Answer to Question #271390 in Analytic Geometry for Wan

Question #271390

Known vectors:

⃗a = (1, 0, 1) , ⃗b = (0, 1, -1) , ⃗c = (0, 0, 1)

Find the angle between:

1. a and b

2. a and c

3. b and c


1
Expert's answer
2021-11-26T11:58:28-0500
ab=1(0)+0(1)+1(1)=1\vec a \cdot \vec b=1(0)+0(1)+1(-1)=-1

ac=1(0)+0(0)+1(1)=1\vec a \cdot \vec c=1(0)+0(0)+1(1)=1

bc=0(0)+1(0)1(1)=1\vec b \cdot \vec c=0(0)+1(0)-1(1)=-1

a=12+02+12=2|\vec a|=\sqrt{1^2+0^2+1^2}=\sqrt{2}

b=02+12+(1)2=2|\vec b|=\sqrt{0^2+1^2+(-1)^2}=\sqrt{2}

c=02+02+12=1|\vec c|=\sqrt{0^2+0^2+1^2}=1

1.

cos(a,b)=abab=12(2)=12\cos \angle(\vec a, \vec b)=\dfrac{\vec a \cdot \vec b}{|\vec a||\vec b|}=\dfrac{-1}{\sqrt{2}(\sqrt{2})}=-\dfrac{1}{2}

(a,b)=120°\angle(\vec a, \vec b)=120\degree

2.

cos(a,c)=acac=12(1)=22\cos \angle(\vec a, \vec c)=\dfrac{\vec a \cdot \vec c}{|\vec a||\vec c|}=\dfrac{1}{\sqrt{2}(1)}=\dfrac{\sqrt{2}}{2}

(a,c)=45°\angle(\vec a, \vec c)=45\degree

3.


cos(b,c)=bcbc=12(1)=22\cos \angle(\vec b, \vec c)=\dfrac{\vec b \cdot \vec c}{|\vec b||\vec c|}=\dfrac{-1}{\sqrt{2}(1)}=-\dfrac{\sqrt{2}}{2}

(b,c)=135°\angle(\vec b, \vec c)=135\degree

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