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Vector Algebra question

2011 · Shift 0 · Q52
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  5. /2011 · Shift 0 · Q52

Vector Algebra question

2011 · Shift 0 · Q52

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a→\overrightarrow aa, b→\overrightarrow bb, c→\overrightarrow cc be three non-zero vectors which are pairwise non-collinear. If a→+3b→\overrightarrow a+3 \overrightarrow ba+3b is collinear with c→\overrightarrow cc and b→+2c→\overrightarrow b+2 \overrightarrow cb+2c is collinear with a→\overrightarrow aa, then a→+b→+6c→\overrightarrow a+\overrightarrow b+6 \overrightarrow ca+b+6c is :
  1. A
    a→+c→\overrightarrow a+\overrightarrow ca+c
  2. B
    c→\overrightarrow cc
  3. C
    a→\overrightarrow aa
  4. D
    0→\overrightarrow 00
View written solutionFree

Correct answer: QUESTION LIKELY HAS A TYPO., IF THE INTENDED EXPRESSION IS $\VEC A+3\VEC B+6\VEC C$, THEN THE CORRECT ANSWER IS $D$ I.E. $\VEC 0$.

  1. Translate the collinearity conditions into equations

Since a⃗+3b⃗\vec a+3\vec ba+3b is collinear with c⃗\vec cc, there exists a scalar λ\lambdaλ such that

a⃗+3b⃗=λc⃗(1)\vec a+3\vec b=\lambda \vec c \qquad (1)a+3b=λc(1)

Similarly, since b⃗+2c⃗\vec b+2\vec cb+2c is collinear with a⃗\vec aa, there exists a scalar μ\muμ such that

b⃗+2c⃗=μa⃗(2)\vec b+2\vec c=\mu \vec a \qquad (2)b+2c=μa(2)

We are given that a⃗,b⃗,c⃗\vec a,\vec b,\vec ca,b,c are pairwise non-collinear and non-zero.

  1. Use (2) to express b⃗\vec bb in terms of a⃗\vec aa and c⃗\vec cc

From (2),

b⃗=μa⃗−2c⃗\vec b=\mu \vec a-2\vec cb=μa−2c

Substitute this into (1):

a⃗+3(μa⃗−2c⃗)=λc⃗\vec a+3(\mu \vec a-2\vec c)=\lambda \vec ca+3(μa−2c)=λc (1+3μ)a⃗−6c⃗=λc⃗(1+3\mu)\vec a-6\vec c=\lambda \vec c(1+3μ)a−6c=λc (1+3μ)a⃗=(λ+6)c⃗(1+3\mu)\vec a=(\lambda+6)\vec c(1+3μ)a=(λ+6)c

Now a⃗\vec aa and c⃗\vec cc are non-collinear, so the only way a scalar multiple of a⃗\vec aa can equal a scalar multiple of c⃗\vec cc is if both coefficients are zero.

Hence,

1+3μ=0,λ+6=01+3\mu=0, \qquad \lambda+6=01+3μ=0,λ+6=0

So,

μ=−13,λ=−6\mu=-\frac13, \qquad \lambda=-6μ=−31​,λ=−6
  1. Find the required vector

From (2), using μ=−13\mu=-\frac13μ=−31​,

b⃗+2c⃗=−13a⃗\vec b+2\vec c=-\frac13\vec ab+2c=−31​a

Multiply by 333:

3b⃗+6c⃗=−a⃗3\vec b+6\vec c=-\vec a3b+6c=−a

Bring all terms to one side:

a⃗+3b⃗+6c⃗=0⃗\vec a+3\vec b+6\vec c=\vec 0a+3b+6c=0

But we need a⃗+b⃗+6c⃗\vec a+\vec b+6\vec ca+b+6c? Let us derive carefully from the equations directly.

From (1), since λ=−6\lambda=-6λ=−6,

a⃗+3b⃗=−6c⃗\vec a+3\vec b=-6\vec ca+3b=−6c

Thus,

a⃗+3b⃗+6c⃗=0⃗\vec a+3\vec b+6\vec c=\vec 0a+3b+6c=0

Now check the expression in the question: a⃗+b⃗+6c⃗\vec a+\vec b+6\vec ca+b+6c.

Using (2):

b⃗=−13a⃗−2c⃗\vec b=-\frac13\vec a-2\vec cb=−31​a−2c

Therefore,

a⃗+b⃗+6c⃗n=a⃗+(−13a⃗−2c⃗)+6c⃗\vec a+\vec b+6\vec c n=\vec a+\left(-\frac13\vec a-2\vec c\right)+6\vec ca+b+6cn=a+(−31​a−2c)+6c =23a⃗+4c⃗=\frac23\vec a+4\vec c=32​a+4c

This is not immediately one of the options. So let us instead solve fully for relations among a⃗,b⃗,c⃗\vec a,\vec b,\vec ca,b,c.

From (1) with λ=−6\lambda=-6λ=−6,

a⃗+3b⃗=−6c⃗\vec a+3\vec b=-6\vec ca+3b=−6c

So,

a⃗=−3b⃗−6c⃗\vec a=-3\vec b-6\vec ca=−3b−6c

Substitute into (2):

b⃗+2c⃗=μ(−3b⃗−6c⃗)\vec b+2\vec c=\mu(-3\vec b-6\vec c)b+2c=μ(−3b−6c)

Using μ=−13\mu=-\frac13μ=−31​,

b⃗+2c⃗=(−13)(−3b⃗−6c⃗)=b⃗+2c⃗\vec b+2\vec c=(-\tfrac13)(-3\vec b-6\vec c)=\vec b+2\vec cb+2c=(−31​)(−3b−6c)=b+2c

which is consistent.

Thus the only definite relation obtained is

a⃗+3b⃗+6c⃗=0⃗.\vec a+3\vec b+6\vec c=\vec 0.a+3b+6c=0.

So the intended expression must match option DDD only if the question intends a⃗+3b⃗+6c⃗\vec a+3\vec b+6\vec ca+3b+6c rather than a⃗+b⃗+6c⃗\vec a+\vec b+6\vec ca+b+6c.

  1. Compare with the options and stored answer

For the expression written in the question,

a⃗+b⃗+6c⃗=23a⃗+4c⃗,\vec a+\vec b+6\vec c=\frac23\vec a+4\vec c,a+b+6c=32​a+4c,

which is not forced to equal any one fixed option among A, B, C, D.

However, from the given conditions we can prove

a⃗+3b⃗+6c⃗=0⃗,\vec a+3\vec b+6\vec c=\vec 0,a+3b+6c=0,

which matches option DDD.

Therefore, the stored answer DDD is consistent only if there is a typo in the question statement and the intended expression was a⃗+3b⃗+6c⃗\vec a+3\vec b+6\vec ca+3b+6c.

So for the question as written, I do not agree with the stored answer.

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