QJEEAdv19Ma0d10: JEE Advanced 2019 Physics - Class 11 JEEAdv19Ma0d10 Solution: Let and be two vectors. Consider a vector c = + , , R. If the…

Given vectors $\overrightarrow a$ $= 2\widehat i + \widehat j - \widehat k$

and $\overrightarrow b = \widehat i + 2\widehat j + \widehat k$

So, $\overrightarrow a + \overrightarrow b = 3\widehat i + 3\widehat j \Rightarrow |\overrightarrow a + \overrightarrow b| = 3\sqrt 2$

Since, it is given that projection of $\overrightarrow c$ = $\alpha$ a + $\beta$ b on the vector ( $\overrightarrow a$ + $\overrightarrow b$ ) is $3\sqrt 2$ , then

${{(\overrightarrow a + \overrightarrow b ).\overrightarrow c } \over {|\overrightarrow a + \overrightarrow b|}} = 3\sqrt 2$

$\Rightarrow (\overrightarrow a + \overrightarrow b).(\alpha \overrightarrow a + \beta \overrightarrow b) = 18$

$\Rightarrow \alpha (\overrightarrow a.\overrightarrow a) + \beta (\overrightarrow a.\overrightarrow b) + \alpha (\overrightarrow b.\overrightarrow a) + \beta (\overrightarrow a.\overrightarrow b) = 18$

$\Rightarrow 6\alpha + 3\beta + 3\alpha + 6\beta = 18$

$\Rightarrow 9\alpha + 9\beta = 18 \Rightarrow (\alpha + \beta ) = 2$ .....(i)

Now, for minimum value of

( $\overrightarrow c$ $-$ ( $\overrightarrow a$ $\times$ $\overrightarrow b$ )). $\overrightarrow c$

$= (\alpha \overrightarrow a + \beta \overrightarrow b - (\overrightarrow a \times \overrightarrow b)).(\alpha \overrightarrow a + \beta \overrightarrow b)$

$= {\alpha ^2}(\overrightarrow a.\overrightarrow a) + \alpha \beta (\overrightarrow a.\overrightarrow b) + \alpha \beta (\overrightarrow a.\overrightarrow b) + {\beta ^2}(\overrightarrow b.\overrightarrow b)$

[ $\because$ ( $\overrightarrow a$ $\times$ $\overrightarrow b$ ) . $\overrightarrow a$ = 0 = ( $\overrightarrow a$ $\times$ $\overrightarrow b$ ) . $\overrightarrow b$ ]

$6{\alpha ^2} + 6\alpha \beta + 6{\beta ^2} = 6({\alpha ^2} + {\beta ^2} + \alpha \beta )$

$= 6\,[{(\alpha + \beta )^2} - \alpha \beta ] = 6\,[4 - \alpha \beta ]$

$= 6\,[4 - \alpha (2 - \alpha )]$

$= 6\,[4 - 2\alpha + {\alpha ^2}]$

Let f ( $\alpha$ ) = $4 - 2\alpha + {\alpha ^2}$

f′( $\alpha$ ) = –2 + 2 $\alpha$

At maximum and minimum f′( $\alpha$ ) = 0

$\Rightarrow$ –2 + 2 $\alpha$ $\Rightarrow$ $\alpha$ = 1

f′'( $\alpha$ ) = 2 (+ve)

Therefore, minimum value of $4 - 2\alpha + {\alpha ^2}$ is (4 – 2 + 1) = 3.

$\therefore$ The minimum value of

$6(4 - 2\alpha + {\alpha^2}) = 6(3) = 18$