Q

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
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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

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