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Vector Algebra question

2022 · 29 Jul · Shift 1 · Q34
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  5. /2022 · 29 Jul · Shift 1 · Q34

Vector Algebra question

2022 · 29 Jul · Shift 1 · Q34

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a^\hat{a}a^ and b^\hat{b}b^ be two unit vectors such that the angle between them is π4\frac{\pi}{4}4π​. If θ\thetaθ is the angle between the vectors (a^+b^)(\hat{a}+\hat{b})(a^+b^) and (a^+2b^+2(a^×b^))(\hat{a}+2 \hat{b}+2(\hat{a} \times \hat{b}))(a^+2b^+2(a^×b^)), then the value of 164 cos⁡2θ164 \,\cos ^{2} \theta164cos2θ is equal to :
  1. A
    90+27290+27 \sqrt{2}90+272​
  2. B
    45+18245+18 \sqrt{2}45+182​
  3. C
    90+3290+3 \sqrt{2}90+32​
  4. D
    54+90254+90 \sqrt{2}54+902​
View written solutionFree

Correct answer: A

  1. Given data
  • a^\hat aa^ and b^\hat bb^ are unit vectors.
  • Angle between them is π4\dfrac{\pi}{4}4π​.

So,

a^⋅b^=cos⁡π4=12\hat a\cdot \hat b = \cos \frac{\pi}{4}=\frac{1}{\sqrt2}a^⋅b^=cos4π​=2​1​

and

∣a^∣=∣b^∣=1.|\hat a|=|\hat b|=1.∣a^∣=∣b^∣=1.

We need the angle θ\thetaθ between

u⃗=a^+b^\vec u=\hat a+\hat bu=a^+b^

and

v⃗=a^+2b^+2(a^×b^).\vec v=\hat a+2\hat b+2(\hat a\times \hat b).v=a^+2b^+2(a^×b^).

Using

cos⁡θ=u⃗⋅v⃗∣u⃗∣∣v⃗∣.\cos\theta=\frac{\vec u\cdot \vec v}{|\vec u||\vec v|}.cosθ=∣u∣∣v∣u⋅v​.
  1. Compute u⃗⋅v⃗\vec u\cdot \vec vu⋅v
u⃗⋅v⃗=(a^+b^)⋅(a^+2b^+2(a^×b^)).\vec u\cdot \vec v=(\hat a+\hat b)\cdot\bigl(\hat a+2\hat b+2(\hat a\times \hat b)\bigr).u⋅v=(a^+b^)⋅(a^+2b^+2(a^×b^)).

Now use the fact that a^×b^\hat a\times \hat ba^×b^ is perpendicular to both a^\hat aa^ and b^\hat bb^, so

a^⋅(a^×b^)=0,b^⋅(a^×b^)=0.\hat a\cdot(\hat a\times\hat b)=0, \qquad \hat b\cdot(\hat a\times\hat b)=0.a^⋅(a^×b^)=0,b^⋅(a^×b^)=0.

Hence,

u⃗⋅v⃗=(a^+b^)⋅(a^+2b^).\vec u\cdot \vec v=(\hat a+\hat b)\cdot(\hat a+2\hat b).u⋅v=(a^+b^)⋅(a^+2b^).

Expand:

=a^⋅a^+2a^⋅b^+b^⋅a^+2b^⋅b^.=\hat a\cdot\hat a+2\hat a\cdot\hat b+\hat b\cdot\hat a+2\hat b\cdot\hat b.=a^⋅a^+2a^⋅b^+b^⋅a^+2b^⋅b^.

Since a^⋅b^=b^⋅a^=12\hat a\cdot\hat b=\hat b\cdot\hat a=\dfrac{1}{\sqrt2}a^⋅b^=b^⋅a^=2​1​,

u⃗⋅v⃗=1+2⋅12+12+2=3+32.\vec u\cdot\vec v=1+2\cdot\frac1{\sqrt2}+\frac1{\sqrt2}+2 =3+\frac{3}{\sqrt2}.u⋅v=1+2⋅2​1​+2​1​+2=3+2​3​.

So,

u⃗⋅v⃗=3+32.\vec u\cdot\vec v=3+\frac{3}{\sqrt2}.u⋅v=3+2​3​.
  1. Compute ∣u⃗∣|\vec u|∣u∣
∣u⃗∣2=(a^+b^)⋅(a^+b^)=∣a^∣2+∣b^∣2+2a^⋅b^.|\vec u|^2=(\hat a+\hat b)\cdot(\hat a+\hat b) =|\hat a|^2+|\hat b|^2+2\hat a\cdot\hat b.∣u∣2=(a^+b^)⋅(a^+b^)=∣a^∣2+∣b^∣2+2a^⋅b^.

Thus,

∣u⃗∣2=1+1+2⋅12=2+2.|\vec u|^2=1+1+2\cdot\frac1{\sqrt2}=2+\sqrt2.∣u∣2=1+1+2⋅2​1​=2+2​.

So,

∣u⃗∣=2+2.|\vec u|=\sqrt{2+\sqrt2}.∣u∣=2+2​​.
  1. Compute ∣v⃗∣|\vec v|∣v∣
v⃗=a^+2b^+2(a^×b^).\vec v=\hat a+2\hat b+2(\hat a\times\hat b).v=a^+2b^+2(a^×b^).

Since a^×b^\hat a\times\hat ba^×b^ is perpendicular to both a^\hat aa^ and b^\hat bb^, we get

∣v⃗∣2=∣a^+2b^∣2+∣2(a^×b^)∣2.|\vec v|^2=|\hat a+2\hat b|^2+|2(\hat a\times\hat b)|^2.∣v∣2=∣a^+2b^∣2+∣2(a^×b^)∣2.

First,

∣a^+2b^∣2=∣a^∣2+4∣b^∣2+4a^⋅b^=1+4+4⋅12=5+22.|\hat a+2\hat b|^2=|\hat a|^2+4|\hat b|^2+4\hat a\cdot\hat b =1+4+4\cdot\frac1{\sqrt2}=5+2\sqrt2.∣a^+2b^∣2=∣a^∣2+4∣b^∣2+4a^⋅b^=1+4+4⋅2​1​=5+22​.

Next,

∣a^×b^∣=∣a^∣∣b^∣sin⁡π4=1⋅1⋅12=12.|\hat a\times\hat b|=|\hat a||\hat b|\sin\frac\pi4=1\cdot1\cdot\frac1{\sqrt2}=\frac1{\sqrt2}.∣a^×b^∣=∣a^∣∣b^∣sin4π​=1⋅1⋅2​1​=2​1​.

Therefore,

∣2(a^×b^)∣2=4(12)2=4⋅12=2.|2(\hat a\times\hat b)|^2=4\left(\frac1{\sqrt2}\right)^2=4\cdot\frac12=2.∣2(a^×b^)∣2=4(2​1​)2=4⋅21​=2.

Hence,

∣v⃗∣2=(5+22)+2=7+22.|\vec v|^2=(5+2\sqrt2)+2=7+2\sqrt2.∣v∣2=(5+22​)+2=7+22​.

So,

∣v⃗∣=7+22.|\vec v|=\sqrt{7+2\sqrt2}.∣v∣=7+22​​.
  1. Compute cos⁡2θ\cos^2\thetacos2θ
cos⁡θ=3+322+2 7+22.\cos\theta=\frac{3+\frac{3}{\sqrt2}}{\sqrt{2+\sqrt2}\,\sqrt{7+2\sqrt2}}.cosθ=2+2​​7+22​​3+2​3​​.

Thus,

cos⁡2θ=(3+32)2(2+2)(7+22).\cos^2\theta=\frac{\left(3+\frac{3}{\sqrt2}\right)^2}{(2+\sqrt2)(7+2\sqrt2)}.cos2θ=(2+2​)(7+22​)(3+2​3​)2​.

Now simplify numerator:

3+32=3(1+12).3+\frac{3}{\sqrt2}=3\left(1+\frac1{\sqrt2}\right).3+2​3​=3(1+2​1​).

So,

(3+32)2=9(1+12)2.\left(3+\frac{3}{\sqrt2}\right)^2=9\left(1+\frac1{\sqrt2}\right)^2.(3+2​3​)2=9(1+2​1​)2.

And

(1+12)2=1+2+12=32+2.\left(1+\frac1{\sqrt2}\right)^2=1+\sqrt2+\frac12=\frac32+\sqrt2.(1+2​1​)2=1+2​+21​=23​+2​.

Hence,

(3+32)2=9(32+2)=272+92.\left(3+\frac{3}{\sqrt2}\right)^2=9\left(\frac32+\sqrt2\right)=\frac{27}{2}+9\sqrt2.(3+2​3​)2=9(23​+2​)=227​+92​.

Now denominator:

(2+2)(7+22)=14+42+72+4=18+112.(2+\sqrt2)(7+2\sqrt2)=14+4\sqrt2+7\sqrt2+4=18+11\sqrt2.(2+2​)(7+22​)=14+42​+72​+4=18+112​.

Thus,

cos⁡2θ=272+9218+112.\cos^2\theta=\frac{\frac{27}{2}+9\sqrt2}{18+11\sqrt2}.cos2θ=18+112​227​+92​​.

Multiply numerator and denominator by 222:

cos⁡2θ=27+18236+222.\cos^2\theta=\frac{27+18\sqrt2}{36+22\sqrt2}.cos2θ=36+222​27+182​​.

Therefore,

164cos⁡2θ=164⋅27+18236+222.164\cos^2\theta=164\cdot\frac{27+18\sqrt2}{36+22\sqrt2}.164cos2θ=164⋅36+222​27+182​​.

Since 164=4(36+222)/(9+?)164=4(36+22\sqrt2)/(9+? )164=4(36+222​)/(9+?) it is simpler to rationalize directly:

164cos⁡2θ=164(27+182)36+222.164\cos^2\theta=\frac{164(27+18\sqrt2)}{36+22\sqrt2}.164cos2θ=36+222​164(27+182​)​.

Factor denominator:

36+222=2(18+112).36+22\sqrt2=2(18+11\sqrt2).36+222​=2(18+112​).

So,

164cos⁡2θ=82(27+182)18+112.164\cos^2\theta=\frac{82(27+18\sqrt2)}{18+11\sqrt2}.164cos2θ=18+112​82(27+182​)​.

Rationalize:

82(27+182)18+112⋅18−11218−112.\frac{82(27+18\sqrt2)}{18+11\sqrt2}\cdot\frac{18-11\sqrt2}{18-11\sqrt2}.18+112​82(27+182​)​⋅18−112​18−112​​.

Denominator becomes

182−(112)2=324−242=82.18^2-(11\sqrt2)^2=324-242=82.182−(112​)2=324−242=82.

So the 828282 cancels, giving

164cos⁡2θ=(27+182)(18−112).164\cos^2\theta=(27+18\sqrt2)(18-11\sqrt2).164cos2θ=(27+182​)(18−112​).

Now expand:

=486−2972+3242−396=486-297\sqrt2+324\sqrt2-396=486−2972​+3242​−396 =90+272.=90+27\sqrt2.=90+272​.
  1. Match with options
164cos⁡2θ=90+272164\cos^2\theta=90+27\sqrt2164cos2θ=90+272​

which is Option A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

So they agree.

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