Introduction
In three-dimensional analytical geometry, two non-parallel lines intersect if and only if there exists a unique common point satisfying both equations simultaneously. When two distinct lines intersect, they are coplanar and uniquely define a single plane containing them, whose normal vector is orthogonal to the direction vectors of both lines.
1. Proof of Intersection
Let the equations of the given lines in symmetric form be equated to parameters \(\lambda\) and \(\mu\):
- Line \(L_1\): \(\frac{x - 1}{2} = \frac{y + 1}{-3} = \frac{z + 10}{8} = \lambda\)
- Line \(L_2\): \(\frac{x - 4}{1} = \frac{y + 3}{-4} = \frac{z + 1}{7} = \mu\)
Any arbitrary point on \(L_1\) can be represented as \(P(2\lambda + 1, -3\lambda - 1, 8\lambda - 10)\) and any arbitrary point on \(L_2\) as \(Q(\mu + 4, -4\mu - 3, 7\mu - 1)\).
For the lines to intersect, points \(P\) and \(Q\) must coincide for some real values of \(\lambda\) and \(\mu\):
- Equating x-coordinates: \(2\lambda + 1 = \mu + 4 \implies 2\lambda - \mu = 3\quad\text{--- (i)}\)
- Equating y-coordinates: \(-3\lambda - 1 = -4\mu - 3 \implies 3\lambda - 4\mu = 2\quad\text{--- (ii)}\)
Multiplying equation (i) by 4 gives \(8\lambda - 4\mu = 12\). Subtracting equation (ii) from this yields:
\((8\lambda - 4\mu) - (3\lambda - 4\mu) = 12 - 2 \implies 5\lambda = 10 \implies \lambda = 2\).
Substituting \(\lambda = 2\) into equation (i): \(2(2) - \mu = 3 \implies \mu = 1\).
Now, substitute \(\lambda = 2\) and \(\mu = 1\) into the expressions for the z-coordinate:
- For \(L_1\): \(z_1 = 8(2) - 10 = 6\)
- For \(L_2\): \(z_2 = 7(1) - 1 = 6\)
Since \(z_1 = z_2 = 6\), the values of \(\lambda\) and \(\mu\) satisfy all three coordinate equations consistently. Hence, the two lines intersect.
2. Point of Intersection
Substituting \(\lambda = 2\) into the coordinates of \(P\):
- \(x = 2(2) + 1 = 5\)
- \(y = -3(2) - 1 = -7\)
- \(z = 8(2) - 10 = 6\)
Thus, the coordinates of the point of intersection are \((5, -7, 6)\).
3. Equation of the Plane Containing the Lines
The direction vectors of lines \(L_1\) and \(L_2\) are \(\vec{d}_1 = 2\hat{i} - 3\hat{j} + 8\hat{k}\) and \(\vec{d}_2 = 1\hat{i} - 4\hat{j} + 7\hat{k}\) respectively.
The normal vector \(\vec{n}\) to the plane containing both lines is given by the vector cross product \(\vec{d}_1 \times \vec{d}_2\):
\(\vec{n} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & -3 & 8 \\ 1 & -4 & 7 \end{vmatrix} = \hat{i}[(-3)(7) - (8)(-4)] - \hat{j}[(2)(7) - (8)(1)] + \hat{k}[(2)(-4) - (-3)(1)]\)
\(\vec{n} = \hat{i}(-21 + 32) - \hat{j}(14 - 8) + \hat{k}(-8 + 3) = 11\hat{i} - 6\hat{j} - 5\hat{k}\).
The equation of a plane passing through \((x_0, y_0, z_0) = (5, -7, 6)\) with normal ratios \((11, -6, -5)\) is:
\(11(x - 5) - 6(y + 7) - 5(z - 6) = 0\)
\(11x - 55 - 6y - 42 - 5z + 30 = 0 \implies 11x - 6y - 5z - 67 = 0\).
Conclusion
The mutual consistency of the parameters across all three coordinates rigorously verifies the intersection of the two lines at (5, -7, 6). The plane containing both lines is uniquely determined by their common point and normal vector, yielding the Cartesian equation 11x - 6y - 5z - 67 = 0.