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QJEEM22Ma1428: JEE Main 2022 Physics - Class 11 JEEM22Ma1428 Solution: Let be three coplanar concurrent vectors such that angles…

Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c} be three coplanar concurrent vectors such that angles between any two of them is same. If the product of their magnitudes is 14 and (a⃗×b⃗)⋅(b⃗×c⃗)+(b⃗×c⃗)⋅(c⃗×a⃗)+(c⃗×a⃗)⋅(a⃗×b⃗)=168(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=168 , then ∣a⃗∣+∣b⃗∣+∣c⃗∣|\vec{a}|+|\vec{b}|+|\vec{c}| is equal to :

Multiple Choice Options

  • A 10
  • B 14
  • C 16
  • D 18

Step-by-Step Verified Solution

∣a⃗∣∣b⃗∣∣c⃗∣=14|\vec{a}||\vec{b}||\vec{c}|=14

amp;a⃗∧b⃗=b⃗∧c⃗=c⃗∧a⃗=θ=2π3amp;a⃗⋅b⃗=−12∣a⃗∣∣b⃗∣amp;b⃗⋅c⃗=−12∣b⃗∣∣c⃗∣amp;c⃗⋅a⃗=−12∣c⃗∣∣a⃗∣\begin{aligned}& \vec{a} \wedge \vec{b}=\vec{b} \wedge \vec{c}=\vec{c} \wedge \vec{a}=\theta=\frac{2 \pi}{3} \\\\& \vec{a} \cdot \vec{b}=-\frac{1}{2}|\vec{a}||\vec{b}| \\\\& \vec{b} \cdot \vec{c}=-\frac{1}{2}|\vec{b}||\vec{c}| \\\\& \vec{c} \cdot \vec{a}=-\frac{1}{2}|\vec{c}||\vec{a}|\end{aligned}

Now,

amp;amp;(a⃗×b⃗)⋅(b⃗×c⃗)+(b⃗×c⃗)⋅(c⃗×a⃗)+(c⃗×a⃗)⋅(a⃗×b⃗)amp;=168...(i)amp;(a⃗×b⃗)⋅(b⃗×c⃗)=(a⃗⋅b⃗)(b⃗⋅c⃗)−(a⃗⋅c⃗)∣b⃗∣2amp;=14∣b⃗∣2∣a⃗∣∣c⃗∣+12∣a⃗∣∣b⃗∣2∣c⃗∣amp;=34∣a⃗∣∣b⃗∣2∣c⃗∣...(ii)\begin{aligned}& \begin{aligned}&(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})+(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})+(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b}) \\\\&=168 \quad\quad...(i)\\\\&(\vec{a} \times \vec{b}) \cdot(\vec{b} \times \vec{c})=(\vec{a} \cdot \vec{b})(\vec{b} \cdot \vec{c})-(\vec{a} \cdot \vec{c})|\vec{b}|^{2} \\\\&= \frac{1}{4}|\vec{b}|^{2}|\vec{a}||\vec{c}|+\frac{1}{2}|\vec{a}||\vec{b}|^{2}|\vec{c}| \\\\&= \frac{3}{4}|\vec{a}||\vec{b}|^{2}|\vec{c}| \quad\quad...(ii)\end{aligned}\end{aligned}

Similarly (b⃗×c⃗)⋅(c⃗×a⃗)=34∣a⃗∣∣b⃗∣∣c⃗∣2...(iii)(\vec{b} \times \vec{c}) \cdot(\vec{c} \times \vec{a})=\frac{3}{4}|\vec{a}||\vec{b}||\vec{c}|^{2} \quad\quad...(iii)

(c⃗×a⃗)⋅(a⃗×b⃗)=34∣a⃗∣2∣b⃗∣∣c⃗∣...(iv)(\vec{c} \times \vec{a}) \cdot(\vec{a} \times \vec{b})=\frac{3}{4}|\vec{a}|^{2}|\vec{b}||\vec{c}| \quad\quad...(iv)

Substitute (ii), (iii), (iv) in (i)

34∣a⃗∣∣b⃗∣∣c⃗∣[∣a⃗∣+∣b⃗∣+∣c⃗∣]=168\frac{3}{4}|\vec{a}||\vec{b}||\vec{c}|[|\vec{a}|+|\vec{b}|+|\vec{c}|]=168

34×14[∣a⃗∣+∣b⃗∣+∣c⃗∣]=168\frac{3}{4} \times 14[|\vec{a}|+|\vec{b}|+|\vec{c}|]=168

∣a⃗∣+∣b⃗∣+∣c⃗∣=16|\vec{a}|+|\vec{b}|+|\vec{c}|=16

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