Question

If\[{z_1} = 1 + i\], \[{z_2} = - 2 + 3i\]and\[{z_3} = ai/3\], where \[{i^2} = - 1\], are collinear then the value of a is [AMU\[2001\]]
A) \[ - 1\]
B) \[3\]
C) \[4\]
D) \[5\]

Answer 1

Hint: in this question we have to find the value of a. In order to find this we have to use the concept that collinear points are lie in same line therefore it will never form triangle. As these points are not forming area therefore area of triangle is zero.


Formula Used:Area of triangle is given by
\[A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|\]
Where
A is area of triangle
\[{a_i}\] is a real part of vertices
\[{b_i}\] is a imaginary part of vertices



Complete step by step solution:Given: \[{z_1} = 1 + i\], \[{z_2} = - 2 + 3i\]and\[{z_3} = ai/3\] these are collinear
Now area of triangle is given by
\[A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|\]
Where
A is area of triangle
\[{a_i}\] is a real part of vertices
\[{b_i}\] is a imaginary part of vertices
Expand determinant to get area of triangle
\[A = \dfrac{1}{2}({a_1}({b_2} - {b_3}) - {a_2}({b_1} - {b_3}) + {a_3}({b_1} - {b_2}))\]
\[A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}{{a_1}}&{{b_1}}&1\\{{a_2}}&{{b_2}}&1\\{{a_3}}&{{b_3}}&1\end{array}} \right|\]
\[A = \dfrac{1}{2}\left| {\begin{array}{*{20}{c}}1&1&1\\{ - 2}&3&1\\0&{\dfrac{a}{3}}&1\end{array}} \right|\]
Expand the determinant
\[A = \dfrac{1}{2}(1(3 - \dfrac{a}{3}) + 2(1 - \dfrac{a}{3}) + 0(1 - 3))\]
\[A = \dfrac{1}{2}(3 - \dfrac{a}{3} + 2 - \dfrac{{2a}}{3})\]
\[A = \dfrac{1}{2}(5 - a)\]
We know that area of triangle is zero
\[\dfrac{1}{2}(5 - a) = 0\]
\[5 - a = 0\]
\[a = 5\]



Option ‘D’ is correct



Note: We have to remember that collinear point is a point which lies in same line. If all points lie in same line then it never form area so area of triangle is zero for collinear points.Complex number is a number which is a combination of real and imaginary number. So in combination number question we have to represent number as a combination of real and its imaginary part. Imaginary part is known as iota. Square of iota is equal to negative one.