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Vector Algebra question

2017 · Shift 2 · Q34
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  5. /2017 · Shift 2 · Q34

Vector Algebra question

2017 · Shift 2 · Q34

JEE AdvancedMathematicsVector AlgebraMCQ+3 / −1
Let O be the origin and OX→\overrightarrow{OX}OX, OY→\overrightarrow{OY}OY, OZ→\overrightarrow{OZ}OZ be three unit vectors in the directions of the sides QR→\overrightarrow{QR}QR​, RP→\overrightarrow{RP}RP, PQ→\overrightarrow{PQ}PQ​ respectively, of a triangle PQR.|OX→×OY→\overrightarrow{OX} \times \overrightarrow{OY}OX×OY| = ?
  1. A
    sin(P + Q)
  2. B
    sin(P + R)
  3. C
    sin(Q + R)
  4. D
    sin2R
View written solutionFree

Correct answer: A

  1. Interpret the given vectors

Let

OX⃗, OY⃗, OZ⃗\vec{OX},\ \vec{OY},\ \vec{OZ}OX, OY, OZ

be unit vectors along the directions of

QR⃗, RP⃗, PQ⃗\vec{QR},\ \vec{RP},\ \vec{PQ}QR​, RP, PQ​

respectively.

We need to find

∣OX⃗×OY⃗∣.|\vec{OX}\times \vec{OY}|.∣OX×OY∣.

Since OX⃗\vec{OX}OX and OY⃗\vec{OY}OY are unit vectors, we use:

∣a⃗×b⃗∣=∣a⃗∣ ∣b⃗∣sin⁡θ=sin⁡θ|\vec{a}\times \vec{b}| = |\vec{a}|\,|\vec{b}|\sin\theta = \sin\theta∣a×b∣=∣a∣∣b∣sinθ=sinθ

where θ\thetaθ is the angle between them.

So our task reduces to finding the angle between the directions of QR⃗\vec{QR}QR​ and RP⃗\vec{RP}RP.

  1. Angle between QR⃗\vec{QR}QR​ and RP⃗\vec{RP}RP

In triangle PQRPQRPQR, the interior angle at vertex RRR is

∠QRP=R.\angle QRP = R.∠QRP=R.

Now:

  • RQ⃗\vec{RQ}RQ​ and RP⃗\vec{RP}RP make angle RRR.
  • But QR⃗\vec{QR}QR​ is opposite in direction to RQ⃗\vec{RQ}RQ​.

Therefore, the angle between QR⃗\vec{QR}QR​ and RP⃗\vec{RP}RP is

π−R.\pi - R.π−R.

Hence,

∣OX⃗×OY⃗∣=sin⁡(π−R).|\vec{OX}\times \vec{OY}| = \sin(\pi - R).∣OX×OY∣=sin(π−R).

Using

sin⁡(π−R)=sin⁡R,\sin(\pi - R)=\sin R,sin(π−R)=sinR,

we get

∣OX⃗×OY⃗∣=sin⁡R.|\vec{OX}\times \vec{OY}| = \sin R.∣OX×OY∣=sinR.
  1. Express in terms of the options

Since in a triangle,

P+Q+R=π,P+Q+R=\pi,P+Q+R=π,

so

P+Q=π−R.P+Q=\pi-R.P+Q=π−R.

Therefore,

sin⁡(P+Q)=sin⁡(π−R)=sin⁡R.\sin(P+Q)=\sin(\pi-R)=\sin R.sin(P+Q)=sin(π−R)=sinR.

Thus,

∣OX⃗×OY⃗∣=sin⁡(P+Q).|\vec{OX}\times \vec{OY}| = \sin(P+Q).∣OX×OY∣=sin(P+Q).
  1. Check options
  • A: sin⁡(P+Q)=sin⁡R\sin(P+Q)=\sin Rsin(P+Q)=sinR ✅
  • B: sin⁡(P+R)=sin⁡(π−Q)=sin⁡Q\sin(P+R)=\sin(\pi-Q)=\sin Qsin(P+R)=sin(π−Q)=sinQ ❌
  • C: sin⁡(Q+R)=sin⁡(π−P)=sin⁡P\sin(Q+R)=\sin(\pi-P)=\sin Psin(Q+R)=sin(π−P)=sinP ❌
  • D: sin⁡2R\sin 2Rsin2R ❌

So the correct option is A.

  1. Comparison with stored answer

Stored correct answer: A

Derived answer: A

They agree.

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