Find the angle between the vectors `vec a = 3 hat i +4 hat j + hat k` and `vec b = 2 hat i +3 hat j - hat k`

Answer

`a to s; b to q; c to p; d to r`

Explanation

a. Lines `x= 2t+1, y=t+2, z=-t-3 or (x-1)/(2) = (y-2)/(1)=(z+3)/(-1)`, which is along the vector
`2hati+hatj-hatk`. Vector `hati+ 3hatj+5hatk` is perpendicular to the line.
b. Normals to the planes `x+y+z-3=0 and 2x-y+3z=0` are `vec(n_(1))= hati+hatj+hatk and vec(n_2) = 2hati-hatj+3hatk`
Then the vector along the line of intersection of planes is `vec(n_1) xx vec(n_2) = |{:(hati,,hatj,,hatk),(1,,1,,1),(2,,-1,,3):}|= 4hati-hatj-3hatk`
c. The shortest distance between the lines `(x)/(2)= (y)/(-3)= (z)/(-1) and vecr= (3hati-hatj+hatk)+ t(hati+hatj-2hatk)` occurs along the vector
`(2hati-3hatj-hatk)xx(hati+hatj-2hatk)= |{:(hati,,hatj,,hatk), (2,,-3,,-1), (1,,1,,-2):}| = 7hati +3hatj+5hatk`
d. Normal to the plane `vecr=-hati+4hatj-6hatk+lamda(hati+3hatj-2hatk)+mu(-hati+2hatj-5hatk)` is
`|{:(hati,,hatj,,hatk),(1,,3,,-2),(-1,,2,,-5):}|= -11hati+ 7hatk + 5hatk`