JEE MAIN - Mathematics (2019 - 10th January Morning Slot - No. 4)
Let $$\overrightarrow a = 2\widehat i + {\lambda _1}\widehat j + 3\widehat k,\,\,$$ $$\overrightarrow b = 4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k,$$ and $$\overrightarrow c = 3\widehat i + 6\widehat j + \left( {{\lambda _3} - 1} \right)\widehat k$$ be three vectors such that $$\overrightarrow b = 2\overrightarrow a $$ and $$\overrightarrow a $$ is perpendicular to $$\overrightarrow c $$. Then a possible value of $$\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right)$$ is :
(1, 5, 1)
(1, 3, 1)
$$\left( { - {1 \over 2},4,0} \right)$$
$$\left( {{1 \over 2},4, - 2} \right)$$
Explanation
Given $$\overrightarrow b = 2\overrightarrow a $$
$$ \therefore $$ $$4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k = 4\widehat i + 2{\lambda _1}\widehat j + 6\widehat k$$
$$ \Rightarrow 3 - {\lambda _2} = 2{\lambda _1} \Rightarrow 2{\lambda _1} + {\lambda _2} = 3\,\,...(1)$$
Given $$\overrightarrow a $$ is perpendicular to $$\overrightarrow c $$
$$ \therefore $$ $$\overrightarrow a .\overrightarrow c = 0$$
$$ \Rightarrow 6 + 6{\lambda _1} + 3\left( {{\lambda _3} - 1} \right) = 0$$
$$ \Rightarrow 2{\lambda _1} + {\lambda _3} = - 1\,\,\,\,\,\,\,\,\,\,...(2)$$
Now $$\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right) = \left( {{\lambda _1},3 - 2{\lambda _1}, - 1 - 2{\lambda _1}} \right)$$
By checking each option you can see,
when $${\lambda _1}$$ = $$ - {1 \over 2}$$
then $${\lambda _2}$$ = $$3 - 2{\lambda _1}$$ = 3 + 1 = 4
and $${\lambda _3}$$ = $$-1 - 2{\lambda _1}$$ = - 1 + 1 = 0
$$ \therefore $$ $$4\widehat i + \left( {3 - {\lambda _2}} \right)\widehat j + 6\widehat k = 4\widehat i + 2{\lambda _1}\widehat j + 6\widehat k$$
$$ \Rightarrow 3 - {\lambda _2} = 2{\lambda _1} \Rightarrow 2{\lambda _1} + {\lambda _2} = 3\,\,...(1)$$
Given $$\overrightarrow a $$ is perpendicular to $$\overrightarrow c $$
$$ \therefore $$ $$\overrightarrow a .\overrightarrow c = 0$$
$$ \Rightarrow 6 + 6{\lambda _1} + 3\left( {{\lambda _3} - 1} \right) = 0$$
$$ \Rightarrow 2{\lambda _1} + {\lambda _3} = - 1\,\,\,\,\,\,\,\,\,\,...(2)$$
Now $$\left( {{\lambda _1},{\lambda _2},{\lambda _3}} \right) = \left( {{\lambda _1},3 - 2{\lambda _1}, - 1 - 2{\lambda _1}} \right)$$
By checking each option you can see,
when $${\lambda _1}$$ = $$ - {1 \over 2}$$
then $${\lambda _2}$$ = $$3 - 2{\lambda _1}$$ = 3 + 1 = 4
and $${\lambda _3}$$ = $$-1 - 2{\lambda _1}$$ = - 1 + 1 = 0
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