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Home  >>  TN XII Math  >>  Vector Algebra
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Verify $(\overrightarrow{a}\times\overrightarrow{b}) \times (\overrightarrow{c}\times\overrightarrow{d})=[\overrightarrow{a} \overrightarrow{b} \overrightarrow{d} ] \overrightarrow{c} - [\overrightarrow{a} \overrightarrow{b} \overrightarrow{c} ] \overrightarrow{d}$, For $\overrightarrow{a}=\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{k},\overrightarrow{ b}=\overrightarrow{2i}+\overrightarrow{k}, \overrightarrow{c}=\overrightarrow{2i}+\overrightarrow{j}+\overrightarrow{k}, \overrightarrow{d}=\overrightarrow{i}+\overrightarrow{j}+\overrightarrow{2k}.$

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  • If $ \overrightarrow a = a_1\overrightarrow i + a_2 \overrightarrow j+a_3 \overrightarrow k,\: \: \overrightarrow b = b_1 \overrightarrow i+b_2\overrightarrow j + b_3 \overrightarrow k$ then $ \overrightarrow a.\overrightarrow b = a_1b_1+a_2b_2+a_3b_3$
  • If $ \overrightarrow a = a_1\overrightarrow i+a_2\overrightarrow j+a_3\overrightarrow k, \: \overrightarrow b = b_1\overrightarrow i+b_2\overrightarrow j+b_3\overrightarrow k$ then $ \overrightarrow a$ x $ \overrightarrow b = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}$
LHS $ \overrightarrow a \times \overrightarrow b = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 1 & 1 & 1 \\ 2 & 0 & 1 \end{vmatrix}=(1-0)\overrightarrow i-(1-2)\overrightarrow j+(0-2)\overrightarrow k$
$ = \overrightarrow i+\overrightarrow j-2\overrightarrow k$
$ \overrightarrow c \times \overrightarrow d = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 2 & 1 & 1 \\ 1 & 1 & 2 \end{vmatrix} = (2-1)\overrightarrow i-(4-1)\overrightarrow j+(2-1)\overrightarrow k$
$ = \overrightarrow i-3\overrightarrow j+\overrightarrow k$
$( \overrightarrow a \times \overrightarrow b) \times (\overrightarrow c \times \overrightarrow d) = \begin{vmatrix}\overrightarrow i & \overrightarrow j & \overrightarrow k \\ 1 & 1 & -2 \\ 1 & -3 & 1 \end{vmatrix} = (1-6)\overrightarrow i-(1+2)\overrightarrow j+(-3-1)\overrightarrow k$
$ = -5\overrightarrow i-3\overrightarrow j-4\overrightarrow k$
RHS $ [\overrightarrow a\: \overrightarrow b\: \overrightarrow d] = \begin{vmatrix} 1 & 1 & 1 \\ 2 & 0 & 1 \\ 1 & 1 & 2 \end{vmatrix} = 1(0-1)-1(4-1)+1(2-0)$
$ = -1-3+2=-2$
$ [ \overrightarrow a \: \overrightarrow b\: \overrightarrow c] = \begin{vmatrix}1 & 1 & 1 \\ 2 & 0 & 1 \\ 2 & 1 & 1 \end{vmatrix}= 1(0-1)-1(2-2)+1(2-0)$
$ -1+2=1$
$ \overrightarrow a\: \overrightarrow b \:\overrightarrow d]\overrightarrow c - [\overrightarrow a\: \overrightarrow b\: \overrightarrow c]\overrightarrow d = -2\overrightarrow c-\overrightarrow d$
$ = -4\overrightarrow i-2\overrightarrow j-2\overrightarrow k-\overrightarrow i-\overrightarrow j-2\overrightarrow k = -5\overrightarrow i-3\overrightarrow j-4\overrightarrow k$
LHS = RHS

 

answered Jun 10, 2013 by thanvigandhi_1
edited Jun 23, 2013 by thanvigandhi_1
 

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