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Question: The points with position vectors \(60\mathbf{i} + 3\mathbf{j}\), \(40\mathbf{i} - 8\mathbf{j},\) , \...

The points with position vectors 60i+3j60\mathbf{i} + 3\mathbf{j}, 40i−8j,40\mathbf{i} - 8\mathbf{j}, , ai−52ja\mathbf{i} - 52\mathbf{j} are collinear, if a=a =

A

– 40

B

40

C

20

D

None of these

Answer

– 40

Explanation

Solution

If the given points be A, B,C, then AB→=k(BC→)\overset{\rightarrow}{AB} = k(\overset{\rightarrow}{BC})

⇒−20i−11j=k[(a−40)i−44j]\Rightarrow - 20\mathbf{i} - 11\mathbf{j} = k\left\lbrack (a - 40)\mathbf{i} - 44\mathbf{j} \right\rbrack

On comparing, −11=−44k⇒k=14- 11 = - 44k \Rightarrow k = \frac{1}{4}

And −20=14(a−40)⇒a=−40.- 20 = \frac{1}{4}(a - 40) \Rightarrow a = - 40.