Let a^ and b^ be two unit vectors such that the angle between them is π4. If θ is the angle between the vectors a^ +b^ and a^ +2b^+2a^ ×b^, then the value of 164 cos2θ is equal to:

a^·b^=|a^||b^|cosϕ=1·1·cosπ4=12Now, cosθ=(a^+b^)·(a^+2b^+2(a^×b^))|a^+b^||a^+2b^+2(a^×b^)|cos2⁡θ=|a^|2+2a^⋅b^+0+b^⋅a^+2|b^|2+02((a^+b^)⋅(a^+b^)(a^⋅a^+2a^⋅b^+0+2b^⋅a^+4b^⋅b^+0+0+0+4|a^|2|b^|2sin2⁡ϕ=1+22+12+22(a^⋅a^+a^⋅b^+b^⋅a^+b^⋅b^)1+22+22+4+42 =3+3221+12+12+17+42=9(2+1)22(2+2)(7+22) =9(2+1)2×(7−22)22(2+1)(7+22)(7−22)=9(2+1)(7−22)22⋅72−(22)2 =922⋅72+7−4−2249−8=9223+5241 164cos2⁡θ=164×9(52+3)22×41=18522+32=90+272

Let a^ and b^ be two unit vectors such that the angle between them is π4. If θ is the angle between the vectors a^ +b^ and a^ +2b^+2a^ ×b^, then the value of 164 cos2θ is equal to:

a^·b^=|a^||b^|cosϕ=1·1·cosπ4=12Now, cosθ=(a^+b^)·(a^+2b^+2(a^×b^))|a^+b^||a^+2b^+2(a^×b^)|cos2⁡θ=|a^|2+2a^⋅b^+0+b^⋅a^+2|b^|2+02((a^+b^)⋅(a^+b^)(a^⋅a^+2a^⋅b^+0+2b^⋅a^+4b^⋅b^+0+0+0+4|a^|2|b^|2sin2⁡ϕ=1+22+12+22(a^⋅a^+a^⋅b^+b^⋅a^+b^⋅b^)1+22+22+4+42 =3+3221+12+12+17+42=9(2+1)22(2+2)(7+22) =9(2+1)2×(7−22)22(2+1)(7+22)(7−22)=9(2+1)(7−22)22⋅72−(22)2 =922⋅72+7−4−2249−8=9223+5241 164cos2⁡θ=164×9(52+3)22×41=18522+32=90+272

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Let $\hat{a} and \hat{b}$ be two unit vectors such that the angle between them is $\frac{π}{4}$ . If $θ$ is the angle between the vectors $(\hat{a} + \hat{b}) and (\hat{a} + 2 \hat{b} + 2 (\hat{a} \times \hat{b}))$ , then the value of 164 $\cos^{2} θ$ is equal to:

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$90 + 27 \sqrt{2}$

$45 + 18 \sqrt{2}$

$90 + 3 \sqrt{2}$

$54 + 90 \sqrt{2}$

answer is A.

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Detailed Solution

$\hat{a} \cdot \hat{b} = | \hat{a} | | \hat{b} | cosϕ = 1 \cdot 1 \cdot \cos \frac{π}{4} = \frac{1}{\sqrt{2}}$
Now, $cosθ = \frac{(\hat{a} + \hat{b}) \cdot (\hat{a} + 2 \hat{b} + 2 (\hat{a} \times \hat{b}))}{| \hat{a} + \hat{b} | | \hat{a} + 2 \hat{b} + 2 (\hat{a} \times \hat{b}) |}$
$\begin{aligned} \cos^{2} θ = \frac{{[| \hat{a} |^{2} + 2 \hat{a} \cdot \hat{b} + 0 + \hat{b} \cdot \hat{a} + 2 | \hat{b} |^{2} + 0]}^{2}}{((\hat{a} + \hat{b}) \cdot (\hat{a} + \hat{b}) (\hat{a} \cdot \hat{a} + 2 \hat{a} \cdot \hat{b} + 0 + 2 \hat{b} \cdot \hat{a} + 4 \hat{b} \cdot \hat{b} + 0} \\ + 0 + 0 + 4 | \hat{a} |^{2} | \hat{b} |^{2} \sin^{2} ϕ) \end{aligned}$
$= \frac{{(1 + \frac{2}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 2)}^{2}}{(\hat{a} \cdot \hat{a} + \hat{a} \cdot \hat{b} + \hat{b} \cdot \hat{a} + \hat{b} \cdot \hat{b}) (1 + \frac{2}{\sqrt{2}} + \frac{2}{\sqrt{2}} + 4 + \frac{4}{2})} = \frac{{(3 + \frac{3}{\sqrt{2}})}^{2}}{(1 + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}} + 1) (7 + \frac{4}{\sqrt{2}})} = \frac{9 (\sqrt{2} + 1)^{2}}{2 (2 + \sqrt{2}) (7 + 2 \sqrt{2})} = \frac{9 (\sqrt{2} + 1)^{2} \times (7 - 2 \sqrt{2})}{2 \sqrt{2} (\sqrt{2} + 1) (7 + 2 \sqrt{2}) (7 - 2 \sqrt{2})} = \frac{9 (\sqrt{2} + 1) (7 - 2 \sqrt{2})}{2 \sqrt{2} \cdot (7^{2} - (2 \sqrt{2})^{2})} = \frac{9}{2 \sqrt{2}} \cdot (\frac{7 \sqrt{2} + 7 - 4 - 2 \sqrt{2}}{49 - 8}) = \frac{9}{2 \sqrt{2}} (\frac{3 + 5 \sqrt{2}}{41}) \begin{array}{l} 164 \cos^{2} θ = \frac{164 \times 9 (5 \sqrt{2} + 3)}{2 \sqrt{2} \times 41} = 18 (\frac{5 \sqrt{2}}{\sqrt{2}} + \frac{3}{\sqrt{2}}) \\ = 90 + 27 \sqrt{2} \end{array}$