JEE MAIN - Mathematics (2018 - 16th April Morning Slot - No. 18)
Let $$\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow c = \widehat j - \widehat k$$ and a vector $$\overrightarrow b $$ be such that $$\overrightarrow a \times \overrightarrow b = \overrightarrow c $$ and $$\overrightarrow a .\overrightarrow b = 3.$$ Then $$\left| {\overrightarrow b } \right|$$ equals :
$${{11} \over 3}$$
$${{11} \over {\sqrt 3 }}$$
$$\sqrt {{{11} \over 3}} $$
$${{\sqrt {11} } \over 3}$$
Explanation
$$ \because $$ $$\overrightarrow a $$ $$=$$ $$\widehat i + \widehat j + \widehat k \Rightarrow \left| {\overrightarrow a } \right| = \sqrt 3 $$
& $$\overrightarrow c = \widehat j - \widehat k \Rightarrow \left| {\overrightarrow c } \right|\sqrt 2 $$
Now, $$\overrightarrow a $$ $$ \times $$ $$\overrightarrow b $$ = $$\overrightarrow c $$ (Given)
$$ \Rightarrow $$ $$\left| {\vec a} \right|\left| {\vec b} \right|\sin \theta = \left| {\overrightarrow c } \right|$$
$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \sqrt 2 $$
also $$\overrightarrow a .\overrightarrow b = 3$$
$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta = 3$$
Dividing [i] by [iii], we get
tan$$\theta $$ = $${{\sqrt 2 } \over 3}$$
$$ \therefore $$ sin$$\theta $$ $$=$$ $${{\sqrt 2 } \over {\sqrt {11} }}$$
Substituting value of sin$$\theta $$ in [i] we get
$$\sqrt 3 \left| {\overrightarrow b } \right|{{\sqrt 2 } \over {\sqrt {11} }} = \sqrt 2 $$
$$\left| {\overrightarrow b } \right| = {{\sqrt {11} } \over {\sqrt 3 }}$$
& $$\overrightarrow c = \widehat j - \widehat k \Rightarrow \left| {\overrightarrow c } \right|\sqrt 2 $$
Now, $$\overrightarrow a $$ $$ \times $$ $$\overrightarrow b $$ = $$\overrightarrow c $$ (Given)
$$ \Rightarrow $$ $$\left| {\vec a} \right|\left| {\vec b} \right|\sin \theta = \left| {\overrightarrow c } \right|$$
$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\sin \theta = \sqrt 2 $$
also $$\overrightarrow a .\overrightarrow b = 3$$
$$ \Rightarrow $$ $$\left| {\overrightarrow a } \right|\left| {\overrightarrow b } \right|\cos \theta = 3$$
Dividing [i] by [iii], we get
tan$$\theta $$ = $${{\sqrt 2 } \over 3}$$
$$ \therefore $$ sin$$\theta $$ $$=$$ $${{\sqrt 2 } \over {\sqrt {11} }}$$
Substituting value of sin$$\theta $$ in [i] we get
$$\sqrt 3 \left| {\overrightarrow b } \right|{{\sqrt 2 } \over {\sqrt {11} }} = \sqrt 2 $$
$$\left| {\overrightarrow b } \right| = {{\sqrt {11} } \over {\sqrt 3 }}$$
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