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Vector Algebra question

2014 · Shift 1 · Q31
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  5. /2014 · Shift 1 · Q31

Vector Algebra question

2014 · Shift 1 · Q31

JEE AdvancedMathematicsVector AlgebraNumerical+3 / −1
Let a→  ,  b→\overrightarrow a \,\,,\,\,\overrightarrow ba,b and c→\overrightarrow cc be three non-coplanar unit vectors such that the angle between every pair of them is π3.{\pi \over 3}.3π​. If a→×b→+b→×c→=pa→+qb→+rc→,\overrightarrow a \times \overrightarrow b + \overrightarrow b \times \overrightarrow c = p\overrightarrow a + q\overrightarrow b + r\overrightarrow c ,a×b+b×c=pa+qb+rc, where p,qp,qp,q and rrr are scalars, then the value of p2+2q2+r2q2{{{p^2} + 2{q^2} + {r^2}} \over {{q^2}}}q2p2+2q2+r2​ is
Numerical answer
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Correct answer: 4

  1. Given information

We have three non-coplanar unit vectors a⃗,b⃗,c⃗\vec a, \vec b, \vec ca,b,c such that the angle between every pair is π/3\pi/3π/3.

So,

a⃗⋅b⃗=b⃗⋅c⃗=c⃗⋅a⃗=cos⁡π3=12\vec a\cdot \vec b = \vec b\cdot \vec c = \vec c\cdot \vec a = \cos\frac{\pi}{3} = \frac12a⋅b=b⋅c=c⋅a=cos3π​=21​

and

∣a⃗∣=∣b⃗∣=∣c⃗∣=1.|\vec a|=|\vec b|=|\vec c|=1.∣a∣=∣b∣=∣c∣=1.

We are given

a⃗×b⃗+b⃗×c⃗=pa⃗+qb⃗+rc⃗.\vec a\times \vec b + \vec b\times \vec c = p\vec a + q\vec b + r\vec c.a×b+b×c=pa+qb+rc.

We need to find

p2+2q2+r2q2.\frac{p^2+2q^2+r^2}{q^2}.q2p2+2q2+r2​.
  1. Use dot product with a⃗,b⃗,c⃗\vec a, \vec b, \vec ca,b,c

Let

X⃗=a⃗×b⃗+b⃗×c⃗=pa⃗+qb⃗+rc⃗.\vec X = \vec a\times \vec b + \vec b\times \vec c = p\vec a + q\vec b + r\vec c.X=a×b+b×c=pa+qb+rc.

We now take dot product with each of a⃗,b⃗,c⃗\vec a,\vec b,\vec ca,b,c.

Dot with a⃗\vec aa

a⃗⋅(a⃗×b⃗)=0\vec a\cdot(\vec a\times \vec b)=0a⋅(a×b)=0

and

a⃗⋅(b⃗×c⃗)=[a⃗ b⃗ c⃗],\vec a\cdot(\vec b\times \vec c)=[\vec a\ \vec b\ \vec c],a⋅(b×c)=[a b c],

where [a⃗ b⃗ c⃗][\vec a\ \vec b\ \vec c][a b c] denotes the scalar triple product.

Thus,

a⃗⋅X⃗=[a⃗ b⃗ c⃗].\vec a\cdot \vec X = [\vec a\ \vec b\ \vec c].a⋅X=[a b c].

From the RHS,

a⃗⋅(pa⃗+qb⃗+rc⃗)=p+q2+r2.\vec a\cdot(p\vec a+q\vec b+r\vec c)=p+\frac q2+\frac r2.a⋅(pa+qb+rc)=p+2q​+2r​.

Hence,

p+q2+r2=T(1)p+\frac q2+\frac r2 = T \qquad (1)p+2q​+2r​=T(1)

where T=[a⃗ b⃗ c⃗]T=[\vec a\ \vec b\ \vec c]T=[a b c].

Dot with b⃗\vec bb

b⃗⋅(a⃗×b⃗)=0,\vec b\cdot(\vec a\times \vec b)=0,b⋅(a×b)=0, b⃗⋅(b⃗×c⃗)=0.\vec b\cdot(\vec b\times \vec c)=0.b⋅(b×c)=0.

So,

b⃗⋅X⃗=0.\vec b\cdot\vec X=0.b⋅X=0.

From RHS,

b⃗⋅(pa⃗+qb⃗+rc⃗)=p2+q+r2.\vec b\cdot(p\vec a+q\vec b+r\vec c)=\frac p2+q+\frac r2.b⋅(pa+qb+rc)=2p​+q+2r​.

Therefore,

p2+q+r2=0.(2)\frac p2+q+\frac r2=0. \qquad (2)2p​+q+2r​=0.(2)

Dot with c⃗\vec cc

c⃗⋅(a⃗×b⃗)=[c⃗ a⃗ b⃗]=T,\vec c\cdot(\vec a\times \vec b)=[\vec c\ \vec a\ \vec b]=T,c⋅(a×b)=[c a b]=T,

(using cyclic permutation), and

c⃗⋅(b⃗×c⃗)=0.\vec c\cdot(\vec b\times \vec c)=0.c⋅(b×c)=0.

So,

c⃗⋅X⃗=T.\vec c\cdot\vec X=T.c⋅X=T.

From RHS,

c⃗⋅(pa⃗+qb⃗+rc⃗)=p2+q2+r.\vec c\cdot(p\vec a+q\vec b+r\vec c)=\frac p2+\frac q2+r.c⋅(pa+qb+rc)=2p​+2q​+r.

Hence,

p2+q2+r=T.(3)\frac p2+\frac q2+r=T. \qquad (3)2p​+2q​+r=T.(3)
  1. Solve for p,q,rp,q,rp,q,r in terms of TTT

From (1) and (3):

p+q2+r2=p2+q2+rp+\frac q2+\frac r2 = \frac p2+\frac q2+rp+2q​+2r​=2p​+2q​+r ⇒p2−r2=0\Rightarrow \frac p2 - \frac r2 = 0⇒2p​−2r​=0 ⇒p=r.\Rightarrow p=r.⇒p=r.

Substitute r=pr=pr=p into (2):

p2+q+p2=0\frac p2+q+\frac p2=02p​+q+2p​=0 p+q=0p+q=0p+q=0 q=−p.q=-p.q=−p.

Also r=pr=pr=p, so

p=r=−q.p=r=-q.p=r=−q.

Thus let p=tp=tp=t, then

q=−t,r=t.q=-t, \qquad r=t.q=−t,r=t.
  1. Compute required expression

Now,

p2+2q2+r2=t2+2t2+t2=4t2.p^2+2q^2+r^2=t^2+2t^2+t^2=4t^2.p2+2q2+r2=t2+2t2+t2=4t2.

Also,

q2=t2.q^2=t^2.q2=t2.

Therefore,

p2+2q2+r2q2=4t2t2=4.\frac{p^2+2q^2+r^2}{q^2} = \frac{4t^2}{t^2}=4.q2p2+2q2+r2​=t24t2​=4.
  1. Final answer
4\boxed{4}4​

This matches the stored correct answer.

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