot product of two orthogonal vectors is zero, we will evaluate the values of $\tan \alpha $and then for n = o, we will solve each option and eliminate them one by one unless we get correct option. Hence, we will get value of $\alpha $.

Complete step by step answer:
As, in question it is given that vector $\vec{b}=\left( tan\alpha ,1,\sqrt{4\sin \dfrac{\alpha }{2}} \right)$ and vector $\vec{c}=\left( tan\alpha ,tan\alpha ,\dfrac{-3}{\sqrt{\sin \dfrac{\alpha }{2}}} \right)$ are orthogonal, ten we know that dot product of two orthogonal vectors is zero.
So, $\vec{b}.\vec{c}=0$
Dot product is done as, if we have two vectors say $\vec{a}={{x}_{1}}\hat{i}+{{y}_{1}}\hat{j}+{{z}_{1}}\hat{k}$ and $\vec{b}={{x}_{2}}\hat{i}+{{y}_{2}}\hat{j}+{{z}_{2}}\hat{k}$, then
$\vec{a}.\vec{b}={{x}_{1}}{{x}_{2}}+{{y}_{1}}{{y}_{2}}+{{z}_{1}}{{z}_{2}}$
Writing, $\vec{b}=\left( tan\alpha ,1,\sqrt{4\sin \dfrac{\alpha }{2}} \right)$in vector form we get $\vec{b}=tan\alpha \hat{i}+1\hat{j}+\sqrt{4\sin \dfrac{\alpha }{2}}\hat{k}$ and writing $\vec{c}=\left( tan\alpha ,tan\alpha ,\dfrac{-3}{\sqrt{\sin \dfrac{\alpha }{2}}} \right)$ in vector form we get $\vec{c}=tan\alpha \hat{i}+tan\alpha \hat{j}-\dfrac{3}{\sqrt{\sin \dfrac{\alpha }{2}}}\hat{k}$
So, $\vec{b}.\vec{c}=0$
$\left( tan\alpha \hat{i}-1\hat{j}+\sqrt{4\sin \dfrac{\alpha }{2}}\hat{k} \right).\left( tan\alpha \hat{i}+tan\alpha \hat{j}-\dfrac{3}{\sqrt{\sin \dfrac{\alpha }{2}}}\hat{k} \right)=0$
Using definition of dot product, we get
\[tan\alpha \times tan\alpha +(-1)\times tan\alpha +\sqrt{4\sin \dfrac{\alpha }{2}}\times -\dfrac{3}{\sqrt{\sin \dfrac{\alpha }{2}}}=0\]
On simplifying, we get
\[ta{{n}^{2}}\alpha -tan\alpha -6=0\]
Using quadratic formula $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ , to solve for value of \[tan\alpha \], we get
\[tan\alpha =\dfrac{-(-1)\pm \sqrt{{{(-1)}^{2}}-4(1)(-6)}}{2(1)}\]
On simplifying, we get
\[tan\alpha =\dfrac{1\pm \sqrt{1+24}}{2}\]
\[tan\alpha =\dfrac{1\pm \sqrt{25}}{2}\]
Or, \[tan\alpha =\dfrac{1\pm 5}{2}\]
\[tan\alpha =2,-3\]
Now, we will eliminate option one by one by putting n = 0 and solving them further.
Now, if $\alpha =(4n+1)\pi -{{\tan }^{-1}}(2)$, then
For, n = 0, we have
$\alpha =\pi -{{\tan }^{-1}}(2)$
Taking tan on both sides, we have
$\tan \alpha =\tan [\pi -{{\tan }^{-1}}(2)]$
As we know that $\tan (\pi -\theta )=-\tan \theta $
So, $\tan \alpha =-\tan [{{\tan }^{-1}}(2)]$
Or, $\tan \alpha =-2$
Putting, $\tan \alpha =-2$ in \[ta{{n}^{2}}\alpha +tan\alpha -6=0\], we get
\[{{(-2)}^{2}}-2-6=-4\ne 0\]
So, option ( a ) is not true.
Now, if $\alpha =(4n+2)\pi -{{\tan }^{-1}}(2)$, then
For, n = 0, we have
$\alpha =2\pi -{{\tan }^{-1}}(2)$
Taking tan on both sides, we have
$\tan \alpha =\tan [2\pi -{{\tan }^{-1}}(2)]$
As we know that $\tan (2\pi -\theta )=-\tan \theta $
So, $\tan \alpha =-\tan [{{\tan }^{-1}}(2)]$
Or, $\tan \alpha =-2$
Putting, $\tan \alpha =-2$ in \[ta{{n}^{2}}\alpha +tan\alpha -6=0\], we get
\[{{(-2)}^{2}}-2-6=-4\ne 0\]
So, option ( b ) is not true.
Now, if $\alpha =(4n-1)\pi +{{\tan }^{-1}}(2)$, then
For, n = 0, we have
$\alpha =-\pi +{{\tan }^{-1}}(2)$
Taking tan on both sides, we have
$\tan \alpha =\tan [-\pi +{{\tan }^{-1}}(2)]$
As we know that $\tan (-\pi +\theta )=\tan \theta $
So, $\tan \alpha =\tan [{{\tan }^{-1}}(2)]$
Or, $\tan \alpha =2$
Putting, $\tan \alpha =2$ in \[ta{{n}^{2}}\alpha +tan\alpha -6=0\], we get
\[{{2}^{2}}+2-6=0\]
So, option ( c ) is true.
Now, if $\alpha =(4n-2)\pi -{{\tan }^{-1}}(2)$, then
For, n = 0, we have
$\alpha =-2\pi -{{\tan }^{-1}}(2)$
Taking tan on both sides, we have
$\tan \alpha =\tan [-2\pi -{{\tan }^{-1}}(2)]$
As we know that $\tan (-2\pi -\theta )=-\tan \theta $
So, $\tan \alpha =-\tan [{{\tan }^{-1}}(2)]$
Or, $\tan \alpha =-2$
Putting, $\tan \alpha =-2$ in \[ta{{n}^{2}}\alpha +tan\alpha -6=0\], we get
\[{{2}^{2}}-2-6=-4\ne 0\]
So, option ( d ) is not true.
Hence, the option ( c ) is true.

Note:
To solve this question, one must know the meaning of orthogonal vectors which is perpendicular vectors, and what is a condition of orthogonality in terms of dot product which is $\vec{a}.\vec{b}=0$, for vectors \[\vec{a}\]and $\vec{b}$. One must know the sign scheme of tangent function for different quadrants. While simplifying questions, try not to make calculation mistakes as this will change the option and make the answer wrong.