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Home  >>  TN XII Math  >>  Vector Algebra
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Show that the vectors $\overrightarrow{3i}-\overrightarrow{2j}+\overrightarrow{k},\overrightarrow{i}-\overrightarrow{3j}+\overrightarrow{5k}$ and $\overrightarrow{2i}+\overrightarrow{j}-\overrightarrow{4k}$ form a right angled tringle.

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  • By $ \Delta$ law of vectors if $ \overrightarrow a+\overrightarrow b=\overrightarrow c \: or \: \overrightarrow a+\overrightarrow b=-\overrightarrow c$ then the vectors form the sides of a $ \Delta$
Method 1 Step 1
Let $ \overrightarrow a =3\overrightarrow i-2\overrightarrow j+\overrightarrow k, \: \: \overrightarrow b=\overrightarrow i-3\overrightarrow j+5\overrightarrow k, \: \: \overrightarrow c=2\overrightarrow i+\overrightarrow j-4\overrightarrow k$
Now $ \overrightarrow b+\overrightarrow c=3\overrightarrow i-2\overrightarrow j+\overrightarrow k=\overrightarrow a$
$ \therefore \overrightarrow a,\overrightarrow b,\overrightarrow c$ form the sides of a $ \Delta$
Step 2
Consider $ \overrightarrow a.\overrightarrow b = (3)(1)+(-2)(-3)+(1)(5)$
$ = 3+6+5=14 \neq 0 \neq 0$
$ \overrightarrow b.\overrightarrow c=(1)(2)+(-3)(1)+(5)(-4)$
$ = 2-3-20=-21$
$ \overrightarrow c.\overrightarrow a = (2)(3)+(1)(-2)+(-4)(1)$
$ = 6-2-4=0$
Step 3
$ \overrightarrow c.\overrightarrow a=0 \: \therefore \overrightarrow c \perp \overrightarrow a \Rightarrow \overrightarrow a,\overrightarrow b,\overrightarrow c$ form a rightangled $ \Delta$
Method 2 Step 1
It can be seen that $ \overrightarrow b+\overrightarrow c=\overrightarrow a\: \therefore \overrightarrow a, \overrightarrow b, \overrightarrow c$ form the sides of a $ \Delta$
Step 2
Now $ |\overrightarrow a| = a=\sqrt{9+4+1}=\sqrt{14}$
$\: \: \: \: \: \: \: \: \: \: |\overrightarrow b|=b=\sqrt{1+9+25}=\sqrt{35}$
$\: \: \: \: \: \: \: \: \: \: |\overrightarrow c|=c=\sqrt{4+1+16}=\sqrt{21}$
Step 3
Now $ a^2+c^2=14+21=35=b^2\: \therefore$ by converse of pythogorus theorem, they form a right angled $ \Delta$

 

answered Jun 2, 2013 by thanvigandhi_1
edited Jun 3, 2013 by thanvigandhi_1
 

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