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Vector Algebra question

2022 · 29 Jul · Shift 2 · Q35
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  5. /2022 · 29 Jul · Shift 2 · Q35

Vector Algebra question

2022 · 29 Jul · Shift 2 · Q35

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}a,b,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(a×b)⋅(b×c)+(b×c)⋅(c×a)+(c×a)⋅(a×b)=168, then ∣a⃗∣+∣b⃗∣+∣c⃗∣|\vec{a}|+|\vec{b}|+|\vec{c}|∣a∣+∣b∣+∣c∣ is equal to :
  1. A
    10
  2. B
    14
  3. C
    16
  4. D
    18
View written solutionFree

Correct answer: C

  1. Use the geometry of the vectors

Since a⃗,b⃗,c⃗\vec a,\vec b,\vec ca,b,c are coplanar concurrent vectors and the angle between any two is the same, let that common angle be θ\thetaθ.

Because all three lie in a plane, the only possible equal pairwise angle is θ=120∘\theta=120^\circθ=120∘ (other than the trivial impossible case 0∘0^\circ0∘ for distinct directions).

Hence, ∠(a⃗,b⃗)=∠(b⃗,c⃗)=∠(c⃗,a⃗)=120∘.\angle(\vec a,\vec b)=\angle(\vec b,\vec c)=\angle(\vec c,\vec a)=120^\circ.∠(a,b)=∠(b,c)=∠(c,a)=120∘.

  1. Simplify each dot product of cross products

Use the identity (u⃗×v⃗)⋅(w⃗×x⃗)=(u⃗⋅w⃗)(v⃗⋅x⃗)−(u⃗⋅x⃗)(v⃗⋅w⃗).(\vec u\times \vec v)\cdot(\vec w\times \vec x)=(\vec u\cdot \vec w)(\vec v\cdot \vec x)-(\vec u\cdot \vec x)(\vec v\cdot \vec w).(u×v)⋅(w×x)=(u⋅w)(v⋅x)−(u⋅x)(v⋅w).

Let A=∣a⃗∣,B=∣b⃗∣,C=∣c⃗∣.A=|\vec a|,\quad B=|\vec b|,\quad C=|\vec c|.A=∣a∣,B=∣b∣,C=∣c∣.

Also, a⃗⋅b⃗=ABcos⁡120∘=−AB2,\vec a\cdot \vec b=AB\cos120^\circ=-\frac{AB}{2},a⋅b=ABcos120∘=−2AB​, b⃗⋅c⃗=BCcos⁡120∘=−BC2,\vec b\cdot \vec c=BC\cos120^\circ=-\frac{BC}{2},b⋅c=BCcos120∘=−2BC​, c⃗⋅a⃗=CAcos⁡120∘=−CA2.\vec c\cdot \vec a=CA\cos120^\circ=-\frac{CA}{2}.c⋅a=CAcos120∘=−2CA​.

Now compute:

First term

=(\vec a\cdot \vec b)(\vec b\cdot \vec c)- (\vec a\cdot \vec c)(\vec b\cdot \vec b).$$ Substitute values: $$=( -\tfrac{AB}{2})( -\tfrac{BC}{2})-( -\tfrac{AC}{2})(B^2) =\frac{AB^2C}{4}+\frac{AB^2C}{2} =\frac{3}{4}AB^2C.$$ ### Second term Similarly, $$(\vec b\times \vec c)\cdot(\vec c\times \vec a)=\frac{3}{4}ABC^2.$$ ### Third term Similarly, $$(\vec c\times \vec a)\cdot(\vec a\times \vec b)=\frac{3}{4}A^2BC.$$ Therefore the given sum is $$\frac34\left(AB^2C+ABC^2+A^2BC\right)=168.$$ Factor $ABC$: $$\frac34\,ABC(A+B+C)=168.$$ --- 3. **Use the product of magnitudes** Given $$ABC=14.$$ So, $$\frac34\cdot 14\,(A+B+C)=168.$$ That is, $$\frac{42}{4}(A+B+C)=168$$ $$\frac{21}{2}(A+B+C)=168.$$ Hence, $$A+B+C=168\cdot \frac{2}{21}=16.$$ So, $$|\vec a|+|\vec b|+|\vec c|=16.$$ --- 4. **Check options** - A: $10$ - B: $14$ - C: $16$ - D: $18$ Therefore the correct option is $$\boxed{\text{C}}.$$
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