1.

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

Answer»

`(pi)/(2)`
`(pi)/(4)`
`cos^(-1)((10)/(sqrt(102)))`
`cos^(-1)((1)/(3))`

Solution :WE aregiventhe points `A (1,2,3),B(-1,0,0)` and`C (0,1,2)` Also , itgiventhat` angleABC `is theanglebetweenthevectorsBAand BChereBA= PVofofA -PV of B
`=( 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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