Class 12 Maths, Worked in Full Past papersGet updates
CBSE Board Examination 2026 · Section D · 5 marks

Q35: Are the lines parallel? If not, find their point of intersection

Board 2026Ch 11 · Three Dimensional Geometry
The question

Check whether the lines given by x−12=y−23=z−34\dfrac{x - 1}{2} = \dfrac{y - 2}{3} = \dfrac{z - 3}{4} and x−45=y−12=z\dfrac{x - 4}{5} = \dfrac{y - 1}{2} = z are parallel or not. If parallel, find the distance between them, otherwise find their point of intersection, if the lines are intersecting.

Exam answer

Direction ratios of the first line are 2,3,42, 3, 4 and of the second line are 5,2,15, 2, 1⁠.

Why?

A line in symmetric form

x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}

passes through (x1,y1,z1)(x_1, y_1, z_1) and has direction ratios a,b,ca, b, c⁠, the denominators.

Watch out: zz on its own means z−01\frac{z - 0}{1}⁠, so the third ratio of the second line is 1.

As 25≠32\frac25 \ne \frac32⁠, the direction ratios are not proportional. Hence the lines are not parallel.

Why?

Two lines are parallel exactly when their direction ratios are proportional:

a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

One unequal pair is enough to rule it out. Here 25\frac25⁠, 32\frac32 and 41\frac41 are all different.

Any point on the first line is (2λ+1,3λ+2,4λ+3)(2\lambda + 1, 3\lambda + 2, 4\lambda + 3)⁠.

Why?

Set each fraction equal to a parameter λ\lambda⁠:

x−12=y−23=z−34=λ\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4} = \lambda

Then x−1=2λx - 1 = 2\lambda⁠, so x=2λ+1x = 2\lambda + 1⁠, and the same for yy and zz⁠. Each value of λ\lambda gives one point of the line.

Any point on the second line is (5μ+4,2μ+1,μ)(5\mu + 4, 2\mu + 1, \mu)⁠.

Why?

Use a different parameter μ\mu for the second line. The two lines need not reach the common point at the same parameter value.

The lines intersect if, for some λ\lambda and μ\mu⁠,

2λ+1=5μ+4…(i)3λ+2=2μ+1…(ii)4λ+3=μ…(iii)\begin{aligned} 2\lambda + 1 &= 5\mu + 4 \quad\ldots(\text{i}) \\ 3\lambda + 2 &= 2\mu + 1 \quad\ldots(\text{ii}) \\ 4\lambda + 3 &= \mu \quad\ldots(\text{iii}) \end{aligned}
Why?

A common point has the same xx⁠, the same yy and the same zz on both lines. That gives three equations in only two unknowns.

Bring the λ\lambda and μ\mu terms to the left and the numbers to the right in (i) and (ii):

2λ−5μ=4−1=33λ−2μ=1−2=−1\begin{aligned} 2\lambda - 5\mu &= 4 - 1 = 3 \\ 3\lambda - 2\mu &= 1 - 2 = -1 \end{aligned}

Multiply every term:

3(2λ−5μ)=3(3)6λ−15μ=92(3λ−2μ)=2(−1)6λ−4μ=−2\begin{aligned} 3(2\lambda - 5\mu) &= 3(3) \\ 6\lambda - 15\mu &= 9 \\[4pt] 2(3\lambda - 2\mu) &= 2(-1) \\ 6\lambda - 4\mu &= -2 \end{aligned}

From (i) and (ii), 2λ−5μ=32\lambda - 5\mu = 3 and 3λ−2μ=−13\lambda - 2\mu = -1⁠. Multiplying the first by 3 and the second by 2,

6λ−15μ=96λ−4μ=−2\begin{aligned} 6\lambda - 15\mu &= 9 \\ 6\lambda - 4\mu &= -2 \end{aligned}
Why?

Making the λ\lambda coefficients equal (both 6) lets us eliminate λ\lambda by subtracting. In the next line, −4μ−(−15μ)=11μ-4\mu - (-15\mu) = 11\mu and −2−9=−11-2 - 9 = -11⁠.

(6λ−4μ)−(6λ−15μ)=−2−96λ−6λ−4μ+15μ=−1111μ=−11\begin{aligned} &(6\lambda - 4\mu) - (6\lambda - 15\mu) \\ &= -2 - 9 \\[4pt] &6\lambda - 6\lambda - 4\mu + 15\mu = -11 \\ &11\mu = -11 \end{aligned}

Subtracting the first from the second, 11μ=−1111\mu = -11⁠, so μ=−1\mu = -1⁠.

Then 2λ=3+5μ=−22\lambda = 3 + 5\mu = -2⁠, so λ=−1\lambda = -1⁠.

Why?

Divide both sides by 11: μ=−1111=−1\mu = \frac{-11}{11} = -1⁠.

Put μ=−1\mu = -1 into 2λ−5μ=32\lambda - 5\mu = 3⁠:

2λ−5(−1)=32λ+5=32λ=3−5=−2λ=−22=−1\begin{aligned} 2\lambda - 5(-1) &= 3 \\ 2\lambda + 5 &= 3 \\ 2\lambda &= 3 - 5 = -2 \\ \lambda &= \frac{-2}{2} = -1 \end{aligned}

In (iii): 4(−1)+3=−1=μ4(-1) + 3 = -1 = \mu⁠. So (iii) is also satisfied, and the lines intersect.

Why?

Two equations in two unknowns can almost always be solved, so (i) and (ii) alone prove nothing. The third equation decides: if it holds, the lines meet; if it fails, the lines are skew (not parallel and not meeting).

Putting λ=−1\lambda = -1 in the point of the first line: x=−2+1=−1x = -2 + 1 = -1⁠, y=−3+2=−1y = -3 + 2 = -1⁠, z=−4+3=−1z = -4 + 3 = -1⁠.

Why?

With λ=−1\lambda = -1⁠:

x=2(−1)+1=−2+1=−1y=3(−1)+2=−3+2=−1z=4(−1)+3=−4+3=−1\begin{aligned} x &= 2(-1) + 1 = -2 + 1 = -1 \\ y &= 3(-1) + 2 = -3 + 2 = -1 \\ z &= 4(-1) + 3 = -4 + 3 = -1 \end{aligned}

Check with the second line, μ=−1\mu = -1⁠: (5(−1)+4,2(−1)+1,−1)(5(-1) + 4, 2(-1) + 1, -1)⁠, which is again (−1,−1,−1)(-1, -1, -1)⁠.

5(−1)+4=−5+4=−12(−1)+1=−2+1=−1\begin{aligned} 5(-1) + 4 &= -5 + 4 = -1 \\ 2(-1) + 1 &= -2 + 1 = -1 \end{aligned}

∴\therefore The lines intersect at the point (−1,−1,−1)(-1, -1, -1)⁠.

Where this comes from

Chapter 11, Three Dimensional Geometry: Two lines: parallel or intersecting.

Question type: Perpendicular or parallel + the foot of a perpendicular (combines 2 types)

How to spot it: click to see

Type 7. Perpendicular or parallel: test the direction ratios

Spot it: "Show that the lines are perpendicular (or parallel)", or "find kk so that the lines are at right angles".

Here: First, 2,3,42, 3, 4 and 5,2,15, 2, 1 are not in proportion (25≠32\frac25 \ne \frac32⁠), so the lines are not parallel.

Type 10. Beyond the book, but in CBSE papers: the foot of a perpendicular

Spot it: "Find the foot of the perpendicular from a point to a line", "the image of a point in a line", or "do the lines meet?". This book has none of these, but recent CBSE sample papers ask all three.

Here: Not parallel, so check whether they meet: set (2λ+1,3λ+2,4λ+3)(2\lambda + 1, 3\lambda + 2, 4\lambda + 3) equal to (5μ+4,2μ+1,μ)(5\mu + 4, 2\mu + 1, \mu)⁠, solve two equations for λ=μ=−1\lambda = \mu = -1⁠, and check the third.

Quick lesson · Worked NCERT solutions · Interactive lesson

Found an error?

Tell us if something on this page is wrong or unclear. Send feedback on this question →

Three Dimensional Geometry in past papers: question 3 of 35 · see them all