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If the vertices A, B, C of a DeltaABC have position vectors (1, 2, 3), (-1, 0, 0), (0, 1, 2) respectively, then angleABC angleABC is the angle between the vectors BA and BC), is equal to |
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Answer» `(pi)/(2)` `=( HATI + 2 hatj+3 hatk )- (-hati+ 0 hatj + 0 hatk )` `=[ hati-(-hati )+(2hatj -0 )+ (3 hatk -0 )] = 2 hati+ 2 hatj+ 3hatk` ` IMPLIES| BA|= sqrt((2) ^(2)+(2)^(2)+(3)^(2))=sqrt(4+4+9)=sqrt(17)` `implies BC=PVofC-PV of B ` `=(0 hati +1 hatj + 2 hatk)-(-hati +0hatj+0hatk)=hati +hatj+2hatk` `implies |BC|=sqrt((1)^(2) +(1)^(2)+(2)^(2))=sqrt(1+1+4)= sqrt(6)` Now , `BA.BC =(2 hati + 2 hatj + 3 hatk ).(hati + hatj + 2 hatk )` `=2xx1+2xx1+3xx2=10` ` thereforecos theta= (BA.BC)/(|BA||BC|)implies cos (angle ABC)=(10)/(sqrt(17)sqrt(6))` `impliesangleABC= cos ^(-1) ((10)/(sqrt(102)))` |
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