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If AB =3hati-2hatj+2hatk and BC=-hatj-2hatk are adjacent sides of a parallelogram, then angle its diagonals can be |
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Answer» `cos^(-1)((2)/(SQRT(13)))` `=(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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