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

Respuesta :

Answer:

a) [tex](-\frac{\sqrt{24}}{6} ,\frac{\sqrt{24}}{12} ,-\frac{\sqrt{24}}{12})[/tex]

b) [tex](-1,1,2)[/tex]

c) [tex]-2x+y-z=1[/tex]

Step-by-step explanation:

a)

AB, OB, OA are vectors.

[tex]AB = OB - OA = (-3, 2, 1) - (1, 0, 3) = (-4, 2, -2)\\||AB|| = \sqrt{(-4)^2+2^2+(-2)^2}=\sqrt{24}\\[/tex]

Normalizing AB: [tex]\frac{AB}{||AB||} = (\frac{-4}{\sqrt{24}},\frac{2}{\sqrt{24}}, \frac{-2}{\sqrt{24}})=(-\frac{\sqrt{24}}{6} ,\frac{\sqrt{24}}{12} ,-\frac{\sqrt{24}}{12})[/tex]

b)

OM, OA, OB are vectors.

[tex]OM = \frac{1}{2}OA+\frac{1}{2}OB=(\frac{-3+1}{2} ,\frac{2+0}{2}, \frac{1+3}{2} )=(-1,1,2)[/tex]

So, [tex]M = (-1,1,2)[/tex]

c)

[tex]-4x+2y-2z=(-4)(-1)+(2)(1)+(-2)(2)=4+2-4=2[/tex]

Hence,

[tex]-2x+y-z=1[/tex]