Show that lines $\dfrac{x+1}{-10} = \dfrac{y+3}{-1} = \dfrac{z-4}{1}$ and $\dfrac{x+10}{-1} = \dfrac{y+1}{-3} = \dfrac{z-1}{4}$ intersect each other. Find the co-ordinates of their point of intersection.

Question

Accepted Answer

Let the common ratio on the first line, $\frac{x+1}{-10}=\frac{y+3}{-1}=\frac{z-4}{1}$, be $\lambda$, giving the general point $(-1-10\lambda,\ -3-\lambda,\ 4+\lambda)$. Let the common ratio on the second line, $\frac{x+10}{-1}=\frac{y+1}{-3}=\frac{z-1}{4}$, be $\mu$, giving the general point $(-10-\mu,\ -1-3\mu,\ 1+4\mu)$.

The lines intersect exactly when these two general points coincide for some $\lambda, \mu$, i.e. when all three coordinates match:
$$-1-10\lambda = -10-\mu,\qquad -3-\lambda = -1-3\mu,\qquad 4+\lambda = 1+4\mu.$$
Simplifying gives $10\lambda - \mu = 9$, $\lambda - 3\mu = -2$, $\lambda - 4\mu = -3$ — three equations in two unknowns, so intersection is not automatic; it holds only if the system is consistent. Checking the determinant formed from the coefficients and constants,
$$\begin{vmatrix}10&-1&9\1&-3&-2\1&-4&-3\end{vmatrix} = 10(9-8) + 1(-3+2) + 9(-4+3) = 10-1-9 = 0,$$ …

Q.Show that lines $\dfrac{x+1}{-10} = \dfrac{y+3}{-1} = \dfrac{z-4}{1}$ and $\dfrac{x+10}{-1} = \dfrac{y+1}{-3} = \dfrac{z-1}{4}$ intersect each other. Find the co-ordinates of their point of intersection.

Concept understanding — Cartesian Equations of a Line through a Point with Given Direction Ratios

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