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Vectors, EXT1* V1 2022 HSC 11d

A triangle is formed in three-dimensional space with vertices `A(1,-1,2)`, `B(0,2,-1)`  and `C(2,1,1)`.

Find the size of `/_ABC`, giving your answer to the nearest degree.   (3 marks)

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`33°`

Show Worked Solution

`vec(BA)=((1),(-1),(2))-((0),(2),(-1))=((1),(-3),(3))`

`abs(vec(BA))=sqrt(1^2+3^2+3^2)=sqrt19`
 

`vec(BC)=((2),(1),(1))-((0),(2),(-1))=((2),(-1),(2))`

`abs(vec(BC))=sqrt(2^2+1^2+2^2)=sqrt9=3`
 

`vec(BA)*vec(BC)=1xx2+ -3xx-1+3xx2=11`

`cos/_ABC=(vec(BA)*vec(BC))/(abs{vec(BA)}abs{vec(BC)})=11/(3sqrt19)`

`:./_ABC=cos^(-1)(11/(3sqrt19))=32.733…=33°\ \ text{(nearest degree)}`

Filed Under: Operations With Vectors Tagged With: Band 4, smc-7286-20-Angles Between Vectors, smc-7286-70-3D Vectors, syllabus-2027

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