Question #164832

Show that the lines

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

are concurrent.


1
Expert's answer
2021-02-24T06:08:29-0500

Solution:

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

we get,

a1=4,b1=3,c1=1a2=2,b2=1,c2=1a3=1,b3=2,c3=1a_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:

a1b1c1a2b2c2a3b3c3=0\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

=a1b1c1a2b2c2a3b3c3=\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

=431211121=\begin{vmatrix} 4 & 3 & -1 \\ -2 & 1 & -1 \\ 1 & 2 & -1 \\ \end{vmatrix}

=4112132111+(1)2112=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(41)=49+5=99=0=4(-1+2)-3(2+1)-1(-4-1) \\ =4-9+5 \\ =9-9 \\ =0

== RHS

Thus, above lines are concurrent.



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