Mathematics: 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

Let a = (1, 0, 3) and b = (−3, 2, 1). (a) (3 points)

Ответ:

thebrain1345

15.01.2022

Mathematics

a) $(-\frac{\sqrt{24}}{6} ,\frac{\sqrt{24}}{12} ,-\frac{\sqrt{24}}{12})$

b) (-1,1,2)

c) -2x+y-z=1

Step-by-step explanation:

AB, OB, OA are vectors.

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

Normalizing AB: $\frac{AB}{||AB||} = (\frac{-4}{\sqrt{24}},\frac{2}{\sqrt{24}}, \frac{-2}{\sqrt{24}})=(-\frac{\sqrt{24}}{6} ,\frac{\sqrt{24}}{12} ,-\frac{\sqrt{24}}{12})$

OM, OA, OB are vectors.

$OM = \frac{1}{2}OA+\frac{1}{2}OB=(\frac{-3+1}{2} ,\frac{2+0}{2}, \frac{1+3}{2} )=(-1,1,2)$

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

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

Hence,

-2x+y-z=1

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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

Ответ:

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