1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

# Show that the points A,B,C with position vectors 2^i−^j+^k,^i−3^j−5^k and 3^i−4^j−4^k respectively, are the vertices of a right-angled triangle. Hence, find the area of the triangle.

Open in App
Solution

## Vertices of ΔABC with position vectors, ¯¯¯¯¯¯¯¯OA=2ˆi−ˆj+ˆk¯¯¯¯¯¯¯¯OB=ˆi−3ˆj−5ˆk¯¯¯¯¯¯¯¯OC=3ˆi−4ˆj−4ˆkthe ΔABC is right angled triangle and find the area of ΔABCthen,−−→AB=−−→OB−−−→OA=ˆ(i−3ˆj−5ˆk)−(2ˆi−ˆj+ˆk)=ˆi−3ˆj−5ˆk−2ˆi+ˆj−ˆk=−ˆi−2ˆj+6ˆk,∴∣∣¯¯¯¯¯¯¯¯AB∣∣=√41−−→BC=−−→OC−−−→OB=(3ˆi−4ˆj−4ˆk)−(ˆi−3ˆj−5ˆk)=3ˆi−4ˆj−4ˆk−ˆi+3ˆj+5ˆk=2ˆi−ˆj+ˆk∴∣∣∣−−→BC∣∣∣=√6−−→CA=−−→OA−−−→OC=(2ˆi−ˆj+ˆk)−(3ˆi−4ˆj−4ˆk)=2ˆi−ˆj+ˆk−3ˆi+4ˆj+4ˆk=−ˆi+3ˆj+5ˆk∴∣∣∣−−→CA∣∣∣=√35 byPathagorastheoram,a2=b2+c2so,∣∣¯¯¯¯¯¯¯¯AB∣∣2=∣∣∣−−→BC∣∣∣2+∣∣∣−−→CA∣∣∣2=6+35=41∣∣¯¯¯¯¯¯¯¯AB∣∣=√41So, ΔABC is right angled triangle at C.and the Area of ΔABC = 12∣∣∣−−→BC∣∣∣×∣∣∣−−→CA∣∣∣=12√6×√35=√2102

Suggest Corrections
0
Join BYJU'S Learning Program
Related Videos
Distance and Displacement
PHYSICS
Watch in App
Explore more
Join BYJU'S Learning Program