Q.13. If the area of the triangle with vertices (1,2,0) , (1,0,2) and (0,x,1) is \sqrt{6} square units,then the value of x is
A. 1
B.2
C.3
D. 4
Answer :- C. 3
Explanation :-
\text{Let } A\equiv(1,2,0), B\equiv(1,0,2) \text{ and } C\equiv(0,x,1) \therefore \quad \overline{\mathrm{AB}}=-2\hat{\mathrm{j}}+2\hat{\mathrm{k}} \text{ and } \overline{\mathrm{AC}}=-\hat{\mathrm{i}}+(x-2)\hat{\mathrm{j}}+\hat{\mathrm{k}} \therefore \quad \text{Area of } \triangle ABC = \frac{1}{2}|\overline{\mathrm{AB}}\times\overline{\mathrm{AC}}| = \sqrt{6} |\overline{\mathrm{AB}}\times\overline{\mathrm{AC}}| = \begin{vmatrix} \hat{\mathrm{i}} & \hat{\mathrm{j}} & \hat{\mathrm{k}} \\ 0 & -2 & 2 \\ -1 & x-2 & 1 \end{vmatrix} = \hat{\mathrm{i}}[-2-2(x-2)] - \hat{\mathrm{j}}(0+2) + \hat{\mathrm{k}}(0-2) = (2-2x)\hat{\mathrm{i}} -2\hat{\mathrm{j}} -2\hat{\mathrm{k}} \frac{1}{2}|\overline{\mathrm{AB}}\times\overline{\mathrm{AC}}| = \sqrt{6} \Rightarrow \frac{1}{2} \sqrt{(2-2x)^2+4+4} = \sqrt{6} \Rightarrow (2-2x)^2 = 16 \Rightarrow 4 - 8x + 4x^2 = 16 \Rightarrow x^2 - 2x - 3 = 0 \Rightarrow (x-3)(x+1) = 0 \Rightarrow x = 3 \text{ or } -1