JEE MAIN - Mathematics (2021 - 16th March Evening Shift - No. 3)
Let $$\overrightarrow a $$ = $$\widehat i$$ + 2$$\widehat j$$ $$-$$ 3$$\widehat k$$ and $$\overrightarrow b = 2\widehat i$$ $$-$$ 3$$\widehat j$$ + 5$$\widehat k$$. If $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ $$\overrightarrow r $$,
$$\overrightarrow r $$ . $$\left( {\alpha \widehat i + 2\widehat j + \widehat k} \right)$$ = 3 and $$\overrightarrow r \,.\,\left( {2\widehat i + 5\widehat j - \alpha \widehat k} \right)$$ = $$-$$1, $$\alpha$$ $$\in$$ R, then the
value of $$\alpha$$ + $${\left| {\overrightarrow r } \right|^2}$$ is equal to :
$$\overrightarrow r $$ . $$\left( {\alpha \widehat i + 2\widehat j + \widehat k} \right)$$ = 3 and $$\overrightarrow r \,.\,\left( {2\widehat i + 5\widehat j - \alpha \widehat k} \right)$$ = $$-$$1, $$\alpha$$ $$\in$$ R, then the
value of $$\alpha$$ + $${\left| {\overrightarrow r } \right|^2}$$ is equal to :
13
11
9
15
Explanation
Given $$\overrightarrow r $$ $$\times$$ $$\overrightarrow a $$ = $$\overrightarrow b $$ $$\times$$ $$\overrightarrow r $$
$$ \Rightarrow $$ $$\overrightarrow r \times \overrightarrow a = - \overrightarrow r \times \overrightarrow b $$
$$\overrightarrow r \times (\overrightarrow a + \overrightarrow b ) = 0$$
$$\overrightarrow r ||(\overrightarrow a + \overrightarrow b )$$
$$\overrightarrow r = \lambda (\overrightarrow a + \overrightarrow b )$$
$$(\overrightarrow a + \overrightarrow b = 3\widehat i - \widehat j + 2\widehat k)$$
$$ \because $$ $$\overrightarrow r \,.\,(2\widehat i + 5\widehat j - \alpha \widehat k) = - 1$$
$$\lambda \left[ {3\widehat i - \widehat j + 2\widehat k} \right]\,.\,\left[ {2\widehat i + 5\widehat j - \alpha \widehat k} \right] = - 1$$
$$ \Rightarrow \lambda (6 - 5 - 2\alpha ) = - 1$$
$$\lambda (1 - 2\alpha ) = - 1$$ .... (1)
$$\overrightarrow r \,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3$$
$$\lambda (3\widehat i - \widehat j + 2\widehat k)\,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3$$
$$ \Rightarrow \lambda [3\alpha - 2 + 2] = 3 \Rightarrow \lambda \alpha = 1$$ .... (2)
From (1) & (2)
$$\lambda \left[ {1 - {2 \over \lambda }} \right] = - 1$$
$$\lambda - 2 = - 1 \Rightarrow \lambda = 1\,\alpha = 1$$
$$\overrightarrow r = 3\widehat i - \widehat j + 2\widehat k$$
$$\alpha + |\overrightarrow r {|^2} = 1 + 14 = 15$$
$$ \Rightarrow $$ $$\overrightarrow r \times \overrightarrow a = - \overrightarrow r \times \overrightarrow b $$
$$\overrightarrow r \times (\overrightarrow a + \overrightarrow b ) = 0$$
$$\overrightarrow r ||(\overrightarrow a + \overrightarrow b )$$
$$\overrightarrow r = \lambda (\overrightarrow a + \overrightarrow b )$$
$$(\overrightarrow a + \overrightarrow b = 3\widehat i - \widehat j + 2\widehat k)$$
$$ \because $$ $$\overrightarrow r \,.\,(2\widehat i + 5\widehat j - \alpha \widehat k) = - 1$$
$$\lambda \left[ {3\widehat i - \widehat j + 2\widehat k} \right]\,.\,\left[ {2\widehat i + 5\widehat j - \alpha \widehat k} \right] = - 1$$
$$ \Rightarrow \lambda (6 - 5 - 2\alpha ) = - 1$$
$$\lambda (1 - 2\alpha ) = - 1$$ .... (1)
$$\overrightarrow r \,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3$$
$$\lambda (3\widehat i - \widehat j + 2\widehat k)\,.\,(\alpha \widehat i + 2\widehat j + \widehat k) = 3$$
$$ \Rightarrow \lambda [3\alpha - 2 + 2] = 3 \Rightarrow \lambda \alpha = 1$$ .... (2)
From (1) & (2)
$$\lambda \left[ {1 - {2 \over \lambda }} \right] = - 1$$
$$\lambda - 2 = - 1 \Rightarrow \lambda = 1\,\alpha = 1$$
$$\overrightarrow r = 3\widehat i - \widehat j + 2\widehat k$$
$$\alpha + |\overrightarrow r {|^2} = 1 + 14 = 15$$
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