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Vector Algebra question

2024 · 5 Apr · Shift 2 · Q49
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  5. /2024 · 5 Apr · Shift 2 · Q49

Vector Algebra question

2024 · 5 Apr · Shift 2 · Q49

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Consider three vectors a⃗,b⃗,c⃗\vec{a}, \vec{b}, \vec{c}a,b,c. Let ∣a⃗∣=2,∣b⃗∣=3|\vec{a}|=2,|\vec{b}|=3∣a∣=2,∣b∣=3 and a⃗=b⃗×c⃗\vec{a}=\vec{b} \times \vec{c}a=b×c. If α∈[0,π3]\alpha \in\left[0, \frac{\pi}{3}\right]α∈[0,3π​] is the angle between the vectors b⃗\vec{b}b and c⃗\vec{c}c, then the minimum value of 27∣c⃗−a⃗∣227|\vec{c}-\vec{a}|^227∣c−a∣2 is equal to:
  1. A
    124
  2. B
    110
  3. C
    121
  4. D
    105
View written solutionFree

Correct answer: A

  1. Use the cross product condition

Given

a⃗=b⃗×c⃗,∣a⃗∣=2,∣b⃗∣=3\vec a = \vec b \times \vec c, \quad |\vec a|=2, \quad |\vec b|=3a=b×c,∣a∣=2,∣b∣=3

and if α\alphaα is the angle between b⃗\vec bb and c⃗\vec cc, then

∣a⃗∣=∣b⃗∣∣c⃗∣sin⁡α.|\vec a| = |\vec b||\vec c|\sin\alpha.∣a∣=∣b∣∣c∣sinα.

So,

2=3∣c⃗∣sin⁡α⇒∣c⃗∣=23sin⁡α.2 = 3|\vec c|\sin\alpha \quad\Rightarrow\quad |\vec c| = \frac{2}{3\sin\alpha}.2=3∣c∣sinα⇒∣c∣=3sinα2​.
  1. Find a⃗⋅c⃗\vec a \cdot \vec ca⋅c

Since

a⃗=b⃗×c⃗,\vec a = \vec b \times \vec c,a=b×c,

we know that a⃗\vec aa is perpendicular to both b⃗\vec bb and c⃗\vec cc. Hence,

a⃗⋅c⃗=0.\vec a \cdot \vec c = 0.a⋅c=0.
  1. Compute ∣c⃗−a⃗∣2|\vec c-\vec a|^2∣c−a∣2

Using

∣c⃗−a⃗∣2=∣c⃗∣2+∣a⃗∣2−2a⃗⋅c⃗,|\vec c-\vec a|^2 = |\vec c|^2 + |\vec a|^2 - 2\vec a\cdot \vec c,∣c−a∣2=∣c∣2+∣a∣2−2a⋅c,

and a⃗⋅c⃗=0\vec a\cdot \vec c=0a⋅c=0, we get

∣c⃗−a⃗∣2=∣c⃗∣2+∣a⃗∣2.|\vec c-\vec a|^2 = |\vec c|^2 + |\vec a|^2.∣c−a∣2=∣c∣2+∣a∣2.

Thus,

∣c⃗−a⃗∣2=(23sin⁡α)2+22=49sin⁡2α+4.|\vec c-\vec a|^2 = \left(\frac{2}{3\sin\alpha}\right)^2 + 2^2 = \frac{4}{9\sin^2\alpha} + 4.∣c−a∣2=(3sinα2​)2+22=9sin2α4​+4.

Therefore,

27∣c⃗−a⃗∣2=27(49sin⁡2α+4)=12sin⁡2α+108.27|\vec c-\vec a|^2 = 27\left(\frac{4}{9\sin^2\alpha}+4\right) = \frac{12}{\sin^2\alpha}+108.27∣c−a∣2=27(9sin2α4​+4)=sin2α12​+108.
  1. Minimize the expression

We are given

α∈[0,π3].\alpha \in \left[0,\frac{\pi}{3}\right].α∈[0,3π​].

Since a⃗=b⃗×c⃗\vec a=\vec b\times\vec ca=b×c and ∣a⃗∣=2≠0|\vec a|=2\neq 0∣a∣=2=0, we must have sin⁡α≠0\sin\alpha \neq 0sinα=0, so effectively

α∈(0,π3].\alpha \in \left(0,\frac{\pi}{3}\right].α∈(0,3π​].

Now, to minimize

12sin⁡2α+108,\frac{12}{\sin^2\alpha}+108,sin2α12​+108,

we need to maximize sin⁡2α\sin^2\alphasin2α.

On (0,π3]\left(0,\frac{\pi}{3}\right](0,3π​], the maximum value of sin⁡α\sin\alphasinα occurs at

α=π3,sin⁡π3=32.\alpha = \frac{\pi}{3}, \quad \sin\frac{\pi}{3} = \frac{\sqrt3}{2}.α=3π​,sin3π​=23​​.

So,

sin⁡2α=34.\sin^2\alpha = \frac{3}{4}.sin2α=43​.

Hence,

27∣c⃗−a⃗∣min⁡2=123/4+108=16+108=124.27|\vec c-\vec a|^2_{\min} = \frac{12}{3/4}+108 = 16+108=124.27∣c−a∣min2​=3/412​+108=16+108=124.
  1. Check options
  • A: 124124124 ✅
  • B: 110110110
  • C: 121121121
  • D: 105105105

So the correct option is A.

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