Let a̲ = α1 î + α2 ĵ + α3 k̂, b̲ = β1 î + β2 ĵ + β3 k̂ and c̲ = γ1 î + γ2 ĵ + γ3 k̂, |a̲| = 22, a̲ makes an angle π3 with the plane of b̲, c̲ and angle between b̲, c̲ is π6 , then α1 α2 α3 β1 β2 β3 γ1 γ2 γ3 n is equal to (n is even natural number)

Given in the question that a̲ = α1 î + α2 ĵ + α3 k̂, b̲ = β1 î + β2 ĵ + β3 k̂ and c̲ = γ1 î + γ2 ĵ + γ3 k̂, |a̲| = 22 , a̲ makes an angle π3 with the plane of b̲, c̲ and angle between b̲, c̲ is π6 where a̲, b̲, c̲ are 3 vectors.α1 α2 α3 β1 β2 β3 γ1 γ2 γ3 n can be written as a̲ b̲ c̲ nAs we know from the basic concept that a̲ b̲ c̲ = a̲ × b̲ . c̲ ⇒ a̲ ⋅ b̲ × c̲normal of the plane containing b̲, c̲ is b̲ × c̲ .given that a̲ makes an angle π3 with the plane of b̲ × c̲ .So a̲ will make an angle π6 with the normal of the plane containing b̲, c̲ that is b̲ × c̲.As we know that a̲ ⋅ b̲ = a̲ b̲ cosθ where θ is the angle between the normal of the plane containing of b̲ and a̲.We also know that a̲ × b̲ = a̲ b̲ sinθ n̂ where θ is the angle between a̲, b̲ and n̂ is the magnitude of the unit vector along the normal which will be equal to 1.Hence the simplified form of α1 α2 α3 β1 β2 β3 γ1 γ2 γ3 n will be⇒ a̲ b̲ c̲ n⇒ a̲ ⋅ b̲ × c̲n ⇒ a̲ ⋅b̲ × c̲coscos π6 n Since, a̲ makes an angle π6 with the normal of the plane containing b̲, c̲ that is b̲ × c̲.a̲ ⋅b̲ × c̲coscos π6 n⇒ 22 b̲ × c̲ 32nSince, a̲ = 22 is given in the question and we know that cos π6 = 32 This can be further simplified as22 b̲ × c̲ 32n⇒ 6 b̲ × c̲ n⇒ 6 b̲ c̲sinsin π6 n̂n Since, we know that n̂ = 1 and sin π6 = 12 6 b̲ c̲sinsin π6 n̂n⇒ 6 b̲ × c̲ 12n⇒ 32 b̲ c̲ nWe can conclude that α1 α2 α3 β1 β2 β3 γ1 γ2 γ3 n = 3 b̲ c̲2nHence, option (2) is correct.

Let a̲ = α1 î + α2 ĵ + α3 k̂, b̲ = β1 î + β2 ĵ + β3 k̂ and c̲ = γ1 î + γ2 ĵ + γ3 k̂, |a̲| = 22, a̲ makes an angle π3 with the plane of b̲, c̲ and angle between b̲, c̲ is π6 , then α1 α2 α3 β1 β2 β3 γ1 γ2 γ3 n is equal to (n is even natural number)

Given in the question that a̲ = α1 î + α2 ĵ + α3 k̂, b̲ = β1 î + β2 ĵ + β3 k̂ and c̲ = γ1 î + γ2 ĵ + γ3 k̂, |a̲| = 22 , a̲ makes an angle π3 with the plane of b̲, c̲ and angle between b̲, c̲ is π6 where a̲, b̲, c̲ are 3 vectors.α1 α2 α3 β1 β2 β3 γ1 γ2 γ3 n can be written as a̲ b̲ c̲ nAs we know from the basic concept that a̲ b̲ c̲ = a̲ × b̲ . c̲ ⇒ a̲ ⋅ b̲ × c̲normal of the plane containing b̲, c̲ is b̲ × c̲ .given that a̲ makes an angle π3 with the plane of b̲ × c̲ .So a̲ will make an angle π6 with the normal of the plane containing b̲, c̲ that is b̲ × c̲.As we know that a̲ ⋅ b̲ = a̲ b̲ cosθ where θ is the angle between the normal of the plane containing of b̲ and a̲.We also know that a̲ × b̲ = a̲ b̲ sinθ n̂ where θ is the angle between a̲, b̲ and n̂ is the magnitude of the unit vector along the normal which will be equal to 1.Hence the simplified form of α1 α2 α3 β1 β2 β3 γ1 γ2 γ3 n will be⇒ a̲ b̲ c̲ n⇒ a̲ ⋅ b̲ × c̲n ⇒ a̲ ⋅b̲ × c̲coscos π6 n Since, a̲ makes an angle π6 with the normal of the plane containing b̲, c̲ that is b̲ × c̲.a̲ ⋅b̲ × c̲coscos π6 n⇒ 22 b̲ × c̲ 32nSince, a̲ = 22 is given in the question and we know that cos π6 = 32 This can be further simplified as22 b̲ × c̲ 32n⇒ 6 b̲ × c̲ n⇒ 6 b̲ c̲sinsin π6 n̂n Since, we know that n̂ = 1 and sin π6 = 12 6 b̲ c̲sinsin π6 n̂n⇒ 6 b̲ × c̲ 12n⇒ 32 b̲ c̲ nWe can conclude that α1 α2 α3 β1 β2 β3 γ1 γ2 γ3 n = 3 b̲ c̲2nHence, option (2) is correct.

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Let $\underset{̲}{a}$ = α1 $\hat{i}$ + α2 $\hat{j}$ + α3 $\hat{k}$ , $\underset{̲}{b}$ = β1 $\hat{i}$ + β2 $\hat{j}$ + β3 $\hat{k}$ and $\underset{̲}{c}$ = γ1 $\hat{i}$ + γ2 $\hat{j}$ + γ3 $\hat{k}$ , | $\underset{̲}{a}$ | = 2 $\sqrt{2}$ , $\underset{̲}{a}$ makes an angle $\frac{π}{3}$ with the plane of $\underset{̲}{b}$ , $\underset{̲}{c}$ and angle between $\underset{̲}{b}$ , $\underset{̲}{c}$ is $\frac{π}{6}$ , then ${|α_{1} α_{2} α_{3} β_{1} β_{2} β_{3} γ_{1} γ_{2} γ_{3}|}^{n}$ is equal to (n is even natural number)

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{[\frac{|\underset{̲}{b}| |\underset{̲}{a}|}{6}]}^{\frac{n}{2}}

{[\frac{\sqrt{3} |\underset{̲}{b}| |\underset{̲}{c}|}{\sqrt{2}}]}^{n}

{\frac{|\underset{̲}{b}| |\underset{̲}{c}|}{\sqrt{3} 2^{n}}}^{\frac{n}{2}}

{[\frac{\sqrt{2} |\underset{̲}{b}| |\underset{̲}{\underset{̲}{a}}|}{\sqrt{3}}]}^{n}

answer is B.

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Detailed Solution

Given in the question that

\underset{̲}{a}

= α1

\hat{i}

+ α2

\hat{j}

+ α3

\hat{k}

\underset{̲}{b}

= β1

\hat{i}

+ β2

\hat{j}

+ β3

\hat{k}

and

\underset{̲}{c}

= γ1

\hat{i}

+ γ2

\hat{j}

+ γ3

\hat{k}

, |

\underset{̲}{a}

| = 2

\sqrt{2}

\underset{̲}{a}

makes an angle

\frac{π}{3}

with the plane of

\underset{̲}{b}

\underset{̲}{c}

and angle between

\underset{̲}{b}

\underset{̲}{c}

\frac{π}{6}

where

\underset{̲}{a}

\underset{̲}{b}

\underset{̲}{c}

are 3 vectors.

{|α_{1} α_{2} α_{3} β_{1} β_{2} β_{3} γ_{1} γ_{2} γ_{3}|}^{n}

can be written as

{[\underset{̲}{a} \underset{̲}{b} \underset{̲}{c}]}^{n}

As we know from the basic concept that

[\underset{̲}{a} \underset{̲}{b} \underset{̲}{c}]

(\underset{̲}{a} \times \underset{̲}{b})

\underset{̲}{c}

⇒

\underset{̲}{a}

⋅

(\underset{̲}{b} \times \underset{̲}{c})

normal of the plane containing

\underset{̲}{b}

\underset{̲}{c}

\underset{̲}{b} \times \underset{̲}{c}

.
given that

\underset{̲}{a}

makes an angle

\frac{π}{3}

with the plane of

\underset{̲}{b} \times \underset{̲}{c}

.
So

\underset{̲}{a}

will make an angle

\frac{π}{6}

with the normal of the plane containing

\underset{̲}{b}

\underset{̲}{c}

that is

\underset{̲}{b} \times \underset{̲}{c}

.
As we know that

\underset{̲}{a}

⋅

\underset{̲}{b}

|\underset{̲}{a}|

|\underset{̲}{b}|

cosθ where θ is the angle between the normal of the plane containing of

\underset{̲}{b}

and

\underset{̲}{a}

.
We also know that

\underset{̲}{a} \times \underset{̲}{b}

|\underset{̲}{a}|

|\underset{̲}{b}|

sinθ

|\hat{n}|

where θ is the angle between

\underset{̲}{a}

\underset{̲}{b}

and

|\hat{n}|

is the magnitude of the unit vector along the normal which will be equal to 1.
Hence the simplified form of

{|α_{1} α_{2} α_{3} β_{1} β_{2} β_{3} γ_{1} γ_{2} γ_{3}|}^{n}

will be
⇒

{[\underset{̲}{a} \underset{̲}{b} \underset{̲}{c}]}^{n}

⇒

{(\underset{̲}{a} \cdot (\underset{̲}{b} \times \underset{̲}{c}))}^{n}

⇒

{(|\underset{̲}{a}| \cdot |\underset{̲}{b} \times \underset{̲}{c}| \cos \cos \frac{π}{6})}^{n}

Since,

\underset{̲}{a}

makes an angle

\frac{π}{6}

with the normal of the plane containing

\underset{̲}{b}

\underset{̲}{c}

that is

\underset{̲}{b} \times \underset{̲}{c}

{(|\underset{̲}{a}| \cdot |\underset{̲}{b} \times \underset{̲}{c}| \cos \cos \frac{π}{6})}^{n}

⇒

{(2 \sqrt{2} |\underset{̲}{b} \times \underset{̲}{c}| \frac{\sqrt{3}}{2})}^{n}

Since,

|\underset{̲}{a}|

= 2

\sqrt{2}

is given in the question and we know that cos

\frac{π}{6}

\frac{\sqrt{3}}{2}

This can be further simplified as

{(2 \sqrt{2} |\underset{̲}{b} \times \underset{̲}{c}| \frac{\sqrt{3}}{2})}^{n}

⇒

{(\sqrt{6} |\underset{̲}{b} \times \underset{̲}{c}|)}^{n}

⇒

{(\sqrt{6} |\underset{̲}{b}| |\underset{̲}{c}| \sin \sin \frac{π}{6} |\hat{n}|)}^{n}

Since, we know that

|\hat{n}|

= 1 and sin

\frac{π}{6}

\frac{1}{2}

{(\sqrt{6} |\underset{̲}{b}| |\underset{̲}{c}| \sin \sin \frac{π}{6} |\hat{n}|)}^{n}

⇒

{(\sqrt{6} |\underset{̲}{b} \times \underset{̲}{c}| \frac{1}{2})}^{n}

⇒

{(\frac{\sqrt{3}}{\sqrt{2}} |\underset{̲}{b}| |\underset{̲}{c}|)}^{n}

We can conclude that

{|α_{1} α_{2} α_{3} β_{1} β_{2} β_{3} γ_{1} γ_{2} γ_{3}|}^{n}

{[\frac{\sqrt{3} |\underset{̲}{b}| |\underset{̲}{c}|}{2}]}^{n}

Hence, option (2) is correct.

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Let a̲ = α1 î + α2 ĵ + α3 k̂, b̲ = β1 î + β2 ĵ + β3 k̂ and c̲ = γ1 î + γ2 ĵ + γ3 k̂, |a̲| = 22, a̲ makes an angle π3 with the plane of b̲, c̲ and angle between b̲, c̲ is π6 , then α1 α2 α3 β1 β2 β3 γ1 γ2 γ3 n is equal to (n is even natural number)

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