QJEEAdv10Maf287: JEE Advanced 2010 Mathematics - Class 12 JEEAdv10Maf287 Solution: Match the statement in Column- with the values in Column-…

(A) Equation of the line passing through origin is

${x \over a} = {y \over b} = {z \over c}$

$\therefore$ $$\left|\begin{array}{ccc}2& 1& -1\\ 1& -2& 1\\ a& b& c\end{array}\right|=0$$

$\Rightarrow a( - 1) - b(3) + c( - 5) = 0$

$\Rightarrow - a - 3b - 5c = 0$

$\Rightarrow a + 3b + 5c = 0$ ....... (i)

Also, $$\left|\begin{array}{ccc}\frac{8}{3}& -3& 1\\ 2& -1& 1\\ a& b& c\end{array}\right|=0$$

$\therefore$ $a( - 2) - b\left( {{2 \over 3}} \right) + c\left( {{{10} \over 3}} \right) = 0$

$\Rightarrow 2a + {{2b} \over 3} - {{10c} \over 3} = 0$

$3a + b - 5c = 0$ ...... (ii)

From Eqs. (i) and (ii),

${a \over { - 20}} = {b \over {20}} = {c \over { - 8}}$

${a \over 5} = {b \over { - 5}} = {c \over 4}$

Equation of line is

${x \over 5} = {y \over { - 5}} = {z \over 4} = \lambda$ (say) ....... (iii)

Also, ${{x - 2} \over 1} = {{y - 1} \over 2} = {{z + 1} \over 1} = {k_1}$ (say) ...... (iv)

Now, ${{x - {8 \over 3}} \over 2} = {{y + 3} \over { - 1}} = {{z - 1} \over 1} = {k_2}$ (say) ...... (v)

Point on (iii) is $(5\lambda ,\, - 5\lambda ,\, + 4\lambda )$

Point on (iv) is $(2 + {k_1},\,1 - 2{k_1},\, - 1 + {k_1})$

Point on (v) is $\left( {{8 \over 3} + 2{k_2},\, - 3 - {k_2},\,1 + {k_2}} \right)$

On solving, $2 + {k_1} + 1 - 2{k_1} = 0$

$- {k_1} + 3 = 0$

${k_1} = 3$

$P \equiv (5,\, - 5,\,2)$

Again, for Q

${8 \over 3} + 2{k_2} - 3 - {k_2} = 0$

${k_2} - {1 \over 3} = 0$

${k_2} = {1 \over 3}$

$Q \equiv \left( {{{10} \over 3},{{ - 10} \over 3},{4 \over 3}} \right)$

Now, $PQ = \sqrt {{{\left( {{5 \over 3}} \right)}^2} + {{\left( {{5 \over 3}} \right)}^2} + {{\left( {{2 \over 3}} \right)}^2}}$

$= {{\sqrt {54} } \over 3}$

$P{Q^2} = {d^2} = {{54} \over 9} = 6$

(B) ${\tan ^{ - 1}}\left( {{{x + 3 - x + 3} \over {1 + ({x^2} - 9)}}} \right) = {\tan ^{ - 1}}\left( {{3 \over 4}} \right)$

$\Rightarrow {6 \over {{x^2} - 8}} = {3 \over 4}$

$\Rightarrow 3{x^2} = 48$

$\Rightarrow x = \, \pm \,4$

(C) $\left( {\overrightarrow b - \overrightarrow a } \right)\,.\,\left( {\overrightarrow b + {{\overrightarrow a - \mu \overrightarrow b } \over 4}} \right) = 0$

$\Rightarrow \left( {\overrightarrow b - \overrightarrow a } \right)\,.\,\left( {4\overrightarrow b + \overrightarrow a - \mu \overrightarrow b } \right) = 0$

$\left( {4 - \mu } \right){\overrightarrow b ^2} - {\overrightarrow a ^2} = 0$ ...... (i)

Also, $2\left| {\overrightarrow b + {{\overrightarrow a - \mu \overrightarrow b } \over 4}} \right| = \left| {\overrightarrow b - \overrightarrow a } \right|$

$\Rightarrow 2\left| {{{(4 - \mu )\overrightarrow b + \overrightarrow a } \over 4}} \right| = \left| {\overrightarrow b - \overrightarrow a } \right|$