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If the vectors of a trianlge are `A(hati+hatj+2hatk), B(3hati-hatj+2hati) and C(2hati-hatj+hatk)` the area of triangle isA. `2sqrt3` sq unitsB. `sqrt3` sq unitsC. `2sqrt3` sq unitsD. 3 sq units |
Answer» Correct Answer - D Now , `AB=(3hat(i)-hat(j)+2hat(k))-(hat(i)+hat(j)+2hat(k))=2hat(i)-2hat(j)` and `AC=(2hat(i)-hat(j)+hat(k))-(hat(i)+hat(j)+2hat(k))=hat(i)-2hat(j)-hat(k)` `(ABxxAC)=|{:(hat(i),hat(j),hat(k)),(2,-2,0),(1,-2,-1):}|` `=hat(i)(2-0)-hat(j)(-2-0)+hat(k)(-4+2)` `=2hat(i)+2hat(j)+2hat(k)` `therefore` Area of triangle `AB=(1)/(2)|ABxxAC|` `=(1)/(2)sqrt((2)^2+(2)^2+(-2)^2)` `=(2)/(2)sqrt(1+1+1)=sqrt(3)` sq units. |
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