# 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.

Toolbox:
• 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 $\Delta$

edited Jun 3, 2013