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Home  >>  TN XII Math  >>  Vector Algebra
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If $ \overrightarrow{a}= \overrightarrow{2i}+ \overrightarrow{3j}- \overrightarrow{5k}, \overrightarrow{b}= -\overrightarrow{1}+ \overrightarrow{j}+ \overrightarrow{2k}$ and $ \overrightarrow{c}= \overrightarrow{4i}- \overrightarrow{2j}+ \overrightarrow{3k}, $ Show that $( \overrightarrow{a}\times \overrightarrow{b})\times \overrightarrow{c} \neq \overrightarrow{a} \times( \overrightarrow{b}\times \overrightarrow{c})$

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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}$
  • Vector triple product $(\overrightarrow a \times \overrightarrow b) \times \overrightarrow c=(\overrightarrow a.\overrightarrow c)\overrightarrow b-(\overrightarrow b.\overrightarrow c)\overrightarrow a$ $\overrightarrow a \times (\overrightarrow b \times \overrightarrow c)= (\overrightarrow a.\overrightarrow c)\overrightarrow b - (\overrightarrow a.\overrightarrow b)\overrightarrow c$
Method 1
$ \overrightarrow a \times \overrightarrow b = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 2 & 3 & -5 \\ -1 & 1 & 2 \end{vmatrix} = (6+5)\overrightarrow i-(4-5)\overrightarrow j+(2+3)\overrightarrow k$
$ = 11\overrightarrow i+\overrightarrow j+5\overrightarrow k$
$(\overrightarrow a \times \overrightarrow b) \times \overrightarrow c = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 11 & 1 & 5 \\ 4 & -2 & 3 \end{vmatrix} = (3+10)\overrightarrow i-(33-20)\overrightarrow j+(-22-4)\overrightarrow k$
$ 13\overrightarrow i-13\overrightarrow j-26\overrightarrow k $ (i)
$ \overrightarrow b \times \overrightarrow c = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ -1 & 1 & 2 \\ 4 & -2 & 3 \end{vmatrix} = (3+4)\overrightarrow i-(-3-8)\overrightarrow j+(2-4)\overrightarrow k$
$ = 7\overrightarrow i+11\overrightarrow j-2\overrightarrow k$
$ \overrightarrow a \times ( \overrightarrow b \times \overrightarrow c) = \begin{vmatrix} \overrightarrow i & \overrightarrow j & \overrightarrow k \\ 2 & 3 & -5 \\ 7 & 11 & -2 \end{vmatrix} = (-6+55)\overrightarrow i-(-4+35)\overrightarrow j+(22-21)\overrightarrow k$
$ = -49\overrightarrow i-31\overrightarrow j+\overrightarrow k$ (ii)
From (i) and (ii) $ (\overrightarrow a \times \overrightarrow b)\times \overrightarrow c \neq \overrightarrow a \times (\overrightarrow b \times \overrightarrow c)$
Method 2
$(\overrightarrow a \times \overrightarrow b) \times \overrightarrow c = (\overrightarrow a.\overrightarrow c)\overrightarrow b-(\overrightarrow b.\overrightarrow c)\overrightarrow a$
$ \overrightarrow a.\overrightarrow c = (2)(4)+3(-2)+(-5)3=8-6-15=-13$
$ \overrightarrow b.\overrightarrow c=(-1)(4)+(1)(-2)+(2)(3)=-4-2+6=0$
$ (\overrightarrow a-\overrightarrow c)\overrightarrow b-(\overrightarrow b.\overrightarrow c)\overrightarrow a = -13\overrightarrow b=13\overrightarrow i-13\overrightarrow j-26\overrightarrow k$
$ \therefore (\overrightarrow a \times \overrightarrow b) \times \overrightarrow c = 13\overrightarrow i-13\overrightarrow j-26\overrightarrow k $ (i)
$ \overrightarrow a \times (\overrightarrow b \times \overrightarrow c)=(\overrightarrow a.\overrightarrow c)\overrightarrow b-(\overrightarrow a.\overrightarrow b)\overrightarrow c$
$ \overrightarrow a.\overrightarrow c=-13$
$ \overrightarrow a.\overrightarrow b=(2)(-1)+3(1)+(-5)(2)=-2+3-10=9$
$ (\overrightarrow a-\overrightarrow c)\overrightarrow b-(\overrightarrow a.\overrightarrow b)\overrightarrow c=-13\overrightarrow b+9\overrightarrow c$
$ = 13\overrightarrow i-13\overrightarrow j-26\overrightarrow k+36\overrightarrow i-18\overrightarrow j+27\overrightarrow k$
$ = 49\overrightarrow i-31\overrightarrow j+\overrightarrow k$
$ \therefore \overrightarrow a \times (\overrightarrow b \times \overrightarrow c)=49\overrightarrow i -31\overrightarrow j+\overrightarrow k$ (ii)
From (i) and (ii) $ (\overrightarrow a \times \overrightarrow b) \times \overrightarrow c \neq \overrightarrow a \times (\overrightarrow b \times \overrightarrow c)$

 

answered Jun 9, 2013 by thanvigandhi_1
edited Jun 23, 2013 by thanvigandhi_1
 

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