If the position vectors of the vertices a, B and C of a `Triangle ABC` be `(1, 2, 3), (-1, 0, 0)` and `(0, 1, 2)` respectively then find `angleABC`.

Options

A. `(pi)/(2)`
B. `(pi)/(4)`
C. `cos^(-1)((10)/(sqrt(102)))`
D. `cos^(-1)((1)/(3))`

Answer

C

Explanation

WE are given the points `A (1,2,3),B (-1,0,0)` and `C (0,1,2)` Also , it given that ` angle ABC ` is the angle between the vectors BA and BC here BA = PV of of A -PV of B
`=( hati + 2 hatj + 3 hatk )- (-hati + 0 hatj + 0 hatk )`
`=[ hati -(-hati )+(2 hatj -0 )+ (3 hatk -0 )] = 2 hati + 2 hatj + 3 hatk`
` implies | BA| = sqrt((2) ^(2)+(2)^(2)+(3)^(2))=sqrt(4+4+9)=sqrt(17)`
`implies BC =PV of C-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`
` therefore cos theta = (BA.BC)/(|BA||BC|)implies cos (angle ABC)=(10)/(sqrt(17)sqrt(6))`
`implies angle ABC = cos ^(-1) ((10)/(sqrt(102)))`