Question

Find the equation of the plane containing the lines:

$\frac{(x-5)}{4}=\frac{(y-7)}{4}=\frac{(z+3)}{-5}$ and $\frac{(x-8)}{7}=\frac{(y-4)}{1}=\frac{(z-5)}{3}$.

Answer 1

Compute the required value:

Given lines:

$\frac{(x-5)}{4}=\frac{(y-7)}{4}=\frac{(z+3)}{-5}$ and $\frac{(x-8)}{7}=\frac{(y-4)}{1}=\frac{(z-5)}{3}$.

If lines $\frac{x-x_1}{a_1}=\frac{y-y_1}{b_1}=\frac{z-z_1}{c_1}$ and $\frac{x-x_2}{a_2}=\frac{y-y_2}{b_2}=\frac{z-z_2}{c_2}$

Lie in the same plane then,

$\left|\begin{array}{ccc}{x}_2-x_1& y_2-y_1& z_2-z_1\\ a_1& b_1& c_1\\ a_2& b_2& c_2\end{array}\right|=0$

Comparing the values we get

$\left|\begin{array}{ccc}8-5& 4-7& 5-\left(-3\right)\\ 4& 4& -5\\ 7& 1& 3\end{array}\right|=\left|\begin{array}{ccc}3& -3& 8\\ 4& 4& -5\\ 7& 1& 3\end{array}\right|$

$$\begin{gathered} =3(12+5)+3(12+35)+8(4-28) \\ =51+141-192 \\ =0 \end{gathered}$$

Therefore, the lines are co-planar.

Now the equation of the plane is given by

$\left|\begin{array}{ccc}x-5& y-7& z+3\\ 4& 4& -5\\ 7& 1& 3\end{array}\right|=0$

$$\begin{gathered} \Rightarrow 17(x-5)-47(y-7)-24(z+3)=0 \\ \Rightarrow 17x-47y-24z+172=0 \end{gathered}$$

Hence, the required equation of the plane is $17x-47y-24z+172=0$.