JEE Main 2022 — Mathematics Vectors & 3D Geometry

Given,

$$|(\hat{a}+\hat{b})+2(\hat{a}\times \hat{b})|=2,\theta \in (0,\pi )$$∣(a^+b^)+2(a^×b^)∣=2,θ∈(0,π)

Squaring both side we get,

$$((\hat{a}+\hat{b})+2(\hat{a}\times \hat{b}))\cdot ((\hat{a}+\hat{b})+2(\hat{a}\times \hat{b}))=4$$((a^+b^)+2(a^×b^))⋅((a^+b^)+2(a^×b^))=4

$${|\hat{a}+\hat{b}|}^{2}+4{|(\hat{a}\times \hat{b})|}^{2}+0=4$$∣a^+b^∣2+4∣(a^×b^)∣2+0=4

Let the angle be $\theta$θ between $\hat{a}$a^ and $\hat{b}$b^

$$\Rightarrow {|\hat{a}|}^{2}+{|\hat{b}|}^{2}+2|\hat{a}||\hat{b}|\mathrm{cos}\theta +4(|\hat{a}||\hat{b}|\mathrm{sin}\theta )=4$$⇒∣a^∣2+∣b^∣2+2∣a^∣∣b^∣cosθ+4(∣a^∣∣b^∣sinθ)=4

$$\Rightarrow 2+2\mathrm{cos}\theta +4{\mathrm{sin}}^{2}\theta =4$$⇒2+2cosθ+4sin2θ=4

$$\Rightarrow 2+2\mathrm{cos}\theta -4{\mathrm{cos}}^{2}\theta =0$$⇒2+2cosθ−4cos2θ=0

Let $\mathrm{cos}\theta =t$cosθ=t then

$$\Rightarrow 2{t}^{2}-t-1=0$$⇒2t2−t−1=0

$$\Rightarrow 2{t}^{2}-2t+t-1=0$$⇒2t2−2t+t−1=0

$$\Rightarrow 2t(t-1)+(t-1)=0$$⇒2t(t−1)+(t−1)=0

$$(2t+1)(t-1)=0$$(2t+1)(t−1)=0

So, $t=-\frac{1}{2}$t=−21​ or $t=1$t=1

Then, $\mathrm{cos}\theta =-\frac{1}{2}$cosθ=−21​ or $\mathrm{cos}\theta =1$cosθ=1 which is not possible as $0\in (0,\pi )$0∈(0,π)

So, $\theta =\frac{2\pi }{3}$θ=32π​

Now,

$${S}_{1}\Rightarrow 2|a\times b|=2\mathrm{sin}(\frac{2\pi }{3})=2\times \frac{\sqrt{3}}{2}=\sqrt{3}$$S1​⇒2∣a×b∣=2sin(32π​)=2×23​​=3​

And $|\hat{a}-\hat{b}|=\sqrt{1+1-2\mathrm{cos}(\frac{2\pi }{3})}$∣a^−b^∣=1+1−2cos(32π​)​

$$=\sqrt{2-2\times (-\frac{1}{2})}$$=2−2×(−21​)​

$$=\sqrt{3}$$=3​

${S}_{1}$S1​ is correct.

${S}_{2}$S2​ projection of $\hat{a}$a^ on $(\hat{a}+\hat{b})$(a^+b^)

$$\Rightarrow \frac{\hat{a}.(\hat{a}+\hat{b})}{\mid \hat{a}+\hat{b})}=\frac{1+\mathrm{cos}(\frac{2\pi }{3})}{\sqrt{2+2\mathrm{cos}\frac{2\pi }{3}}}$$⇒∣a^+b^)a^.(a^+b^)​=2+2cos32π​​1+cos(32π​)​

$$\Rightarrow \frac{1-\frac{1}{2}}{\sqrt{1}}=\frac{1}{2}$$⇒1​1−21​​=21​

Which is true, So both are true.