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KCET2015MathematicsBinomial Theorem

If direction cosines of a vector of magnitude 3 are ( 2 3 ,- 1 3 , 2 3 ) and (a>0 ), then vector is

Options

  1. A(2 i+j+2 k )
  2. B(2 i-j+2 k )
  3. C(i-2 j+2 k )
  4. D(i+2 j+2 k )

Correct answer

B. (2 i-j+2 k )

Step-by-step solution

Given that, direction cosines are ( 2 3 , -a 3 , 2 3 ) We know that, ( l²+m²+n²=1 ) So, ( ( 2 3 )²+ ( -a 3 )²+ ( 2 3 )²=1 ) ( 4 9 + a² 9 + 4 9 =1 ) ( a²+8 9 =1 a²=1 a= 1 ) Vector is given by [ V =| V | ( array l i +m j +n k array ) ] Given magnitude of the vector is ( 3 ). So, required vector is ( 3 ( 2 3 i - 1 3 j + 2 3 k )=2 i - j +2 k )

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