# Show that the points whose position vectors $\overrightarrow{4i}-\overrightarrow{3j}+\overrightarrow{k}, \overrightarrow{2i}-\overrightarrow{4j}+\overrightarrow{5k} ,\overrightarrow{i}-\overrightarrow{j}$ 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 the points A, B, C have position vectors $\overrightarrow {OA}=4\overrightarrow i-3\overrightarrow j+\overrightarrow k, \: \overrightarrow {OB}=2\overrightarrow i-4\overrightarrow j+5\overrightarrow k, \: \overrightarrow {OC}=\overrightarrow i-\overrightarrow j.$
Then the sides of the $\Delta$ ABC formed by them are given by
$\overrightarrow {AB}=\overrightarrow {OB}-\overrightarrow {OA}=(2\overrightarrow i-4\overrightarrow j+5\overrightarrow k)-(4\overrightarrow i-3\overrightarrow j+\overrightarrow k)$
$= -2\overrightarrow i-\overrightarrow j+4\overrightarrow k$
$\overrightarrow {BC}=\overrightarrow {OC}-\overrightarrow {OB}=-\overrightarrow i+3\overrightarrow j-5\overrightarrow k$
$\overrightarrow {CA}=\overrightarrow {OA}-\overrightarrow {OC}=3\overrightarrow i-2\overrightarrow j+\overrightarrow k$
Step 2
Now $\overrightarrow {AB}.\overrightarrow {BC}=(-2)(-1)+(-1)(3)+(4)(-5)$
$= 2-3-20=-21 \neq 0$
$\overrightarrow {BC}.\overrightarrow{CA}=(-1)(3)+(3)(-2)+(-5)(1)$
$=-3-6-5=-14 \neq 0$
$\overrightarrow {CA}.\overrightarrow {AB}=(3)(-2)+(-2)(-1)+(1)(4)$
$-6+2+4=0$
Step 3
$\overrightarrow {CA}.\overrightarrow {AB}=0 \Rightarrow {CA} \perp \overrightarrow {AB} \Rightarrow \Delta ABC$ is right angled at A.
Method 2
Step 1 and 2 as above.
Step 3
$|\overrightarrow {AB}| = \sqrt{4+1+16}=\sqrt{21}=AB$
$|\overrightarrow {BC}| = \sqrt{1+9+25}=\sqrt{35}=BC$
$|\overrightarrow {CA}| = \sqrt{9+4+1}=\sqrt{14}=CA$
$AB^2+CA^2=21+14=35=BC^2$
$\therefore \Delta ABC$  is rightangled at  A, by converse of pythogorus theorem.

edited Jun 20, 2013