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BITSAT 2013
Exam
BITSAT
Year
2013
Subject
Mathematics
Chapter
Vector Algebra

BITSAT 2013 · Mathematics · Vector Algebra

If vectors 2 i-j+k, i+2 j-3 k and 3 i+aj+5 k are coplanar, then the...

If vectors 2i−j+k,i+2j−3k2 \mathrm{i}-\mathrm{j}+\mathrm{k}, \mathrm{i}+2 \mathrm{j}-3 \mathrm{k} and 3i+aj+5k3 \mathrm{i}+\mathrm{aj}+5 \mathrm{k} are coplanar, then the value of a is

Options

  1. A2
  2. B-2
  3. C-1
  4. D-4

Correct answer

D. -4

Step-by-step solution

If given vectors are coplanar, then there exists two scalar quantities x\mathrm{x} and y\mathrm{y} such that 2i^−j^+k^=x(i^+2j^−3k^)+y(3i^+aj^+5k^)2 \hat{i}-\hat{j}+\hat{k}=x(\hat{i}+2 \hat{j}-3 \hat{k})+y(3 \hat{i}+a \hat{j}+5 \hat{k}) Comparing coefficient of i^,j^\hat{i}, \hat{j} and k^\hat{k} on both sides of (1) we get x+3y=2,2x+ay=−1,−3x+5y=1…(2)x+3 y=2,2 x+a y=-1,-3 x+5 y=1 \ldots(2) Solving first and third equations, we get x=1/2,y=1/2\mathrm{x}=1 / 2, \mathrm{y}=1 / 2 Since the vectors are coplanar, therefore these values of xx and yy will satisfy the equation 2x+ay=−1∴2(1/2)+a(1/2)=−1⇒a=−42 x+a y=-1\therefore 2(1 / 2)+a(1 / 2)=-1 \Rightarrow a=-4
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