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CBSE Sample Paper 2025-26 · Section C · 3 marks

Q29: Distance from (2,−1,3)(2, -1, 3) to a line parallel to the z-axis, or a point of intersection and its distance from the z-axis

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

(a) Find the distance of the point (2,−1,3)(2, -1, 3) from the line

r⃗=(2i^−j^+2k^)+μ(3i^+6j^+2k^)\begin{aligned} \vec r = (2\hat i - \hat j + 2\hat k) & \\ + \mu(3\hat i + 6\hat j + 2\hat k) & \end{aligned}

measured parallel to the z-axis.

OR

(b) Find the point of intersection of the line r⃗=(3i^+k^)+μ(i^+j^+k^)\vec r = (3\hat i + \hat k) + \mu(\hat i + \hat j + \hat k) and the line through (2,−1,1)(2, -1, 1) parallel to the z-axis. How far is this point from the z-axis?

Exam answer

(a)

The line through (2,−1,3)(2, -1, 3) parallel to the z-axis is

r⃗=(2i^−j^+3k^)+λk^\vec r = (2\hat i - \hat j + 3\hat k) + \lambda\hat k
Why?

"Measured parallel to the z-axis" means: go from the point along a line parallel to the z-axis until you hit the given line, and measure that piece.

A line through the point a⃗\vec a in the direction b⃗\vec b is r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠. The z-axis has direction k^\hat k⁠.

Any point on this line is P(2,−1,3+λ)P(2, -1, 3 + \lambda)⁠.

Any point on the given line is

Q(2+3μ,−1+6μ,2+2μ)Q(2 + 3\mu, -1 + 6\mu, 2 + 2\mu)
Why?

Collect the i^\hat i⁠, j^\hat j and k^\hat k parts of each vector equation. For the given line:

x=2+3μy=−1+6μz=2+2μ\begin{aligned} x &= 2 + 3\mu \\ y &= -1 + 6\mu \\ z &= 2 + 2\mu \end{aligned}

Each value of μ\mu (or λ\lambda⁠) gives one point of the line.

For the point of intersection, Q=PQ = P⁠:

2+3μ=2⇒μ=0−1+6μ=−1(true for μ=0)3+λ=2+2μ=2⇒λ=−1\begin{aligned} 2 + 3\mu &= 2 \Rightarrow \mu = 0 \\ -1 + 6\mu &= -1\quad(\text{true for } \mu = 0) \\ 3 + \lambda &= 2 + 2\mu = 2 \Rightarrow \lambda = -1 \end{aligned}
Why?

Two points are the same when their xx⁠, yy and zz coordinates are all equal. That gives three equations in two unknowns.

Solve two of them, then check the third. Here the yy-equation also holds for μ=0\mu = 0⁠, so the lines really do meet.

Putting μ=0\mu = 0 in QQ gives the point below.

2+3μ=2⇒3μ=0⇒μ=03+λ=2⇒λ=2−3=−1Q=(2+3(0),−1+6(0),2+2(0))=(2,−1,2)\begin{aligned} 2 + 3\mu = 2 &\Rightarrow 3\mu = 0 \Rightarrow \mu = 0 \\ 3 + \lambda = 2 &\Rightarrow \lambda = 2 - 3 = -1 \\[4pt] Q &= \big(2 + 3(0), -1 + 6(0), \\ &\qquad 2 + 2(0)\big) \\ &= (2, -1, 2) \end{aligned}

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

Distance=02+02+(3−2)2=1\begin{aligned} \text{Distance} &= \sqrt{0^2 + 0^2 + (3 - 2)^2} \\ &= 1 \end{aligned}
Why?

The distance dd between (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2) is given by

d2=(x2−x1)2+(y2−y1)2+(z2−z1)2\begin{aligned} d^2 &= (x_2 - x_1)^2 + (y_2 - y_1)^2 \\ &\quad + (z_2 - z_1)^2 \end{aligned}

The points (2,−1,3)(2, -1, 3) and (2,−1,2)(2, -1, 2) differ only in zz⁠, by 1.

d2=(2−2)2+(−1−(−1))2+(2−3)2=02+02+(−1)2=0+0+1=1d=1=1\begin{aligned} d^2 &= (2 - 2)^2 + \big(-1 - (-1)\big)^2 \\ &\quad + (2 - 3)^2 \\ &= 0^2 + 0^2 + (-1)^2 \\ &= 0 + 0 + 1 = 1 \\ d &= \sqrt1 = 1 \end{aligned}

∴\therefore The required distance is 1 unit.

(b)

The line through (2,−1,1)(2, -1, 1) parallel to the z-axis is

r⃗=(2i^−j^+k^)+λk^\vec r = (2\hat i - \hat j + \hat k) + \lambda\hat k
Why?

A line through the point a⃗\vec a in the direction b⃗\vec b is r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠. Parallel to the z-axis means the direction is k^\hat k⁠.

Any point on this line is P(2,−1,1+λ)P(2, -1, 1 + \lambda)⁠.

Any point on the given line is Q(3+μ,μ,1+μ)Q(3 + \mu, \mu, 1 + \mu)⁠.

Why?

The given line is r⃗=(3+μ)i^+μj^+(1+μ)k^\vec r = (3 + \mu)\hat i + \mu\hat j + (1 + \mu)\hat k⁠, since 3i^+k^3\hat i + \hat k has no j^\hat j part.

For the point of intersection, Q=PQ = P⁠:

3+μ=2⇒μ=−1μ=−1(same)1+λ=1+μ=0⇒λ=−1\begin{aligned} 3 + \mu &= 2 \Rightarrow \mu = -1 \\ \mu &= -1\quad(\text{same}) \\ 1 + \lambda &= 1 + \mu = 0 \Rightarrow \lambda = -1 \end{aligned}
Why?

Equate the xx⁠, yy and zz coordinates. The xx- and yy-equations both give μ=−1\mu = -1⁠, so they agree and the lines meet.

Putting μ=−1\mu = -1 in QQ⁠: (3−1,−1,1−1)=(2,−1,0)(3 - 1, -1, 1 - 1) = (2, -1, 0)⁠.

The point of intersection is (2,−1,0)(2, -1, 0)⁠.

Its distance from the z-axis is x2+y2\sqrt{x^2 + y^2}⁠:

22+(−1)2=5\sqrt{2^2 + (-1)^2} = \sqrt5
Why?

The nearest point of the z-axis to (x,y,z)(x, y, z) is (0,0,z)(0, 0, z)⁠, at the same height. The distance between them is

x2+y2+02=x2+y2\sqrt{x^2 + y^2 + 0^2} = \sqrt{x^2 + y^2}

Here x=2x = 2⁠, y=−1y = -1⁠: 4+1=54 + 1 = 5⁠.

22+(−1)2=4+1=5\begin{aligned} 2^2 + (-1)^2 &= 4 + 1 = 5 \end{aligned}

∴\therefore The point of intersection is (2,−1,0)(2, -1, 0)⁠, and it is 5\sqrt5 units from the z-axis.

Where this comes from

Chapter 11, Three Dimensional Geometry: Lines in space: intersection and distance.

Question type: Write the line + the foot of a perpendicular (combines 2 types)

How to spot it: click to see

Type 4. Write the line: a point and a direction

Spot it: "Find the equation of the line through the point … parallel to (or in the direction of) …".

Here: First, the measuring line goes through the given point parallel to the z-axis, so its direction is k^\hat k⁠. (a) r⃗=2i^−j^+3k^+λk^\vec r = 2\hat i - \hat j + 3\hat k + \lambda\hat k⁠. (b) r⃗=2i^−j^+k^+λk^\vec r = 2\hat i - \hat j + \hat k + \lambda\hat k⁠.

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: Then find where it meets the given line by setting the general points equal. (a) μ=0\mu = 0 gives (2,−1,2)(2, -1, 2)⁠, 1 unit from (2,−1,3)(2, -1, 3)⁠. (b) μ=−1\mu = -1 gives (2,−1,0)(2, -1, 0)⁠, which is 5\sqrt5 from the z-axis.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 10, Vector Algebra.

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