JEE MAIN - Mathematics (2020 - 7th January Evening Slot - No. 15)
Let $$\overrightarrow a $$
, $$\overrightarrow b $$
and $$\overrightarrow c $$
be three unit vectors such that
$$\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 $$. If $$\lambda = \overrightarrow a .\vec b + \vec b.\overrightarrow c + \overrightarrow c .\overrightarrow a $$ and
$$\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a $$, then the ordered pair, $$\left( {\lambda ,\overrightarrow d } \right)$$ is equal to :
$$\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 $$. If $$\lambda = \overrightarrow a .\vec b + \vec b.\overrightarrow c + \overrightarrow c .\overrightarrow a $$ and
$$\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a $$, then the ordered pair, $$\left( {\lambda ,\overrightarrow d } \right)$$ is equal to :
$$\left( {{3 \over 2},3\overrightarrow a \times \overrightarrow c } \right)$$
$$\left( { - {3 \over 2},3\overrightarrow c \times \overrightarrow b } \right)$$
$$\left( { - {3 \over 2},3\overrightarrow a \times \overrightarrow b } \right)$$
$$\left( {{3 \over 2},3\overrightarrow b \times \overrightarrow c } \right)$$
Explanation
$$\overrightarrow a + \vec b + \overrightarrow c = \overrightarrow 0 $$
$$ \Rightarrow $$ $${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2}$$ = 0
$$ \Rightarrow $$ $${{{\left| {\overrightarrow a } \right|}^2}}$$ + $${{{\left| {\overrightarrow b } \right|}^2}}$$ + $${{{\left| {\overrightarrow c } \right|}^2}}$$ +
$${2\left( {\overrightarrow a .\overrightarrow b } \right)}$$ + $${2\left( {\overrightarrow b .\overrightarrow c } \right)}$$ + $${2\left( {\overrightarrow c .\overrightarrow {a} } \right)}$$ = 0
$$ \Rightarrow $$ 3 + $${2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right)}$$ = 0
$$ \Rightarrow $$ 3 + 2$$\lambda $$ = 0
$$ \Rightarrow $$ $$\lambda $$ = $$ - {3 \over 2}$$
$$\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a $$
= $${\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \left( { - \overrightarrow a - \overrightarrow b } \right) + \left( { - \overrightarrow a - \overrightarrow b } \right) \times \overrightarrow a }$$
= $${\overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow a + }$$ $${\overrightarrow b \times \overrightarrow b - \overrightarrow a \times \overrightarrow a - \overrightarrow b \times \overrightarrow a }$$
= $${\overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow b }$$
= $${3\left( {\overrightarrow a \times \overrightarrow b } \right)}$$
$$ \Rightarrow $$ $${\left| {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right|^2}$$ = 0
$$ \Rightarrow $$ $${{{\left| {\overrightarrow a } \right|}^2}}$$ + $${{{\left| {\overrightarrow b } \right|}^2}}$$ + $${{{\left| {\overrightarrow c } \right|}^2}}$$ +
$${2\left( {\overrightarrow a .\overrightarrow b } \right)}$$ + $${2\left( {\overrightarrow b .\overrightarrow c } \right)}$$ + $${2\left( {\overrightarrow c .\overrightarrow {a} } \right)}$$ = 0
$$ \Rightarrow $$ 3 + $${2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right)}$$ = 0
$$ \Rightarrow $$ 3 + 2$$\lambda $$ = 0
$$ \Rightarrow $$ $$\lambda $$ = $$ - {3 \over 2}$$
$$\overrightarrow d = \overrightarrow a \times \vec b + \vec b \times \overrightarrow c + \overrightarrow c \times \overrightarrow a $$
= $${\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \left( { - \overrightarrow a - \overrightarrow b } \right) + \left( { - \overrightarrow a - \overrightarrow b } \right) \times \overrightarrow a }$$
= $${\overrightarrow a \times \overrightarrow b - \overrightarrow b \times \overrightarrow a + }$$ $${\overrightarrow b \times \overrightarrow b - \overrightarrow a \times \overrightarrow a - \overrightarrow b \times \overrightarrow a }$$
= $${\overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow b }$$
= $${3\left( {\overrightarrow a \times \overrightarrow b } \right)}$$
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