JEE Advance - Mathematics (2019 - Paper 2 Offline - No. 9)
Let $$\overrightarrow a = 2\widehat i + \widehat j - \widehat k$$ and $$\overrightarrow b = \widehat i + 2\widehat j + \widehat k$$ be two vectors. Consider a vector c = $$\alpha $$$$\overrightarrow a$$ + $$\beta $$$$\overrightarrow b$$, $$\alpha $$, $$\beta $$ $$ \in $$ R. If the projection of $$\overrightarrow c$$ on the vector ($$\overrightarrow a$$ + $$\overrightarrow b$$) is $$3\sqrt 2 $$, then the
minimum value of ($$\overrightarrow c$$ $$-$$($$\overrightarrow a$$ $$ \times $$ $$\overrightarrow b$$)).$$\overrightarrow c$$ equals ................
minimum value of ($$\overrightarrow c$$ $$-$$($$\overrightarrow a$$ $$ \times $$ $$\overrightarrow b$$)).$$\overrightarrow c$$ equals ................
Answer
18
Explanation
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
$$6(4 - 2\alpha + {\alpha^2}) = 6(3) = 18$$
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
$$6(4 - 2\alpha + {\alpha^2}) = 6(3) = 18$$
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