JEE MAIN - Mathematics (2021 - 18th March Evening Shift - No. 5)
Let $$\overrightarrow a $$ and $$\overrightarrow b $$ be two non-zero vectors perpendicular to each other and $$|\overrightarrow a | = |\overrightarrow b |$$. If $$|\overrightarrow a \times \overrightarrow b | = |\overrightarrow a |$$, then the angle between the vectors $$\left( {\overrightarrow a + \overrightarrow b + \left( {\overrightarrow a \times \overrightarrow b } \right)} \right)$$ and $${\overrightarrow a }$$ is equal to :
$${\sin ^{ - 1}}\left( {{1 \over {\sqrt 6 }}} \right)$$
$${\cos ^{ - 1}}\left( {{1 \over {\sqrt 2 }}} \right)$$
$${\sin ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$$
$${\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$$
Explanation
$$\overrightarrow a $$ is perpendicular to $$\overrightarrow b $$
$$ \therefore $$ $$\overrightarrow a $$ . $$\overrightarrow b $$ = 0
Given, | $$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$ | = | $$\overrightarrow a $$ |
and | $$\overrightarrow a $$ | = | $$\overrightarrow b $$ |
$$ \therefore $$ | $$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$ | = | $$\overrightarrow a $$ | = | $$\overrightarrow b $$ | = k(assume)
Now, angle between $$\overrightarrow a $$ and ($$\overrightarrow a $$ + $$\overrightarrow b $$ + ($$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$))
$$\cos \theta = {{\overrightarrow a ((\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ))} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$$
$$ = {{|\overrightarrow a {|^2} + \,\overrightarrow a .\,\overrightarrow b + \overrightarrow a (\overrightarrow a \times \overrightarrow b )} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$$
[Note $$\overrightarrow a .(\overrightarrow a \times \overrightarrow b ) = [\overrightarrow a \overrightarrow a \overrightarrow b ] = 0$$]
$$ = {{|\overrightarrow a {|^2} + 0 + 0} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$$
Now, $$|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2}$$
$$ = |\overrightarrow a {|^2} + |\overrightarrow a {|^2} + |\overrightarrow a \times \overrightarrow b {|^2} + $$
$$2\overrightarrow a .\overrightarrow b + 2\overrightarrow b .(\overrightarrow a \times \overrightarrow b ) + 2\overrightarrow a .(\overrightarrow a \times \overrightarrow b )$$
$$ = {k^2} + {k^2} + {k^2}$$
$$ \therefore $$ $$|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2} = 3{k^2}$$
$$ \Rightarrow |\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )| = \sqrt 3 k$$
$$ \therefore $$ $$\cos \theta = {{{k^2}} \over {k(\sqrt 3 k)}} = {1 \over {\sqrt 3 }}$$
$$ \Rightarrow \theta = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$$
$$ \therefore $$ $$\overrightarrow a $$ . $$\overrightarrow b $$ = 0
Given, | $$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$ | = | $$\overrightarrow a $$ |
and | $$\overrightarrow a $$ | = | $$\overrightarrow b $$ |
$$ \therefore $$ | $$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$ | = | $$\overrightarrow a $$ | = | $$\overrightarrow b $$ | = k(assume)
Now, angle between $$\overrightarrow a $$ and ($$\overrightarrow a $$ + $$\overrightarrow b $$ + ($$\overrightarrow a $$ $$\times$$ $$\overrightarrow b $$))
$$\cos \theta = {{\overrightarrow a ((\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ))} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$$
$$ = {{|\overrightarrow a {|^2} + \,\overrightarrow a .\,\overrightarrow b + \overrightarrow a (\overrightarrow a \times \overrightarrow b )} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$$
[Note $$\overrightarrow a .(\overrightarrow a \times \overrightarrow b ) = [\overrightarrow a \overrightarrow a \overrightarrow b ] = 0$$]
$$ = {{|\overrightarrow a {|^2} + 0 + 0} \over {|\overrightarrow a |.|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )|}}$$
Now, $$|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2}$$
$$ = |\overrightarrow a {|^2} + |\overrightarrow a {|^2} + |\overrightarrow a \times \overrightarrow b {|^2} + $$
$$2\overrightarrow a .\overrightarrow b + 2\overrightarrow b .(\overrightarrow a \times \overrightarrow b ) + 2\overrightarrow a .(\overrightarrow a \times \overrightarrow b )$$
$$ = {k^2} + {k^2} + {k^2}$$
$$ \therefore $$ $$|\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b ){|^2} = 3{k^2}$$
$$ \Rightarrow |\overrightarrow a + \overrightarrow b + (\overrightarrow a \times \overrightarrow b )| = \sqrt 3 k$$
$$ \therefore $$ $$\cos \theta = {{{k^2}} \over {k(\sqrt 3 k)}} = {1 \over {\sqrt 3 }}$$
$$ \Rightarrow \theta = {\cos ^{ - 1}}\left( {{1 \over {\sqrt 3 }}} \right)$$
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