Find the angle between the lines `x-3y-4=0,4y-z+5=0a n dx+3y-11=0,2y=z+6=0.`

Explanation

Line `x-3y-4=0=4y-z+5` is line of intersection of planes `x-3y-4=0 and 4y-z+5=0`.
Vector along the line intersection is `|{:(hati,,hatj,,hatk),(1,,-3,,0),(0,,4,,-1):}|=3hati+hatj+4hatk=veca`
Line `x+3y-11=0=2y-z+6` is line of intersection of planes `x+3y-11=0 and 2y-z+6=0.`
Vector along line of intersection is `|{:(hati,,hatj,,hatk),(1,,3,,0),(0,,2,,-1):}|=-3hati+hatj+2hatk=vecb`
`" "veca*vecb=(3)(-3)+(1)(1)+(4)(2)=0`
Therefore, lines are perpendicular.