CBSE Delhi 2014. Previous Year Question.
Problem
Show that the lines $\dfrac{x+1}{3}=\dfrac{y+3}{5}=\dfrac{z+5}{7}$ and $\dfrac{x-2}{1}=\dfrac{y-4}{3}=\dfrac{z-6}{5}$ intersect.
Key insight. Two lines in 3D intersect only if a single point can be written using both lines’ parametric forms simultaneously. Setting up one equation for each coordinate ($x$, $y$, $z$) gives three equations in two unknown parameters — which is one equation more than needed. If the extra equation turns out to be automatically satisfied by the parameters found from the first two, the lines genuinely meet; if not, they’re skew.
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Approach
Write a general point on each line using a separate parameter for each ($\lambda$ for the first line, $\mu$ for the second). Set the $x$, $y$, and $z$ coordinates equal between the two lines, giving three equations in $\lambda$ and $\mu$. Solve any two of them for $\lambda$ and $\mu$, then substitute into the third to check whether it’s automatically satisfied — confirming intersection if it is.
Solution
Step 1 — Write general points on both lines
For the first line: $(3\lambda-1,\ 5\lambda-3,\ 7\lambda-5)$
For the second line: $(\mu+2,\ 3\mu+4,\ 5\mu+6)$
Step 2 — Set up three equations by equating coordinates
$$3\lambda-1 = \mu+2 \implies 3\lambda-\mu=3 \quad \text{…(i)}$$
$$5\lambda-3 = 3\mu+4 \implies 5\lambda-3\mu=7 \quad \text{…(ii)}$$
$$7\lambda-5 = 5\mu+6 \implies 7\lambda-5\mu=11 \quad \text{…(iii)}$$
Step 3 — Solve equations (i) and (ii) for λ and μ
From (i): $\mu = 3\lambda-3$. Substituting into (ii):
$$5\lambda-3(3\lambda-3) = 7 \implies 5\lambda-9\lambda+9=7 \implies -4\lambda=-2 \implies \lambda=\frac{1}{2}$$
$$\mu = 3\left(\frac{1}{2}\right)-3 = -\frac{3}{2}$$
Step 4 — Check whether equation (iii) is also satisfied
$$7\left(\frac{1}{2}\right)-5\left(-\frac{3}{2}\right) = \frac{7}{2}+\frac{15}{2} = \frac{22}{2}=11 \checkmark$$
Equation (iii) is satisfied automatically — confirming the lines do intersect.
Step 5 — Find the point of intersection
Substituting $\lambda=\dfrac{1}{2}$ into the first line’s general point:
$$\left(3\left(\frac{1}{2}\right)-1,\ 5\left(\frac{1}{2}\right)-3,\ 7\left(\frac{1}{2}\right)-5\right) = \left(\frac{1}{2},\ -\frac{1}{2},\ -\frac{3}{2}\right)$$
Answer
The lines intersect at $\left(\dfrac{1}{2},\ -\dfrac{1}{2},\ -\dfrac{3}{2}\right)$.
Common mistakes
- Using only two of the three coordinate equations and stopping there. Solving any two equations for $\lambda$ and $\mu$ will always succeed (two equations, two unknowns) — but this alone doesn’t prove intersection. The third equation must also be checked; if it fails, the lines are skew, not intersecting.
- Using the same parameter letter for both lines. Since the two lines are independent, they need two different parameters ($\lambda$ and $\mu$) — using the same one implicitly (and incorrectly) forces a relationship between points on the two lines that isn’t actually given.
Practise next
- Show that the lines $\dfrac{x-1}{2}=\dfrac{y+1}{3}=\dfrac{z-2}{4}$ and $\dfrac{x-2}{1}=\dfrac{y+2}{4}=\dfrac{z-1}{5}$ intersect, using the same three-equation method.
Show answer
They meet at $(3,2,6)$. Write the lines as $(1,-1,2)+s(2,3,4)$ and $(2,-2,1)+t(1,4,5)$ and equate coordinates.
The $x$ equation gives $t=2s-1$. Substituting into the $y$ equation, $-1+3s=-2+4(2s-1)$, so $5s=5$ and $s=t=1$.
The third equation is the test: $2+4(1)=6$ and $1+5(1)=6$ agree, so the lines do intersect, at $(3,2,6)$. Equivalently the scalar triple product $\left[\vec{A_2}-\vec{A_1}\ \ \vec d_1\ \ \vec d_2\right]$ is $0$ — two skew lines would satisfy the first two equations and fail the third.
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