Question

Given $A=ai+bj+ck,B=di+3j+4k$ and $C=3i+j-2k$. If the vectors $A,B$ and $C$ form a triangle such that $A=B+C$ and $area\left(\Delta ABC\right)=5\sqrt{6}$, then

• A. $a=-8,b=-4,c=2,d=-11$
• B. $a=-8,b=4,c=-2,d=-11$
• C. $a=-8,b=4,c=2,d=-11$
• D. None of the above

Answer 1

The correct option is C $a=-8,b=4,c=2,d=-11$

Here $A,B,C$ are the vectors which represent the sides of the triangle $ABC$ where

$A=ai+bj+ckB=di+3j+4kC=3i+j-2k$

Given that $A=B+C$

$\therefore ai+bj+ck=(d+3)i+4j+2k\Rightarrow a=d+3,b=4,c=2B\times C=\left|\begin{array}{ccc}i& j& k\\ d& 3& 4\\ 3& 1& -2\end{array}\right|=-10i+(2d+12)j+(d-9)k$

$\therefore$ Area of the $△ABC=\frac{1}{2}|B\times C|=\frac{1}{2}\sqrt{[100+{(2d+12)}^2+{(d-9)}^2]}=5\sqrt{6}$

$\Rightarrow \sqrt{5d^2+30d+325}=10\sqrt{6}\Rightarrow 5d^2+30d+325=600\Rightarrow 5d^2+30d-275=0\Rightarrow d^2+6d-55=0\Rightarrow (d+11)(d-5)=0\Rightarrow d=5,-11$

When $d=5,a=8,b=4,c=2$ and when $d=-11,a=-8,b=4,c=2$