1.

If AB =3hati-2hatj+2hatk and BC=-hatj-2hatk are adjacent sides of a parallelogram, then angle its diagonals can be

Answer»

`cos^(-1)((2)/(SQRT(13)))`
`(pi)/(3)`
`(4pi)/(3)`
`(2PI)/(3)`

SOLUTION :Here , `AC=AB+BC`
`=(3hat(i)-2hat(j)2hat(k))+(-hat(j)-2hat(k))`
`=3hat(i)-3hat(j)`
Also `BC=AD`
and `AB=AD+DB`
`DB=AB=AD`
`=(3hat(i)-2hat(j)+2hat(k))-(-hat(j)-2hat(k))`
`=3hat(i)-hat(j)+4hat(k)`
`THEREFORE` Angle between AC and BD is
`theta = cos^(-1)(AC.BD)/(|AC|.AD|)`
`=cos^(-1).((3hat(i)-3hat(j)).(3hat(i)-hat(j)+4hat(k)))/(sqrt(3^2+(-3)^2)sqrt(3^2+1^2+4^2))`
`=cos^(-1).((9+3)/(sqrt(18)sqrt(20)))=cos^(-1).((12)/(3sqrt(2)sqrt(2)sqrt(3)))`
`=cos^(-1)((2)/(sqrt(13)))` .


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