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Vector Algebra question

2004 · Shift 0 · Q113
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Vector Algebra question

2004 · Shift 0 · Q113

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a→,b→\overrightarrow a ,\overrightarrow ba,b and c→\overrightarrow cc be non-zero vectors such that (a→×b→)×c→=13∣b→∣∣c→∣a→  .\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a \,\,.(a×b)×c=31​​b​​c​a. If θ\thetaθ is the acute angle between the vectors b→{\overrightarrow b }b and c→,{\overrightarrow c },c, then sinθsin\thetasinθ equals :
  1. A
    223{{2\sqrt 2 } \over 3}322​​
  2. B
    23{{\sqrt 2 } \over 3}32​​
  3. C
    23{2 \over 3}32​
  4. D
    13{1 \over 3}31​
View written solutionFree

Correct answer: A

  1. Use the vector triple product identity

We use

(a⃗×b⃗)×c⃗=b⃗(a⃗⋅c⃗)−a⃗(b⃗⋅c⃗).(\vec a \times \vec b) \times \vec c = \vec b(\vec a \cdot \vec c) - \vec a(\vec b \cdot \vec c).(a×b)×c=b(a⋅c)−a(b⋅c).

Given,

(a⃗×b⃗)×c⃗=13∣b⃗∣ ∣c⃗∣ a⃗.(\vec a \times \vec b) \times \vec c = \frac13 |\vec b|\,|\vec c|\,\vec a.(a×b)×c=31​∣b∣∣c∣a.

So,

b⃗(a⃗⋅c⃗)−a⃗(b⃗⋅c⃗)=13∣b⃗∣ ∣c⃗∣ a⃗.\vec b(\vec a \cdot \vec c) - \vec a(\vec b \cdot \vec c)= \frac13 |\vec b|\,|\vec c|\,\vec a.b(a⋅c)−a(b⋅c)=31​∣b∣∣c∣a.
  1. Compare directions

The right-hand side is purely along a⃗\vec aa. Therefore the left-hand side must also be along a⃗\vec aa only.

Now,

b⃗(a⃗⋅c⃗)\vec b(\vec a \cdot \vec c)b(a⋅c)

is along b⃗\vec bb, and

−a⃗(b⃗⋅c⃗)-\vec a(\vec b \cdot \vec c)−a(b⋅c)

is along a⃗\vec aa.

For their sum to be purely along a⃗\vec aa, we must have

a⃗⋅c⃗=0.\vec a \cdot \vec c = 0.a⋅c=0.

Hence,

−a⃗(b⃗⋅c⃗)=13∣b⃗∣ ∣c⃗∣ a⃗.-\vec a(\vec b \cdot \vec c)= \frac13 |\vec b|\,|\vec c|\,\vec a.−a(b⋅c)=31​∣b∣∣c∣a.

Since a⃗≠0⃗\vec a \ne \vec 0a=0, comparing coefficients of a⃗\vec aa gives

−(b⃗⋅c⃗)=13∣b⃗∣ ∣c⃗∣.-(\vec b \cdot \vec c)= \frac13 |\vec b|\,|\vec c|.−(b⋅c)=31​∣b∣∣c∣.

Thus,

b⃗⋅c⃗=−13∣b⃗∣ ∣c⃗∣.\vec b \cdot \vec c = -\frac13 |\vec b|\,|\vec c|.b⋅c=−31​∣b∣∣c∣.
  1. Find the angle between b⃗\vec bb and c⃗\vec cc

Using

b⃗⋅c⃗=∣b⃗∣ ∣c⃗∣cos⁡θ,\vec b \cdot \vec c = |\vec b|\,|\vec c|\cos\theta,b⋅c=∣b∣∣c∣cosθ,

we get

∣b⃗∣ ∣c⃗∣cos⁡θ=−13∣b⃗∣ ∣c⃗∣.|\vec b|\,|\vec c|\cos\theta = -\frac13 |\vec b|\,|\vec c|.∣b∣∣c∣cosθ=−31​∣b∣∣c∣.

So,

cos⁡θ=−13.\cos\theta = -\frac13.cosθ=−31​.

Now,

sin⁡θ=1−cos⁡2θ=1−19=89=223.\sin\theta = \sqrt{1-\cos^2\theta} = \sqrt{1-\frac19} = \sqrt{\frac89} = \frac{2\sqrt2}{3}.sinθ=1−cos2θ​=1−91​​=98​​=322​​.

Since sine is positive for the acute-angle value asked,

sin⁡θ=223.\boxed{\sin\theta = \frac{2\sqrt2}{3}}.sinθ=322​​​.
  1. Check options
  • A: 223\dfrac{2\sqrt2}{3}322​​ ✅
  • B: 23\dfrac{\sqrt2}{3}32​​ ❌
  • C: 23\dfrac{2}{3}32​ ❌
  • D: 13\dfrac{1}{3}31​ ❌

Therefore, the correct option is A.

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