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  • BITSAT The volume of the parallelopiped whose edges are represented by -12i +lambda k , 3j-k , 2i+j -15k is 546. Then the value of lambda

  1.  -5
  2. 6
  3. 1
  4. -3

Answers (1)

Volume of parallelopiped, V = \left ( \vec a \times \vec b \right ). \vec c

Here \vec a = -12i+\lambda k

\vec b = 3j - k

and, \vec c = 2i+j-15k

\vec a \times \vec b = \begin{bmatrix} i & j & k \\ -12 & 0 & \lambda \\ 0& 3 & -1 \end{bmatrix}

\Rightarrow \vec a \times \vec b = -3\lambda i -12j -36k

V = \left ( -3\lambda i -12j -36k \right ).(2i+j-15k) = -6\lambda -12 +540 = -6\lambda +528

Given that V is 546, put it in the above equation.

546 = -6\lambda +528

\Rightarrow \lambda =-3

Option (4) is correct

Posted by

Abhishek Sahu

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