MCQ
The area of triangle whose vertices are $(1,\,2,\,3),\,(2,\,5,\, - 1)$ and $( - 1,\,1,\,2)$ is
- A$150 $ sq. unit
- B$145 $ sq. unit
- ✓$\frac{{\sqrt {155} }}{2}$ sq. unit
- D$\frac{{155}}{2}$ sq. unit
$=\frac{1}{2}\left| \,\left| \,\begin{matrix}
i & j & k \\
{{x}_{2}}-{{x}_{1}} & {{y}_{2}}-{{y}_{1}} & {{z}_{2}}-{{z}_{1}} \\
{{x}_{3}}-{{x}_{1}} & {{y}_{3}}-{{y}_{1}} & {{z}_{3}}-{{z}_{1}} \\
\end{matrix}\, \right|\, \right|$
Here, $({x_1},\,{y_1},\,{z_1}) \equiv (1,\,2,\,3)$, $({x_2},\,{y_2},\,{z_2}) \equiv (2,\,5,\, - 1)$,
$({x_3},\,{y_3},\,{z_3}) \equiv ( - 1,\,1,\,2)$
$ = \frac{1}{2}\left| {\,\left| {\,\begin{array}{*{20}{c}}i&j&k\\1&3&{ - 4}\\{ - 2}&{ - 1}&{ - 1}\end{array}\,} \right|\,} \right|$$ = \frac{1}{2}|( - 7i + 9j + 5k)|$
$ = \frac{1}{2}\sqrt {49 + 81 + 25} $ $ = \frac{{\sqrt {155} }}{2}$ sq. unit.
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$f(x) = log_{10}(4x^3 -12x^2 + 11x -3)$, $x \in \left[ {2,3} \right]$, is
Statement $2$ : A function $f : R \to R$ is discontinuous at $x_0$ if and only if, $\mathop {\lim }\limits_{x \to {x_0}} \,f\left( x \right)$ exists and $\mathop {\lim }\limits_{x \to {x_0}} \,f\left( x \right) \ne f\left( {{x_0}} \right)$