JEE Advance - Mathematics (2018 - Paper 1 Offline - No. 12)
Let a and b be two unit vectors such that a . b = 0. For some x, y$$ \in $$R, let $$\overrightarrow c = x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b $$. If | $$\overrightarrow c $$| = 2 and the vector c is inclined at the same angle $$\alpha $$ to both a and b, then the value of $$8{\cos ^2}\alpha $$ is ..............
Answer
3
Explanation
We have,
$$\overrightarrow c = x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b $$ and $$\overrightarrow a .\overrightarrow b $$ = 0
|$$\overrightarrow a $$| = |$$\overrightarrow b $$| = 1
and |$$\overrightarrow c $$| = 2
Also, given $$\overrightarrow c $$ is inclined on $$\overrightarrow a $$ and $$\overrightarrow b $$ with same angle $$\alpha $$.
$$ \therefore $$ $$\overrightarrow a .\overrightarrow c = x|\overrightarrow a {|^2} + y(\overrightarrow a .\overrightarrow b ) + \overrightarrow a .(\overrightarrow a \times \overrightarrow b )$$
$$|\overrightarrow a ||\overrightarrow c |cos\alpha = x + 0 + 0$$
x = 2cos$$\alpha $$
Similarly,
$$|\overrightarrow b ||\overrightarrow c |cos\alpha = 0 + y + 0$$
$$ \Rightarrow $$ y = 2cos$$\alpha $$
$$|\overrightarrow c {|^2} = {x^2} + {y^2} + |\overrightarrow a \, \times \,\overrightarrow b {|^2}$$
$$4 = 8{\cos ^2}\alpha + |a{|^2}|b{|^2}{\sin ^2}90^\circ $$
$$4 = 8{\cos ^2}\alpha + 1$$
$$ \Rightarrow $$ $$8{\cos ^2}\alpha $$ = 3
$$\overrightarrow c = x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b $$ and $$\overrightarrow a .\overrightarrow b $$ = 0
|$$\overrightarrow a $$| = |$$\overrightarrow b $$| = 1
and |$$\overrightarrow c $$| = 2
Also, given $$\overrightarrow c $$ is inclined on $$\overrightarrow a $$ and $$\overrightarrow b $$ with same angle $$\alpha $$.
$$ \therefore $$ $$\overrightarrow a .\overrightarrow c = x|\overrightarrow a {|^2} + y(\overrightarrow a .\overrightarrow b ) + \overrightarrow a .(\overrightarrow a \times \overrightarrow b )$$
$$|\overrightarrow a ||\overrightarrow c |cos\alpha = x + 0 + 0$$
x = 2cos$$\alpha $$
Similarly,
$$|\overrightarrow b ||\overrightarrow c |cos\alpha = 0 + y + 0$$
$$ \Rightarrow $$ y = 2cos$$\alpha $$
$$|\overrightarrow c {|^2} = {x^2} + {y^2} + |\overrightarrow a \, \times \,\overrightarrow b {|^2}$$
$$4 = 8{\cos ^2}\alpha + |a{|^2}|b{|^2}{\sin ^2}90^\circ $$
$$4 = 8{\cos ^2}\alpha + 1$$
$$ \Rightarrow $$ $$8{\cos ^2}\alpha $$ = 3
Comments (0)
