JEE MAIN - Mathematics (2004 - No. 47)
Let $$\overrightarrow a ,\overrightarrow b $$ and $$\overrightarrow c $$ be non-zero vectors such that $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a \,\,.$$ If $$\theta $$ is the acute angle between the vectors $${\overrightarrow b }$$ and $${\overrightarrow c },$$ then $$sin\theta $$ equals :
$${{2\sqrt 2 } \over 3}$$
$${{\sqrt 2 } \over 3}$$
$${2 \over 3}$$
$${1 \over 3}$$
Explanation
Given $$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
Clearly $$\overrightarrow a $$ and $$\overrightarrow b $$ are noncollinear
$$ \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow a = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
$$\therefore$$ $$\overrightarrow a .\overrightarrow c = 0$$
and $$ - \overrightarrow b .\overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right| \Rightarrow \cos \theta = {{ - 1} \over 3}$$
$$\therefore$$ $$\sin \theta = \sqrt {1 - {1 \over 9}} = {{2\sqrt 2 } \over 3}$$
$$\,\,\,\,\,\,\,\left[ {} \right.$$ $$\theta $$ is acute angle between $$\overrightarrow b $$ and $$\overrightarrow c $$ $$\left. {} \right]$$
Clearly $$\overrightarrow a $$ and $$\overrightarrow b $$ are noncollinear
$$ \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow a = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $$
$$\therefore$$ $$\overrightarrow a .\overrightarrow c = 0$$
and $$ - \overrightarrow b .\overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right| \Rightarrow \cos \theta = {{ - 1} \over 3}$$
$$\therefore$$ $$\sin \theta = \sqrt {1 - {1 \over 9}} = {{2\sqrt 2 } \over 3}$$
$$\,\,\,\,\,\,\,\left[ {} \right.$$ $$\theta $$ is acute angle between $$\overrightarrow b $$ and $$\overrightarrow c $$ $$\left. {} \right]$$
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