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Vector Algebra question

2004 · Shift 0 · Q115
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Vector Algebra question

2004 · Shift 0 · Q115

JEE MainMathematicsVector AlgebraMCQ+4 / −1
Let a→,b→\overrightarrow a ,\overrightarrow ba,b and c→\overrightarrow cc be three non-zero vectors such that no two of these are collinear. If the vector a→+2b→\overrightarrow a + 2\overrightarrow ba+2b is collinear with c→\overrightarrow cc and b→+3c→\overrightarrow b + 3\overrightarrow cb+3c is collinear with a→\overrightarrow aa (λ\lambdaλ being some non-zero scalar) then a→+2b→+6c→\overrightarrow a + 2\overrightarrow b + 6\overrightarrow ca+2b+6c equals to :
  1. A
    0→\overrightarrow{0}0
  2. B
    λb→\lambda \overrightarrow bλb
  3. C
    λc→\lambda \overrightarrow cλc
  4. D
    λa→\lambda \overrightarrow aλa
View written solutionFree

Correct answer: A

  1. Translate the collinearity conditions into equations

Since a→+2b→\overrightarrow a + 2\overrightarrow ba+2b is collinear with c→\overrightarrow cc, there exists a non-zero scalar mmm such that

a→+2b→=mc→.(1)\overrightarrow a + 2\overrightarrow b = m\overrightarrow c. \qquad (1)a+2b=mc.(1)

Since b→+3c→\overrightarrow b + 3\overrightarrow cb+3c is collinear with a→\overrightarrow aa, there exists a non-zero scalar nnn such that

b→+3c→=na→.(2)\overrightarrow b + 3\overrightarrow c = n\overrightarrow a. \qquad (2)b+3c=na.(2)

We need to find

a→+2b→+6c→.\overrightarrow a + 2\overrightarrow b + 6\overrightarrow c.a+2b+6c.
  1. Use the fact that no two vectors are collinear

From (2),

b→+3c→=na→.\overrightarrow b + 3\overrightarrow c = n\overrightarrow a.b+3c=na.

So,

a→=1nb→+3nc→.\overrightarrow a = \frac{1}{n}\overrightarrow b + \frac{3}{n}\overrightarrow c.a=n1​b+n3​c.

Substitute this into (1):

(1nb→+3nc→)+2b→=mc→.\left(\frac{1}{n}\overrightarrow b + \frac{3}{n}\overrightarrow c\right) + 2\overrightarrow b = m\overrightarrow c.(n1​b+n3​c)+2b=mc.

Rearrange:

(1n+2)b→+3nc→=mc→.\left(\frac{1}{n}+2\right)\overrightarrow b + \frac{3}{n}\overrightarrow c = m\overrightarrow c.(n1​+2)b+n3​c=mc.

Thus,

(1n+2)b→+(3n−m)c→=0→.\left(\frac{1}{n}+2\right)\overrightarrow b + \left(\frac{3}{n}-m\right)\overrightarrow c = \overrightarrow 0.(n1​+2)b+(n3​−m)c=0.

Now b→\overrightarrow bb and c→\overrightarrow cc are not collinear, so they are linearly independent in this relation. Therefore both coefficients must be zero:

1n+2=0⇒n=−12.\frac{1}{n}+2=0 \quad \Rightarrow \quad n=-\frac12.n1​+2=0⇒n=−21​.

Also,

3n−m=0⇒m=3n=3(−2)=−6.\frac{3}{n}-m=0 \quad \Rightarrow \quad m=\frac{3}{n}=3\left(-2\right)=-6.n3​−m=0⇒m=n3​=3(−2)=−6.

Hence,

a→+2b→=−6c→.\overrightarrow a + 2\overrightarrow b = -6\overrightarrow c.a+2b=−6c.

So,

a→+2b→+6c→=0→.\overrightarrow a + 2\overrightarrow b + 6\overrightarrow c = \overrightarrow 0.a+2b+6c=0.
  1. Check options
  • A: 0→\overrightarrow 00 ✔️
  • B: λb→\lambda \overrightarrow bλb ✘
  • C: λc→\lambda \overrightarrow cλc ✘
  • D: λa→\lambda \overrightarrow aλa ✘

So the correct option is A.


  1. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

They agree.

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