Problem Statement
Show that the lines $\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ and $\frac{x - 4}{5} = \frac{y - 1}{2} = z$ are coplanar. Also, find the equation of the plane containing them.
Verified Solution & Marking Scheme
Extract Points and Direction Ratios
Line 1: Passes through $A(1, 2, 3)$, direction ratios $\vec{b}_1 = \langle 2, 3, 4 \rangle$.
Line 2: Passes through $B(4, 1, 0)$, direction ratios $\vec{b}_2 = \langle 5, 2, 1 \rangle$.
$\vec{a}_2 - \vec{a}_1 = \langle 4 - 1, \; 1 - 2, \; 0 - 3 \rangle = \langle 3, -1, -3 \rangle$
Evaluate Scalar Triple Product Condition
$\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = \begin{vmatrix} 3 & -1 & -3 \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix}$
$= 3(3 - 8) - (-1)(2 - 20) + (-3)(4 - 15) = 3(-5) + 1(-18) - 3(-11) = -15 - 18 + 33 = 0$
Since the determinant is 0, the lines are coplanar.
Find Equation of Containing Plane
Normal to the plane is $\vec{n} = \vec{b}_1 \times \vec{b}_2 = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 5 & 2 & 1 \end{vmatrix} = \hat{i}(3 - 8) - \hat{j}(2 - 20) + \hat{k}(4 - 15) = -5\hat{i} + 18\hat{j} - 11\hat{k}$.
Plane passing through $(1, 2, 3)$:
$-5(x - 1) + 18(y - 2) - 11(z - 3) = 0 \implies -5x + 5 + 18y - 36 - 11z + 33 = 0$
$5x - 18y + 11z - 2 = 0$