JEE MAIN - Mathematics (2018 - 15th April Morning Slot - No. 13)
If $$\overrightarrow a ,\,\,\overrightarrow b ,$$ and $$\overrightarrow C $$ are unit vectors such that $$\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 ,$$ then $$\left| {\overrightarrow a \times \overrightarrow c } \right|$$ is equal to :
$${{\sqrt {15} } \over 4}$$
$${{1} \over {4}}$$
$${{15} \over {16}}$$
$${{\sqrt {15} } \over 16}$$
Explanation
Given,
$$\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 $$
$$ \Rightarrow $$ $$\overrightarrow a + 2\overrightarrow c = - 2\overrightarrow b $$
Squaring both sides,
$${\left| {\overrightarrow a } \right|^2} + 4\overrightarrow a .\overrightarrow c + 4{\left| {\overrightarrow c } \right|^2} = 4{\left| {\overrightarrow b } \right|^2}$$
$$ \Rightarrow $$ 1 + $$4\overrightarrow a .\overrightarrow c $$ + 4 = 4 [as $$\left| {\overrightarrow a } \right|^2$$ = $$\left| {\overrightarrow b } \right|^2$$ = $$\left| {\overrightarrow c } \right|^2$$ = 1]
$$ \Rightarrow $$ $$\overrightarrow a .\overrightarrow c = - {1 \over 4}$$
$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos \theta = - {1 \over 4}$$
$$ \therefore $$ $$\cos \theta $$ = $$ - {1 \over 4}$$
$$ \therefore $$ $$\sin ^2 \theta $$ = 1 - $$\cos ^2 \theta $$
= 1 - $$1 \over 16$$
= $$15\over 16$$
$$ \therefore $$ sin$$\theta $$ = $${{\sqrt {15} } \over 4}$$
$$ \therefore $$ $$\left| {\overrightarrow a \times \overrightarrow c } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \theta $$
= 1 . 1 . $${{\sqrt {15} } \over 4}$$
= $${{\sqrt {15} } \over 4}$$
$$\overrightarrow a + 2\overrightarrow b + 2\overrightarrow c = \overrightarrow 0 $$
$$ \Rightarrow $$ $$\overrightarrow a + 2\overrightarrow c = - 2\overrightarrow b $$
Squaring both sides,
$${\left| {\overrightarrow a } \right|^2} + 4\overrightarrow a .\overrightarrow c + 4{\left| {\overrightarrow c } \right|^2} = 4{\left| {\overrightarrow b } \right|^2}$$
$$ \Rightarrow $$ 1 + $$4\overrightarrow a .\overrightarrow c $$ + 4 = 4 [as $$\left| {\overrightarrow a } \right|^2$$ = $$\left| {\overrightarrow b } \right|^2$$ = $$\left| {\overrightarrow c } \right|^2$$ = 1]
$$ \Rightarrow $$ $$\overrightarrow a .\overrightarrow c = - {1 \over 4}$$
$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\cos \theta = - {1 \over 4}$$
$$ \therefore $$ $$\cos \theta $$ = $$ - {1 \over 4}$$
$$ \therefore $$ $$\sin ^2 \theta $$ = 1 - $$\cos ^2 \theta $$
= 1 - $$1 \over 16$$
= $$15\over 16$$
$$ \therefore $$ sin$$\theta $$ = $${{\sqrt {15} } \over 4}$$
$$ \therefore $$ $$\left| {\overrightarrow a \times \overrightarrow c } \right| = \left| {\overrightarrow a } \right|\left| {\overrightarrow c } \right|\sin \theta $$
= 1 . 1 . $${{\sqrt {15} } \over 4}$$
= $${{\sqrt {15} } \over 4}$$
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