Q.Show that the lines $\dfrac{x-1}{2} = \dfrac{y-2}{3} = \dfrac{z-3}{4}$ and $\dfrac{x-4}{5} = \dfrac{y-1}{2} = z$ intersect. Also, find their point of intersection.

Concept: Angle Between Lines — but here we only need to check intersection, so we equate parametric forms.

Step 1 – Parametric form

First line:

$\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} = \lambda$

$\Rightarrow x = 1+2\lambda,\; y = 2+3\lambda,\; z = 3+4\lambda$

Second line:

$\frac{x-4}{5} = \frac{y-1}{2} = z = \mu$

$\Rightarrow x = 4+5\mu,\; y = 1+2\mu,\; z = \mu$

Step 2 – Equate coordinates

From $z$: $3+4\lambda = \mu$

From $x$: $1+2\lambda = 4+5\mu$

Substitute $\mu$: $1+2\lambda = 4+5(3+4\lambda)$

$\Rightarrow 1+2\lambda = 4+15+20\lambda$

$\Rightarrow 1+2\lambda = 19+20\lambda$

$\Rightarrow -18 = 18\lambda \Rightarrow \lambda = -1$

Then $\mu = 3+4(-1) = -1$

Step 3 – Verify with $y$

$y$ from first: $2+3(-1) = -1$

$y$ from second: $1+2(-1) = -1$ — consistent.

Thus the lines intersect at the point corresponding to $\lambda = -1$ (or $\mu = -1$).

Concept and Intuition

Two lines in 3D can be parallel, skew (non-parallel and non-intersecting), or intersecting. To check for intersection, we need two conditions:

1. The lines must not be parallel — their direction vectors should not be scalar multiples.
2. There must exist a point that lies on both lines — we find this by equating parametric forms and solving for the parameters.

The key insight: if the lines intersect, the parameters $t$ and $s$ we introduce for each line will give the same coordinates when plugged in. If the system of equations has a consistent solution, the lines meet; if not, they are skew.

Step-by-step solution

1. Write each line in parametric form.

For the first line:

This gives:

For the second line:

This gives:

2. Check if the lines are parallel.

Direction vector of line 1: $\vec{d_1} = (2, 3, 4)$

Direction vector of line 2: $\vec{d_2} = (5, 2, 1)$

Are these scalar multiples? If $\vec{d_1} = k \vec{d_2}$, then $2 = 5k$, $3 = 2k$, $4 = k$. From $4 = k$ we get $k=4$, but then $2 = 5(4) = 20$ — false. So the lines are not parallel, meaning they could intersect or be skew.

3. Equate the parametric forms to find intersection.

If the lines intersect, there exist parameters $t$ and $s$ such that:

4. Solve the system.

From equation (3): $s = 3 + 4t$

Substitute into equation (1):

Now $s = 3 + 4(-1) = 3 - 4 = -1$

5. Verify with the remaining equation.

Check equation (2): $2 + 3(-1) = 2 - 3 = -1$

Right side: $1 + 2(-1) = 1 - 2 = -1$

Both sides equal $-1$, so the solution is consistent.

6. Find the point of intersection.

Using $t = -1$ in the first line:

Using $s = -1$ in the second line as a check:

Both give the same point $(-1, -1, -1)$.

Method: Testing whether two lines intersect (and finding the point)

Use this to decide whether two lines meet and, if so, where.

Steps

Step 1: Parametrise each line with its OWN parameter.

Set the first line's equal ratios to $\lambda$ and the second's to $\mu$ — two DIFFERENT parameters. Using one symbol for both is the classic error, because the lines reach a common point at generally different parameter values.

Step 2: Equate coordinates and solve two of the three equations.

Match $x$, $y$, $z$ to get three equations in $\lambda,\mu$. Solve any two for $\lambda$ and $\mu$.

Step 3: Verify with the third equation.

Substitute $\lambda,\mu$ into the unused equation. If it holds, the lines intersect — put the value back to get the point. If it fails, the lines are skew and there is no intersection.

Common Mistakes

Mistake 1: Using the SAME parameter for both lines.

Why it's wrong: the two lines reach a common point at generally different parameter values, so they need independent parameters $\lambda$ and $\mu$. Correct approach: set line 1 $=\lambda$, line 2 $=\mu$, then equate coordinates.

Mistake 2: Concluding intersection from only two equations.

Why it's wrong: values of $\lambda,\mu$ solving the $x$ and $z$ equations may fail the $y$ equation, in which case the lines are skew. Correct approach: always verify with the third equation before declaring they intersect.

Mistake 3: Misreading $\frac{x-4}{5}=\frac{y-1}{2}=z$.

Why it's wrong: the bare $z$ means $\frac{z-0}{1}$, so $z$ itself equals the parameter. Correct approach: treat the $z$ direction ratio as $1$, giving $z=\mu$.