Question

How do you determine the value(s) of k such that the system of linear equations has the indicated number of solutions: no solutions for $x+2y+kx=6$ and $3x+6y+8z=4$?

Answer 1

Hint: The two linear equations given in the above question are in terms of the three variables; x, y, and z. Therefore we can consider them to be the equations of two planes. The solutions for the given system can be considered as the points of intersection of the two planes. For the given system to have no solution, the normal vectors of the two planes must be parallel to each other. From this condition, we will obtain the required value of k.

Complete step by step solution:
The system of the linear equations is given as
$\begin{align}
  & \Rightarrow x+2y+kx=6.......\left( i \right) \\
 & \Rightarrow 3x+6y+8z=4......\left( ii \right) \\
\end{align}$
The above two equations, being in terms of x, y, and z, can represent two planes. So the solution of the given system will be the points of intersection of the two planes.
Now, according to the question, there is no solution for the given system. This means that the planes will not intersect. For this, the planes need to be parallel. This in turn means that their normal vectors need to be parallel to each other. The normal vectors for the first and the second equation can be respectively written as
$\begin{align}
  & \Rightarrow {{{\vec{n}}}_{1}}=\hat{i}+2\hat{j}+\left( k \right)\hat{k} \\
 & \Rightarrow {{{\vec{n}}}_{1}}=3\hat{i}+6\hat{j}+8\hat{k} \\
\end{align}$
Since the above two vectors are parallel, their components must be proportional so that we can write
$\begin{align}
  & \Rightarrow \dfrac{1}{3}=\dfrac{2}{6}=\dfrac{k}{8} \\
 & \Rightarrow \dfrac{1}{3}=\dfrac{1}{3}=\dfrac{k}{8} \\
\end{align}$
From the above, we can write the equation
$\Rightarrow \dfrac{k}{8}=\dfrac{1}{3}$
Multiplying $8$ both the sides, we finally get
$\Rightarrow k=\dfrac{8}{3}$
Hence, for the given system to have no solution, the value of k is equal to $\dfrac{8}{3}$.

Note: We must be familiar with the concepts of three-dimensional geometry for understanding the solution of the linear equations in three variables. Since the constant term in the two equations were different, the parallel planes will be situated at a distance from each other and thus we must not worry about this condition.