If the area of the triangle whose vertices are \[A\left( { - 1,1,2} \right)\], \[B\left( {1,2,3} \right)\] and \[C\left( {t,1,1} \right)\] is minimum, then find the absolute value of the parameter \[t\].

Hint: Coordinates of the three vertices of a triangle are given. One of the coordinates is dependent on a parameter \[t\]. The value of the parameter is asked for which the area of the triangle will be minimum. So, at first find the area of the triangle using vector method. The obtained area will be a function of \[t\]. Let it be \[A\left( t \right)\] and hence find \[{A^2}\left( t \right)\], \[A\left( t \right)\] being a square root expression. To minimize the area, you have to use second order derivative test. Find the value of \[t\] for which the first order derivative of \[{A^2}\left( t \right)\] is zero and the second order derivative of \[{A^2}\left( t \right)\] is positive. Then take its absolute value to get the answer.

Formula Used:
Area of a triangle having the corresponding sides \[\overrightarrow {AB} \] and \[\overrightarrow {AC} \] is given by \[\dfrac{1}{2}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right|\]
The position vector of a point \[A\left( {x,y,z} \right)\] is \[\overrightarrow {OA} = x\hat i + y\hat j + z\hat k\]
If the position vectors of the points \[A\] and \[B\] are \[\overrightarrow {OA} = {a_1}\hat i + {a_2}\hat j + {a_3}\hat k\] and \[\overrightarrow {OB} = {b_1}\hat i + {b_2}\hat j + {b_3}\hat k\] then \[\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} = \left( {{b_1} - {a_1}} \right)\hat i + \left( {{b_2} - {a_2}} \right)\hat j + \left( {{b_3} - {a_3}} \right)\hat k\]
The cross product of the two vectors \[\overrightarrow {AB} \] and \[\overrightarrow {AC} \] is given by the determinant \[\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\{{a_1}}&{{a_2}}&{{a_3}}\\{{b_1}}&{{b_2}}&{{b_3}}\end{array}} \right|\]
Magnitude of a vector \[\overrightarrow v = a\hat i + b\hat j + c\hat k\] is given by \[\left| {\overrightarrow v } \right| = \sqrt {{a^2} + {b^2} + {c^2}} \]

Complete step-by-step solution:
The given vertices of a triangle are \[A\left( { - 1,1,2} \right)\], \[B\left( {1,2,3} \right)\] and \[C\left( {t,1,1} \right)\]
So, the position vectors of the vertices are \[\overrightarrow {OA} = - \hat i + \hat j + 2\hat k\], \[\overrightarrow {OB} = \hat i + 2\hat j + 3\hat k\] and \[\overrightarrow {OC} = t\hat i + \hat j + \hat k\]
\[\therefore \overrightarrow {AB} = 2\hat i + \hat j + \hat k\] and \[\overrightarrow {AC} = \left( {t + 1} \right)\hat i - \hat k\]
Now, \[\overrightarrow {AB} \times \overrightarrow {AC} = \left| {\begin{array}{*{20}{c}}{\hat i}&{\hat j}&{\hat k}\\2&1&1\\{t + 1}&0&{ - 1}\end{array}} \right|\]
Expanding the determinant, we get
\[\overrightarrow {AB} \times \overrightarrow {AC} = \left\{ {\left( 1 \right)\left( { - 1} \right) - \left( 1 \right)\left( 0 \right)} \right\}\hat i - \left\{ {\left( 2 \right)\left( { - 1} \right) - \left( 1 \right)\left( {t + 1} \right)} \right\}\hat j + \left\{ {\left( 2 \right)\left( 0 \right) - \left( 1 \right)\left( {t + 1} \right)} \right\}\hat k\]
\[ = \left( { - 1 - 0} \right)\hat i - \left( { - 2 - t - 1} \right)\hat j + \left( {0 - t - 1} \right)\hat k\]
\[ = \left( { - 1} \right)\hat i - \left( { - 3 - t} \right)\hat j + \left( { - t - 1} \right)\hat k\]
\[ = - \hat i + \left( {3 + t} \right)\hat j - \left( {t + 1} \right)\hat k\]
So, area of the triangle is \[\dfrac{1}{2}\left| {\overrightarrow {AB} \times \overrightarrow {AC} } \right| = \dfrac{1}{2}\left| { - \hat i + \left( {3 + t} \right)\hat j - \left( {t + 1} \right)\hat k} \right|\]
The magnitude of the vector \[ - \hat i + \left( {3 + t} \right)\hat j - \left( {t + 1} \right)\hat k\] is \[\left| { - \hat i + \left( {3 + t} \right)\hat j - \left( {t + 1} \right)\hat k} \right| = \sqrt {{{\left( { - 1} \right)}^2} + {{\left( {3 + t} \right)}^2} + {{\left( {t + 1} \right)}^2}} \]
Simplify the expression using the formula \[{\left( {a + b} \right)^2} = {a^2} + 2ab + {b^2}\]
\[ = \sqrt {1 + 9 + 6t + {t^2} + {t^2} + 2t + 1} \]
\[ = \sqrt {2{t^2} + 8t + 11} \]
So, area of the triangle is \[\dfrac{1}{2}\sqrt {2{t^2} + 8t + 11} \] square units.
Let \[A\left( t \right) = \dfrac{1}{2}\sqrt {2{t^2} + 8t + 11} \]
Squaring both sides, we get
\[{A^2}\left( t \right) = \dfrac{1}{4}\left( {2{t^2} + 8t + 11} \right)\]
We have to find the value of \[t\] for which the area \[A\left( t \right)\] is minimum.
For the same value of \[t\], the value of \[{A^2}\left( t \right)\] is also minimum.
Differentiating \[{A^2}\left( t \right)\] with respect to \[t\], we get
\[\dfrac{d}{{dt}}\left\{ {{A^2}\left( t \right)} \right\} = \dfrac{d}{{dt}}\left\{ {\dfrac{1}{4}\left( {2{t^2} + 8t + 11} \right)} \right\}\]
Use the formula \[\dfrac{d}{{dx}}\left\{ {cf\left( x \right)} \right\} = c\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\}\], where \[c\] is a constant and \[f\left( x \right)\] is a function of \[x\].
\[ \Rightarrow \dfrac{d}{{dt}}\left\{ {{A^2}\left( t \right)} \right\} = \dfrac{1}{4}\dfrac{d}{{dt}}\left( {2{t^2} + 8t + 11} \right)\]
Use the formula \[\dfrac{d}{{dx}}\left\{ {{f_1}\left( x \right) + {f_2}\left( x \right) + ..... + {f_n}\left( x \right)} \right\} = \dfrac{d}{{dx}}\left\{ {{f_1}\left( x \right)} \right\} + \dfrac{d}{{dx}}\left\{ {{f_2}\left( x \right)} \right\} + ..... + \dfrac{d}{{dx}}\left\{ {{f_n}\left( x \right)} \right\}\]
\[ \Rightarrow \dfrac{d}{{dt}}\left\{ {{A^2}\left( t \right)} \right\} = \dfrac{1}{4}\left\{ {\dfrac{d}{{dt}}\left( {2{t^2}} \right) + \dfrac{d}{{dt}}\left( {8t} \right) + \dfrac{d}{{dt}}\left( {11} \right)} \right\}\]
Use the formula \[\dfrac{d}{{dx}}\left( c \right) = 0\] and \[\dfrac{d}{{dx}}\left\{ {cf\left( x \right)} \right\} = c\dfrac{d}{{dx}}\left\{ {f\left( x \right)} \right\}\], where \[c\] is a constant and \[f\left( x \right)\] is a function of \[x\].
\[ \Rightarrow \dfrac{d}{{dt}}\left\{ {{A^2}\left( t \right)} \right\} = \dfrac{1}{4}\left\{ {2\dfrac{d}{{dt}}\left( {{t^2}} \right) + 8\dfrac{d}{{dt}}\left( t \right) + 0} \right\}\]
Use the formula \[\dfrac{d}{{dx}}\left( {{x^n}} \right) = n{x^{n - 1}}\], where \[n\] is a constant
\[ \Rightarrow \dfrac{d}{{dt}}\left\{ {{A^2}\left( t \right)} \right\} = \dfrac{1}{4}\left\{ {2\left( {2t} \right) + 8\left( 1 \right)} \right\}\]
\[ \Rightarrow \dfrac{d}{{dt}}\left\{ {{A^2}\left( t \right)} \right\} = \dfrac{1}{4}\left( {4t + 8} \right)\]
Taking \[4\] as common from the terms on the right-hand side.
\[ \Rightarrow \dfrac{d}{{dt}}\left\{ {{A^2}\left( t \right)} \right\} = \dfrac{1}{4} \times 4\left( {t + 2} \right)\]
\[ \Rightarrow \dfrac{d}{{dt}}\left\{ {{A^2}\left( t \right)} \right\} = t + 2\]
Differentiating again with respect to \[t\], we get
\[\dfrac{d}{{dt}}\left[ {\dfrac{d}{{dt}}\left\{ {{A^2}\left( t \right)} \right\}} \right] = \dfrac{d}{{dt}}\left( {t + 2} \right)\]
\[ \Rightarrow \dfrac{{{d^2}}}{{d{t^2}}}\left\{ {{A^2}\left( t \right)} \right\} = \dfrac{d}{{dt}}\left( t \right) + \dfrac{d}{{dt}}\left( 2 \right)\]
\[ \Rightarrow \dfrac{{{d^2}}}{{d{t^2}}}\left\{ {{A^2}\left( t \right)} \right\} = 1 + 0 = 1\]
Now, \[\dfrac{d}{{dt}}\left\{ {{A^2}\left( t \right)} \right\} = 0\]
\[ \Rightarrow t + 2 = 0\]
\[ \Rightarrow t = - 2\]
At \[t = - 2\], \[\dfrac{{{d^2}}}{{d{t^2}}}\left\{ {{A^2}\left( { - 2} \right)} \right\} = 1 > 0\]
So, the value of the function \[{A^2}\left( t \right)\] is minimum at \[t = - 2\]
i.e. the value of the function \[A\left( t \right)\] is minimum at \[t = - 2\]
Thus, the area of the triangle is minimum for \[t = - 2\]
Absolute value of \[\left( { - 2} \right)\] is \[\left| { - 2} \right| = 2\]
Hence, the absolute value of \[t\] is \[2\].

Note: Since the vertices are of ordered triplet, so the triangle is in \[3D\] space. Area of such a triangle can’t be found using the formula used in coordinate geometry. It is found using vector method only. In vector method, area of a triangle is half of the magnitude of the cross product of two corresponding sides of the triangle. Many students can’t remember the formula properly and hence they obtain a wrong answer. Maximum or minimum value of a function and the point at which it occurs can be found using second derivative test only. Using first derivative, you can find the critical points only. Absolute value of an integer always refers a whole number. Absolute value of zero is equal to zero.