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CBSE Sample Paper 2025-26 · Section D · 5 marks

Q35: Point of intersection of two lines and the line perpendicular to both

Sample 2025-26Ch 11 · Three Dimensional GeometryCh 10 · Vector Algebra
The question

The two lines x−13=−y\dfrac{x - 1}{3} = -y⁠, z+1=0z + 1 = 0 and −x2=y+12=z+2\dfrac{-x}{2} = \dfrac{y + 1}{2} = z + 2 intersect at a point whose y-coordinate is 1. Find the co-ordinates of their point of intersection. Find the vector equation of a line perpendicular to both the given lines and passing through this point of intersection.

Exam answer

The first line can be written as

x−13=y−0−1=z+10=λ\frac{x - 1}{3} = \frac{y - 0}{-1} = \frac{z + 1}{0} = \lambda
Why?

The symmetric form of a line through (x1,y1,z1)(x_1, y_1, z_1) with direction ratios a,b,ca, b, c is

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

−y-y is written as y−0−1\frac{y - 0}{-1}⁠. The condition z+1=0z + 1 = 0 says zz never changes, so its direction ratio is 0. The "z+10\frac{z + 1}{0}" is only a convention for z=−1z = -1⁠; we never divide by 0.

Any point on it is (3λ+1,−λ,−1)(3\lambda + 1, -\lambda, -1)⁠.

Why?

Set each fraction equal to λ\lambda and solve: x−1=3λx - 1 = 3\lambda⁠, y=−λy = -\lambda⁠, and z=−1z = -1⁠.

For the point where y=1y = 1⁠: −λ=1⇒λ=−1-\lambda = 1 \Rightarrow \lambda = -1⁠. Then x=3(−1)+1=−2x = 3(-1) + 1 = -2 and z=−1z = -1⁠.

The point of intersection is (−2,1,−1)(-2, 1, -1)⁠.

(It lies on the second line: 22=1+12=−1+2=1\frac{2}{2} = \frac{1 + 1}{2} = -1 + 2 = 1⁠.)

Why?

The question tells us the y-coordinate is 1, so one line is enough to find the point. Putting it into the second line is a check that the lines really meet there.

For x=−2x = -2⁠: −x2=22=1\frac{-x}{2} = \frac{2}{2} = 1⁠.

−x2=−(−2)2=22=1y+12=1+12=22=1z+2=−1+2=1\begin{aligned} \frac{-x}{2} &= \frac{-(-2)}{2} = \frac22 = 1 \\ \frac{y + 1}{2} &= \frac{1 + 1}{2} = \frac22 = 1 \\ z + 2 &= -1 + 2 = 1 \end{aligned}

All three are equal, so the point is on the second line too. (Earlier, x=3(−1)+1=−3+1=−2x = 3(-1) + 1 = -3 + 1 = -2⁠.)

The second line is x−2=y+12=z+21\frac{x}{-2} = \frac{y + 1}{2} = \frac{z + 2}{1}⁠. The directions of the two lines are

m⃗=3i^−j^,n⃗=−2i^+2j^+k^\vec m = 3\hat i - \hat j,\quad \vec n = -2\hat i + 2\hat j + \hat k
Why?

The direction ratios are the denominators, once each numerator is written as x−x1x - x_1⁠, y−y1y - y_1 or z−z1z - z_1 with coefficient 1.

So −x2\frac{-x}{2} becomes x−2\frac{x}{-2} and z+2z + 2 becomes z+21\frac{z + 2}{1}⁠. The first line has ratios 3,−1,03, -1, 0⁠; the second has −2,2,1-2, 2, 1⁠.

A line perpendicular to both is parallel to m⃗×n⃗\vec m \times \vec n⁠:

Why?

The cross product of two vectors is perpendicular to each of them. So m⃗×n⃗\vec m \times \vec n is a direction perpendicular to both lines.

m⃗×n⃗=∣i^j^k^3−10−221∣=(−1−0)i^−(3−0)j^+(6−2)k^=−i^−3j^+4k^\begin{aligned} \vec m \times \vec n &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 3 & -1 & 0 \\ -2 & 2 & 1 \end{vmatrix} \\ &= (-1 - 0)\hat i - (3 - 0)\hat j \\ &\quad + (6 - 2)\hat k \\ &= -\hat i - 3\hat j + 4\hat k \end{aligned}
Why?

Expand along the first row with signs +,−,++, -, +⁠:

i^:(−1)(1)−(0)(2)=−1j^:(3)(1)−(0)(−2)=3k^:(3)(2)−(−1)(−2)=4\begin{aligned} \hat i&: (-1)(1) - (0)(2) = -1 \\ \hat j&: (3)(1) - (0)(-2) = 3 \\ \hat k&: (3)(2) - (-1)(-2) = 4 \end{aligned}

Step by step, cover the row and column of each unit vector:

i^:∣−1021∣=(−1)(1)−(0)(2)=−1−0=−1j^:∣30−21∣=(3)(1)−(0)(−2)=3−0=3k^:∣3−1−22∣=(3)(2)−(−1)(−2)=6−2=4\begin{aligned} \hat i&: \begin{vmatrix} -1 & 0 \\ 2 & 1 \end{vmatrix} \\ &= (-1)(1) - (0)(2) \\ &= -1 - 0 = -1 \\[4pt] \hat j&: \begin{vmatrix} 3 & 0 \\ -2 & 1 \end{vmatrix} \\ &= (3)(1) - (0)(-2) \\ &= 3 - 0 = 3 \\[4pt] \hat k&: \begin{vmatrix} 3 & -1 \\ -2 & 2 \end{vmatrix} \\ &= (3)(2) - (-1)(-2) \\ &= 6 - 2 = 4 \end{aligned}

With the signs: +(−1)i^−(3)j^+(4)k^=−i^−3j^+4k^+(-1)\hat i - (3)\hat j + (4)\hat k = -\hat i - 3\hat j + 4\hat k⁠.

Check with dot products, which must be 0: −3+3+0=0-3 + 3 + 0 = 0 with m⃗\vec m⁠, and 2−6+4=02 - 6 + 4 = 0 with n⃗\vec n⁠.

(−1)(3)+(−3)(−1)+(4)(0)=−3+3+0=0(−1)(−2)+(−3)(2)+(4)(1)=2−6+4=0\begin{aligned} &(-1)(3) + (-3)(-1) + (4)(0) \\ &= -3 + 3 + 0 = 0 \\[4pt] &(-1)(-2) + (-3)(2) + (4)(1) \\ &= 2 - 6 + 4 = 0 \end{aligned}

The equation of the required line is

Why?

A line through the point with position vector a⃗\vec a⁠, parallel to b⃗\vec b⁠, is r⃗=a⃗+μb⃗\vec r = \vec a + \mu\vec b⁠.

Here a⃗=−2i^+j^−k^\vec a = -2\hat i + \hat j - \hat k⁠, the point of intersection, and b⃗=m⃗×n⃗\vec b = \vec m \times \vec n⁠.

r⃗=(−2i^+j^−k^)\vec r = (-2\hat i + \hat j - \hat k) + μ(−i^−3j^+4k^)+\ \mu(-\hat i - 3\hat j + 4\hat k)

Where this comes from

Chapter 11, Three Dimensional Geometry: Intersecting lines; a line perpendicular to both.

Question type: From Cartesian to vector form + A line perpendicular to two lines (combines 2 types)

How to spot it: click to see

Type 5. From Cartesian to vector form: read off the point and the direction

Spot it: a line is given as three equal fractions, and you need its vector form, its point or its direction.

Here: First, x−13=−y\frac{x - 1}{3} = -y⁠, z+1=0z + 1 = 0 is not three fractions: write it as x−13=y−1=z+10\frac{x - 1}{3} = \frac{y}{-1} = \frac{z + 1}{0}⁠. Its points are (3λ+1,−λ,−1)(3\lambda + 1, -\lambda, -1)⁠, and y=1y = 1 gives (−2,1,−1)(-2, 1, -1)⁠.

Type 9. A line perpendicular to two lines: use the cross product

Spot it: "Find the line through the point … perpendicular to the two lines …". The CBSE sample papers for 2024-25, 2025-26 and 2026-27 all ask it.

Here: The new line is perpendicular to both, so its direction is (3i^−j^)×(−2i^+2j^+k^)=−i^−3j^+4k^(3\hat i - \hat j)\times(-2\hat i + 2\hat j + \hat k) = -\hat i - 3\hat j + 4\hat k⁠, through (−2,1,−1)(-2, 1, -1)⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 10, Vector Algebra.

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