Question

Represent the equations of lines $l_1$ and $l_2$ in vector form and check whether they are intersecting or not. \[ l_1 : \frac{x + 3}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5} \quad \text{and} \quad l_2 : \frac{x + 1}{-1} = \frac{2 - y}{-2} = \frac{z - 5}{5} \]

Hint:

Always double check the $z$-coordinate equation first if its expressions look identical (here, $5 + 5\lambda = 5 + 5\mu$). It immediately tells you that $\lambda = \mu$, saving you from solving a tedious system of simultaneous equations.

Answer 1

Concept: The standard Cartesian form of a line is $\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}$, which translates to the vector form $\vec{r} = \vec{a} + \lambda \vec{b}$, where $\vec{a} = x_1\hat{i} + y_1\hat{j} + z_1\hat{k}$ is a passing point vector, and $\vec{b} = a\hat{i} + b\hat{j} + c\hat{k}$ is the direction vector. If lines intersect, equating their parametric coordinates must yield consistent values for the scale parameters.

Step 1: Write line $l_1$ and $l_2$ in standard Cartesian and vector form.
Line $l_1$ is already in standard form: \[ \frac{x - (-3)}{-3} = \frac{y - 1}{1} = \frac{z - 5}{5} \implies \vec{r}_1 = (-3\hat{i} + \hat{j} + 5\hat{k}) + \lambda(-3\hat{i} + \hat{j} + 5\hat{k}) \] For line $l_2$, look at the middle fraction $\frac{2 - y}{-2}$. To convert the $y$ term's coefficient to $+1$, multiply the numerator and denominator by $-1$: \[ \frac{2 - y}{-2} = \frac{y - 2}{2} \implies l_2 : \frac{x - (-1)}{-1} = \frac{y - 2}{2} = \frac{z - 5}{5} \] Thus, the vector form of line $l_2$ is: \[ \vec{r}_2 = (-\hat{i} + 2\hat{j} + 5\hat{k}) + \mu(-\hat{i} + 2\hat{j} + 5\hat{k}) \]

Step 2: Define parametric coordinates for generic points on both lines.
Let a general point on line $l_1$ be expressed in terms of parameter $\lambda$: \[ x = -3 - 3\lambda, \quad y = 1 + \lambda, \quad z = 5 + 5\lambda \] Let a general point on line $l_2$ be expressed in terms of parameter $\mu$: \[ x = -1 - \mu, \quad y = 2 + 2\mu, \quad z = 5 + 5\mu \]

Step 3: Equate coordinate equations to solve for parameters $\lambda$ and $\mu$.
Equating the $x$ and $y$ coordinates: \[ -3 - 3\lambda = -1 - \mu \implies 3\lambda - \mu = -2 \quad \text{--- (1)} \] \[ 1 + \lambda = 2 + 2\mu \implies \lambda - 2\mu = 1 \quad \text{--- (2)} \] Multiply equation (1) by 2: \[ 6\lambda - 2\mu = -4 \quad \text{--- (3)} \] Subtract equation (2) from equation (3): \[ (6\lambda - 2\mu) - (\lambda - 2\mu) = -4 - 1 \implies 5\lambda = -5 \implies \lambda = -1 \] Substitute $\lambda = -1$ back into equation (2): \[ -1 - 2\mu = 1 \implies -2\mu = 2 \implies \mu = -1 \]

Step 4: Check consistency by testing values in the $z$-coordinate equation.
Substitute $\lambda = -1$ and $\mu = -1$ into the $z$-coordinate expressions:
• For line $l_1$: $z = 5 + 5(-1) = 0$
• For line $l_2$: $z = 5 + 5(-1) = 0$ Since $0 = 0$, the values of the parameters are perfectly consistent across all three axes. Thus, the two lines do intersect.

Step 5: Find the unique point of intersection.
Substitute $\lambda = -1$ into the coordinate formulas of line $l_1$: \[ x = -3 - 3(-1) = 0, \quad y = 1 + (-1) = 0, \quad z = 5 + 5(-1) = 0 \] Wait, let's re-verify the coordinate parameters: Ah, look at the values! If $\lambda = 0$, then $x=-3, y=1, z=5$. Let's test if that satisfies $l_2$: $\frac{-3+1}{-1} = 2$, $\frac{2-1}{-2} = -0.5 \neq 2$. Let's find why there was a calculation mismatch: From step 3: General point on $l_1$: $(-3 - 3\lambda, 1 + \lambda, 5 + 5\lambda)$. General point on $l_2$: $(-1 - \mu, 2 + 2\mu, 5 + 5\mu)$. Equating $z$: $5 + 5\lambda = 5 + 5\mu \implies \lambda = \mu$. Substitute $\lambda = \mu$ into the $x$ equation: $-3 - 3\lambda = -1 - \lambda \implies -2 = 2\lambda \implies \lambda = -1 \implies \mu = -1$. Now substitute $\lambda = -1$ into the $y$ equation: $y_1 = 1 + (-1) = 0$. Substitute $\mu = -1$ into the $y$ equation of line 2: $y_2 = 2 + 2(-1) = 0$. The coordinates match! $x = -3 -3(-1) = 0$, $y = 0$, $z = 5 + 5(-1) = 0$. Thus, the correct intersection point is the origin $(0, 0, 0)$.