QJEEM20Maafba: JEE Main 2020 Physics - Class 11 JEEM20Maafba Solution: Let the position vectors of points 'A' and 'B' be and ,…

Let, $\overrightarrow a$ = $\widehat i + \widehat j + \widehat k$

and $\overrightarrow b$ = $2\widehat i + \widehat j + 3\widehat k$
$\overrightarrow {OB} = \overrightarrow b$

$\overrightarrow {OP} = {{\overrightarrow a + \lambda \overrightarrow b } \over {1 + \lambda }}$

$\overrightarrow {OA} = \overrightarrow a$

$\therefore$ $\overrightarrow {OB} .\overrightarrow {OP} - 3{\left| {\overrightarrow {OA} \times \overrightarrow {OP} } \right|^2} = 6$

$\Rightarrow$ $\overrightarrow b .{{\left( {\overrightarrow a + \lambda \overrightarrow b } \right)} \over {1 + \lambda }} - 3\left| {\overrightarrow a \times {{\left( {\overrightarrow a + \lambda \overrightarrow b } \right)} \over {1 + \lambda }}} \right|$ = 6

$\Rightarrow$ ${{\overrightarrow b .\overrightarrow a + \lambda \left( {\overrightarrow b .\overrightarrow b } \right)} \over {1 + \lambda }} - 3\left| {{{\overrightarrow a \times \overrightarrow a + \lambda \left( {\overrightarrow a \times \overrightarrow b } \right)} \over {1 + \lambda }}} \right|$ = 6

[ ${\overrightarrow b .\overrightarrow a }$ = ( $2\widehat i + \widehat j + 3\widehat k$ ).( $\widehat i + \widehat j + \widehat k$ )

         = 2 + 1 + 3 = 6

${\overrightarrow b .\overrightarrow b }$ = ( $2\widehat i + \widehat j + 3\widehat k$ ).( $2\widehat i + \widehat j + 3\widehat k$ )

         = 4 + 1 + 9 = 14

$$\overrightarrow{a}\times \overrightarrow{b}=\left|\begin{array}{ccc}\hat{i}& \hat{j}& \hat{k}\\ 1& 1& 1\\ 2& 1& 3\end{array}\right|$$

= (3 - 1) ${\widehat i}$ - (3 - 2) ${\widehat j}$ + (1 - 2) ${\widehat k}$

= 2 ${\widehat i}$ - ${\widehat j}$ - ${\widehat k}$

$\therefore$ $\left| {\overrightarrow a \times \overrightarrow b } \right|$ = $\sqrt {{2^2} + {{\left( { - 1} \right)}^2} + {{\left( { - 1} \right)}^2}}$ = $\sqrt 6$ ]

$\Rightarrow$ ${{6 + 14\lambda } \over {1 + \lambda }} - 3{\left| {{{\lambda \left( {\sqrt 6 } \right)} \over {1 + \lambda }}} \right|^2}$ = 6

$\Rightarrow$ ${{6 + 14\lambda } \over {1 + \lambda }} - {{3{\lambda ^2} \times 6} \over {{{\left( {1 + \lambda } \right)}^2}}}$ = 6

$\Rightarrow$ (14 $\lambda$ + 6)( $\lambda$ + 1)—18 $\lambda$ ² = 6( $\lambda$ + 1)²

$\Rightarrow$ —4 $\lambda$ ² + 20 $\lambda$ + 6 = 6 $\lambda$ ² + 12 $\lambda$ + 6

$\Rightarrow$ 10 $\lambda$ ² — 8 $\lambda$ = 0

$\Rightarrow$ $\lambda$ (10 $\lambda$ — 8) = 0

As given $\lambda$ > 0

$\therefore$ $\lambda$ = ${8 \over {10}}$ = 0.8