Let \vec{a},\vec{b}\; \; and\; \; \vec{c} be three non­-zero vectors such that no two of these are collinear. If the vector \vec{a}+2\vec{b}\; is collinear with \vec{c}\; \; and\; \; \vec{b}\;+3\vec{c}\; is collinear with \vec{a}\; (\lambda being some non-­zero scalar) then \vec{a}+2\vec{b}+6\vec{c} equals

  • Option 1)

    \lambda \vec{c}\;

  • Option 2)

    \; \; \lambda \vec{b}\;

  • Option 3)

    \; \; \lambda \vec{a}

  • Option 4)

    0

 

Answers (1)

As we learnt in 

Collinear Vectors -

Two vectors are said to be collinear if their directed line segments are parallel disregards to their direction. 

- wherein

If \vec{a} and \vec{b} are collinear , then \vec{a}= K\vec{b} where K \epsilon R

 

 \vec{a} +\vec{2b} is collinear with \vec{c}

(\vec{a}+\vec{2b}) =p\vec{c}

Similarily (\vec{b}+\vec{3b}) =q\vec{a}

Eliminating \vec{b}

we get \vec{a} +\vec{2b} -\vec{2b} -\vec{6c} = p\vec{c} -2q\vec{a}

\vec{a}\left ( 1+2q \right )+\vec{c}(-p-6)=0

Thus p= -6, q=\frac{-1}{2}

Now

\vec{a}+\vec{2b} = p\vec{c}

\Rightarrow \vec{a}+\vec{2b}+\vec{6c} = 0


Option 1)

\lambda \vec{c}\;

Incorrect

Option 2)

\; \; \lambda \vec{b}\;

Incorrect

Option 3)

\; \; \lambda \vec{a}

Incorrect

Option 4)

0

Correct

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