Answer to Question #164832 in Abstract Algebra for Suleman

Question

Show that the lines

r·(4i+3j)−1=0, r·(−2i+j)−1=0, r·(i+2j)−1=0 ̃ ̃ ̃ ̃ ̃ ̃ ̃ ̃ ̃

are concurrent.

Answer 1

Solution:

Comparing each of them with $\vec r.(a\hat i+b\hat j)+c=0$

we get,

$a_1=4, b_1=3,c_1=-1 \\ a_2=-2, b_2=1,c_2=-1 \\ a_3=1,b_3=2,c_3=-1$

For these lines to be concurrent, they must satisfy following:

$\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix} =0$

Take LHS

$=\begin{vmatrix} a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3 \\ \end{vmatrix}$

Putting above values

$=\begin{vmatrix} 4 & 3 & -1 \\ -2 & 1 & -1 \\ 1 & 2 & -1 \\ \end{vmatrix}$

$=4\begin{vmatrix} 1 & -1 \\ 2 & -1 \\ \end{vmatrix}-3\begin{vmatrix} -2 & -1 \\ 1 & -1 \\ \end{vmatrix}+(-1)\begin{vmatrix} -2 & 1 \\ 1 & 2 \\ \end{vmatrix}$

$=4(-1+2)-3(2+1)-1(-4-1) \\ =4-9+5 \\ =9-9 \\ =0$

$=$ RHS

Thus, above lines are concurrent.