Show that the vectors `2hat(i)-hat(j)+hat(k),hat(i)-3hat(j)`. And `3hat(i)-4hat(j)-3hat(k)`. Form the vertices of a right angled triangle.

Explanation

Let `2hat(i)-hat(j)+hat(k),hat(i)-3hat(j)-5hat(k) and 3hat(i)-4hat(j)-4hat(k)`
`therefore` Side `vec(AB)=PV of B -PV of A=(hat(i)-3hat(j)-5hat(k))-(2hat(i)-hat(j)+hat(k))`
`=hat(i)-3hat(j)-5hat(k)-2hat(i)+hat(j)-hat(k)=-hat(i)-2hat(j)-6hat(k)`
`|vec(AB)|=sqrt((-1)^2+(-2)^2+(-6)^2)=sqrt(1+4+36)=sqrt(41)`
`vec(BC)=PV of C-PV of B=(3hat(i)-4hat(j)-4hat(k))-(hat(i)-3hat(j)-5hat(k))`
`=3hat(i)-4hat(j)-4hat(k)-hat(i)+3hat(j)+5hat(k)=2hat(i)-hat(j)+hat(k)`
`|vec(BC)|=sqrt(2^2+(-1)^2+1^2)=sqrt(4+1+1)=sqrt(6)`.
and ` vec(AC)=PV of C-PV of A =(3hat(i)-4hat(j)-4hat(k))-(2hat(i)-hat(j)+hat(k))`.
`=3hat(i)-4hat(j)-4hat(k)-2hat(i)+hat(j)-hat(k)=hat(i)-3hat(j)-5hat(k)`
`|vec(AC)|=sqrt(1^2+(-3)^2+(-5)^2)=sqrt(1+9+25)=sqrt(35)`
Now, `|vec(BC)|+|vec(AC)|^2=6+35=41=|vec(AB)|^2`.
Which shows that ABC is right angled triangle.