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QJEEAdv18Maa4f6: JEE Advanced 2018 Physics - Class 11 JEEAdv18Maa4f6 Solution: Let a and b be two unit vectors such that a . b = 0. For some…

Let a and b be two unit vectors such that a . b = 0. For some x, y ∈\in R, let c→=xa→+yb→+a→×b→\overrightarrow c = x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b . If | c→\overrightarrow c | = 2 and the vector c is inclined at the same angle α\alpha to both a and b, then the value of 8cos⁡2α8{\cos ^2}\alpha is ..............

Step-by-Step Verified Solution

We have,

c→=xa→+yb→+a→×b→\overrightarrow c = x\overrightarrow a + y\overrightarrow b + \overrightarrow a \times \overrightarrow b and a→.b→\overrightarrow a .\overrightarrow b = 0

| a→\overrightarrow a | = | b→\overrightarrow b | = 1

and | c→\overrightarrow c | = 2

Also, given c→\overrightarrow c is inclined on a→\overrightarrow a and b→\overrightarrow b with same angle α\alpha .

∴\therefore a→.c→=x∣a→∣2+y(a→.b→)+a→.(a→×b→)\overrightarrow a .\overrightarrow c = x|\overrightarrow a {|^2} + y(\overrightarrow a .\overrightarrow b ) + \overrightarrow a .(\overrightarrow a \times \overrightarrow b )

∣a→∣∣c→∣cosα=x+0+0|\overrightarrow a ||\overrightarrow c |cos\alpha = x + 0 + 0

x = 2cos α\alpha

Similarly,

∣b→∣∣c→∣cosα=0+y+0|\overrightarrow b ||\overrightarrow c |cos\alpha = 0 + y + 0

⇒\Rightarrow y = 2cos α\alpha

∣c→∣2=x2+y2+∣a→ × b→∣2|\overrightarrow c {|^2} = {x^2} + {y^2} + |\overrightarrow a \, \times \,\overrightarrow b {|^2}

4=8cos⁡2α+∣a∣2∣b∣2sin⁡290∘4 = 8{\cos ^2}\alpha + |a{|^2}|b{|^2}{\sin ^2}90^\circ

4=8cos⁡2α+14 = 8{\cos ^2}\alpha + 1

⇒\Rightarrow 8cos⁡2α8{\cos ^2}\alpha = 3

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