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Question

Mathematics Question on Vector Algebra

Three points (2,1,3)(2, -1,3), (3,5,1)(3, -5,1) and (1,11,9)(-1, 11, 9) are

A

Non-collinear

B

Non-coplanar

C

Collinear

D

None of these

Answer

Collinear

Explanation

Solution

Let AA, BB and CC be three points whose coordinates are (2,1,3)(2, -1,3), (3,5,1)(3, -5,1) and (1,11,9)(-1,11,9) respectively, then OA=2i^j^+3k^\overrightarrow{OA} = 2\hat{i} - \hat{j} +3\hat{k}, OB=3i^5j^+k^ \overrightarrow{OB} = 3\hat{i} - 5\hat{j} +\hat{k} and OC=i^11j^+9k^\overrightarrow{OC} = -\hat{i} -11 \hat{j} +9\hat{k} AB=OBOA=(3i^5j^+k^)(2i^j^+3k^)\therefore \overrightarrow{AB}= \overrightarrow{OB} - \overrightarrow{OA} = \left( 3\hat{i} - 5\hat{j} +\hat{k}\right) - \left( 2\hat{i} - \hat{j} +3\hat{k}\right) =i^4j^2k^= \hat{i} - 4\hat{j} - 2\hat{k} AC=OCOA=(i^11j^+9k^)(2i^j^+3k^) \overrightarrow{AC} = \overrightarrow{OC} - \overrightarrow{OA} = \left( -\hat{i} -11 \hat{j} +9\hat{k}\right) - \left( 2\hat{i} - \hat{j} +3\hat{k}\right) =3i^+12j^+6k^= - 3\hat{i} +12 \hat{j} +6\hat{k} AC=3AB\Rightarrow \overrightarrow{AC} =-3 \overrightarrow{AB} Thus, the vector AB\overrightarrow{AB} and AC\overrightarrow{AC} are parallel having the same initial point AA. Hence, the points AA, BB, CC are collinear.