2020-11-29T15:39:06-05:00
If A is the point (1,-1,2),B is the (-1,2,2)and C is the (4,3,0),fine the direction cosines of BA andBC
1
2020-11-30T14:51:23-0500
1) BA
B A ⃗ = ( 1 , − 1 , 2 ) − ( − 1 , 2 , 2 ) = ( 2 , − 3 , 0 ) \vec{BA}=(1,-1,2)-(-1,2,2)=(2,-3,0) B A = ( 1 , − 1 , 2 ) − ( − 1 , 2 , 2 ) = ( 2 , − 3 , 0 )
α = 2 2 2 + 3 2 + 0 2 = 2 13 β = − 3 2 2 + 3 2 + 0 2 = − 3 13 γ = 0 2 2 + 3 2 + 0 2 = 0 \alpha=\frac{2}{\sqrt{2^2+3^2+0^2}}=\frac{2}{\sqrt{13}}\\\beta=\frac{-3}{\sqrt{2^2+3^2+0^2}}=\frac{-3}{\sqrt{13}}\\\gamma=\frac{0}{\sqrt{2^2+3^2+0^2}}=0 α = 2 2 + 3 2 + 0 2 2 = 13 2 β = 2 2 + 3 2 + 0 2 − 3 = 13 − 3 γ = 2 2 + 3 2 + 0 2 0 = 0 2) BC
B C ⃗ = ( 4 , 3 , 0 ) − ( − 1 , 2 , 2 ) = ( 5 , 1 , − 2 ) \vec{BC}=(4,3,0)-(-1,2,2)=(5,1,-2) BC = ( 4 , 3 , 0 ) − ( − 1 , 2 , 2 ) = ( 5 , 1 , − 2 )
α = 5 5 2 + 1 2 + 2 2 = 5 30 β = 1 5 2 + 1 2 + 2 2 = 1 30 γ = − 2 5 2 + 1 2 + 2 2 = − 2 30 \alpha=\frac{5}{\sqrt{5^2+1^2+2^2}}=\frac{5}{\sqrt{30}}\\\beta=\frac{1}{\sqrt{5^2+1^2+2^2}}=\frac{1}{\sqrt{30}}\\\gamma=\frac{-2}{\sqrt{5^2+1^2+2^2}}=\frac{-2}{\sqrt{30}} α = 5 2 + 1 2 + 2 2 5 = 30 5 β = 5 2 + 1 2 + 2 2 1 = 30 1 γ = 5 2 + 1 2 + 2 2 − 2 = 30 − 2
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