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


Expert's answer

a⃗⋅b⃗=1(0)+0(1)+1(−1)=−1\vec a \cdot \vec b=1(0)+0(1)+1(-1)=-1

a⃗⋅c⃗=1(0)+0(0)+1(1)=1\vec a \cdot \vec c=1(0)+0(0)+1(1)=1

b⃗⋅c⃗=0(0)+1(0)−1(1)=−1\vec b \cdot \vec c=0(0)+1(0)-1(1)=-1

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

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

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

1.

cos⁡∠(a⃗,b⃗)=a⃗⋅b⃗∣a⃗∣∣b⃗∣=−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}

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

2.

cos⁡∠(a⃗,c⃗)=a⃗⋅c⃗∣a⃗∣∣c⃗∣=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}

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

3.


cos⁡∠(b⃗,c⃗)=b⃗⋅c⃗∣b⃗∣∣c⃗∣=−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}

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