Find a vector of magnitude √171 which is perpendicular to both of the vectors →a=ˆi+2ˆj−3ˆk and →b=3ˆi−ˆj+2ˆk.
We have →a×→b=∣∣ ∣ ∣∣ˆiˆjˆk12−33−12∣∣ ∣ ∣∣=ˆi−11ˆj−7ˆk ∴ Required vector = √171(±→a×→b|→a×→b|)=√171(±ˆi−11ˆj−7ˆk√1+121+49)=±(ˆi−11ˆj−7ˆk).