Question

Let A = (1, 0, 3) and B = (−3, 2, 1).

(a) (3 points) Normalize −−→AB.
(b) (3 points) Find the midpoint M of the segment AB.
(c) (4 points) Find equation of the plane passing through M and perpendicular to AB

Answer 1

a)
`(-(√(24))/(6) ,(√(24))/(12) ,-(√(24))/(12))`

b)
`(-1,1,2)`

c)
`-2x+y-z=1`

Explanation:

a)

AB, OB, OA are vectors.

`AB = OB - OA = (-3, 2, 1) - (1, 0, 3) = (-4, 2, -2)\\||AB|| = √((-4)^2+2^2+(-2)^2)=√(24)\\`

Normalizing AB:
`(AB)/(||AB||) = ((-4)/(√(24)),(2)/(√(24)), (-2)/(√(24)))=(-(√(24))/(6) ,(√(24))/(12) ,-(√(24))/(12))`

b)

OM, OA, OB are vectors.

`OM = (1)/(2)OA+(1)/(2)OB=((-3+1)/(2) ,(2+0)/(2), (1+3)/(2) )=(-1,1,2)`

So,
`M = (-1,1,2)`

c)

`-4x+2y-2z=(-4)(-1)+(2)(1)+(-2)(2)=4+2-4=2`

Hence,

`-2x+y-z=1`