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Home  >>  TN XII Math  >>  Vector Algebra
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The volume of a parallelopiped whose edges are represented by $-\overrightarrow{12i}+\lambda\overrightarrow{k}, \overrightarrow{3j}-\overrightarrow{k}, \overrightarrow{2i}+\overrightarrow{j}-\overrightarrow{15k} $ is $546;\quad$ find the value of $\lambda$.

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  • $ ( \overrightarrow a \times \overrightarrow b).\overrightarrow c = (\overrightarrow b \times \overrightarrow c)\overrightarrow a = (\overrightarrow c \times \overrightarrow a).\overrightarrow b = [ \overrightarrow a, \overrightarrow b, \overrightarrow c ]$ Also $ [ \overrightarrow a\: \overrightarrow b\: \overrightarrow c] = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}$ = volume of the parallelopiped formed by three non-coplanar vectors $ \overrightarrow a, \overrightarrow b, \overrightarrow c$.
Step 1
The volume of the parallelopiped whose edges are represented by $ \overrightarrow a = -12\overrightarrow i+ \lambda \overrightarrow k, \: \overrightarrow b = 3\overrightarrow j-\overrightarrow k\: and \: \overrightarrow c = 2\overrightarrow i+\overrightarrow j-15\overrightarrow k\: is \: 5+6$
$ [ \overrightarrow a\: \overrightarrow b\: \overrightarrow c]=546$
Step 2
$ \begin{vmatrix} -12 & 0  &   \lambda \\ 0 & 3 & -1 \\ 2 & 1 & -15 \end{vmatrix} = 546$
$ \Rightarrow -12(-45+1)+ \lambda(0-6)=546$
$ \Rightarrow 528-6\lambda=546$
$ \Rightarrow 6\lambda=-18\: or \: \lambda=-3$

 

answered Jun 8, 2013 by thanvigandhi_1
edited Jun 23, 2013 by thanvigandhi_1
 

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