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CBSE Board Examination 2025 · Section D · 5 marks

Q34: Image of A(1, 6, 3) in a line and the line AA′, or a point P at distance 7 from Q(2, 4, −1-1⁠)

Board 2025Ch 11 · Three Dimensional Geometry
The question

(a) Find the image A′ of the point A(1, 6, 3) in the line x1=y−12=z−23\dfrac{x}{1} = \dfrac{y - 1}{2} = \dfrac{z - 2}{3}⁠. Also, find the equation of the line joining A and A′.

OR

(b) Find a point P on the line x+51=y+34=z−6−9\dfrac{x + 5}{1} = \dfrac{y + 3}{4} = \dfrac{z - 6}{-9} such that its distance from point Q(2, 4, −1-1⁠) is 7 units. Also, find the equation of line joining P and Q.

Exam answer

(a)

Let x1=y−12=z−23=λ\frac{x}{1} = \frac{y - 1}{2} = \frac{z - 2}{3} = \lambda⁠.

Any point on the line is M(λ,2λ+1,3λ+2)M(\lambda, 2\lambda + 1, 3\lambda + 2)⁠. Let M be the foot of the perpendicular from A to the line.

Why?

Setting each fraction equal to λ\lambda gives x=λx = \lambda⁠, y=2λ+1y = 2\lambda + 1⁠, z=3λ+2z = 3\lambda + 2⁠. Every point of the line has this form for some λ\lambda⁠, so the foot M does too. We only need to find its λ\lambda⁠.

Direction ratios of AM are λ−1,2λ−5,3λ−1\lambda - 1, 2\lambda - 5, 3\lambda - 1⁠.

Why?

The direction ratios of the join of two points are the differences of their coordinates, "M minus A":

λ−1, (2λ+1)−6,(3λ+2)−3\begin{aligned} &\lambda - 1,\ (2\lambda + 1) - 6, \\ &(3\lambda + 2) - 3 \end{aligned}

AM is perpendicular to the line, whose direction ratios are 1,2,31, 2, 3⁠, so

1(λ−1)+2(2λ−5)+3(3λ−1)=0⇒ 14λ−14=0⇒λ=1\begin{aligned} &1(\lambda - 1) + 2(2\lambda - 5) \\ &\quad + 3(3\lambda - 1) = 0 \\ \Rightarrow\ &14\lambda - 14 = 0 \Rightarrow \lambda = 1 \end{aligned}
Why?

Two lines with direction ratios a1,b1,c1a_1, b_1, c_1 and a2,b2,c2a_2, b_2, c_2 are perpendicular when

a1a2+b1b2+c1c2=0a_1a_2 + b_1b_2 + c_1c_2 = 0

The direction ratios of the given line are its denominators 1,2,31, 2, 3⁠. Expanding: λ−1+4λ−10+9λ−3=14λ−14\lambda - 1 + 4\lambda - 10 + 9\lambda - 3 = 14\lambda - 14⁠.

1(λ−1)=λ−12(2λ−5)=4λ−103(3λ−1)=9λ−3λ+4λ+9λ=14λ−1−10−3=−1414λ−14=014λ=14λ=1\begin{aligned} 1(\lambda - 1) &= \lambda - 1 \\ 2(2\lambda - 5) &= 4\lambda - 10 \\ 3(3\lambda - 1) &= 9\lambda - 3 \\[4pt] \lambda + 4\lambda + 9\lambda &= 14\lambda \\ -1 - 10 - 3 &= -14 \\[4pt] 14\lambda - 14 &= 0 \\ 14\lambda &= 14 \\ \lambda &= 1 \end{aligned}

So the foot of the perpendicular is M(1,3,5)M(1, 3, 5)⁠.

Why?

Put λ=1\lambda = 1 in M(λ,2λ+1,3λ+2)M(\lambda, 2\lambda + 1, 3\lambda + 2)⁠: this gives (1,2+1,3+2)(1, 2 + 1, 3 + 2)⁠.

Let the image be A′(α,β,γ)A'(\alpha, \beta, \gamma)⁠. M is the mid-point of AA′, so

1+α2=1, 6+β2=3, 3+γ2=5\frac{1 + \alpha}{2} = 1,\ \frac{6 + \beta}{2} = 3,\ \frac{3 + \gamma}{2} = 5 ⇒α=1, β=0, γ=7\Rightarrow \alpha = 1,\ \beta = 0,\ \gamma = 7
Why?

The image of A in the line is as far beyond the line as A is in front of it, along the same perpendicular. So the foot M is exactly half-way between A and A′.

The mid-point formula averages each coordinate, so each coordinate of M is the average of those of A and A′. Then 1+α=21 + \alpha = 2⁠, 6+β=66 + \beta = 6 and 3+γ=103 + \gamma = 10⁠.

Multiply each equation by 2, then subtract:

1+α=2⇒α=2−1=16+β=6⇒β=6−6=03+γ=10⇒γ=10−3=7\begin{aligned} 1 + \alpha &= 2 \Rightarrow \alpha = 2 - 1 = 1 \\ 6 + \beta &= 6 \Rightarrow \beta = 6 - 6 = 0 \\ 3 + \gamma &= 10 \Rightarrow \gamma = 10 - 3 = 7 \end{aligned}

The image is A′(1,0,7)A'(1, 0, 7)⁠.

The line AA′ passes through A(1, 6, 3) and has direction ratios 1−1,0−6,7−31 - 1, 0 - 6, 7 - 3⁠, that is 0,−6,40, -6, 4⁠, or 0,−3,20, -3, 2⁠. Its equation is

x−10=y−6−3=z−32\frac{x - 1}{0} = \frac{y - 6}{-3} = \frac{z - 3}{2}
Why?

The 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}

Direction ratios may be divided by a common number, so 0,−6,40, -6, 4 becomes 0,−3,20, -3, 2⁠.

A 00 in a denominator is not a division. It is the accepted way of writing that x−1=0x - 1 = 0⁠, that is x=1x = 1⁠, at every point of the line.

∴ A′(1,0,7)\therefore\ A'(1, 0, 7)⁠, and line AA′: x−10=y−6−3=z−32\frac{x - 1}{0} = \frac{y - 6}{-3} = \frac{z - 3}{2}⁠.

(b)

Let x+51=y+34=z−6−9=λ\frac{x + 5}{1} = \frac{y + 3}{4} = \frac{z - 6}{-9} = \lambda⁠.

Any point on the line is P(λ−5,4λ−3,−9λ+6)P(\lambda - 5, 4\lambda - 3, -9\lambda + 6)⁠.

Why?

As in (a), each fraction equal to λ\lambda gives x=λ−5x = \lambda - 5⁠, y=4λ−3y = 4\lambda - 3⁠, z=−9λ+6z = -9\lambda + 6⁠. The required point P has this form; we find λ\lambda from the distance condition.

PQ=7PQ = 7⁠, so PQ2=49PQ^2 = 49⁠:

(λ−7)2+(4λ−7)2+(−9λ+7)2=49⇒ (λ2−14λ+49)+(16λ2−56λ+49)+(81λ2−126λ+49)=49⇒ 98λ2−196λ+147=49⇒ 98(λ2−2λ+1)=0⇒ (λ−1)2=0⇒λ=1\begin{aligned} &(\lambda - 7)^2 + (4\lambda - 7)^2 \\ &\quad + (-9\lambda + 7)^2 = 49 \\ \Rightarrow\ &(\lambda^2 - 14\lambda + 49) \\ &\quad + (16\lambda^2 - 56\lambda + 49) \\ &\quad + (81\lambda^2 - 126\lambda + 49) = 49 \\ \Rightarrow\ &98\lambda^2 - 196\lambda + 147 = 49 \\ \Rightarrow\ &98(\lambda^2 - 2\lambda + 1) = 0 \\ \Rightarrow\ &(\lambda - 1)^2 = 0 \Rightarrow \lambda = 1 \end{aligned}
Why?

Use the distance formula. Squaring first avoids the square root. The coordinate differences of P and Q(2,4,−1)(2, 4, -1) are

(λ−5)−2=λ−7(4λ−3)−4=4λ−7(−9λ+6)+1=−9λ+7\begin{aligned} (\lambda - 5) - 2 &= \lambda - 7 \\ (4\lambda - 3) - 4 &= 4\lambda - 7 \\ (-9\lambda + 6) + 1 &= -9\lambda + 7 \end{aligned}

Expand each square with (p+q)2=p2+2pq+q2(p + q)^2 = p^2 + 2pq + q^2⁠, then collect: 1+16+81=981 + 16 + 81 = 98⁠, 14+56+126=19614 + 56 + 126 = 196⁠, and 3×49=1473\times49 = 147⁠.

(λ−7)2=λ2−2(λ)(7)+72=λ2−14λ+49(4λ−7)2=(4λ)2−2(4λ)(7)+72=16λ2−56λ+49(−9λ+7)2=(−9λ)2+2(−9λ)(7)+72=81λ2−126λ+49\begin{aligned} (\lambda - 7)^2 &= \lambda^2 - 2(\lambda)(7) + 7^2 \\ &= \lambda^2 - 14\lambda + 49 \\[4pt] (4\lambda - 7)^2 &= (4\lambda)^2 - 2(4\lambda)(7) + 7^2 \\ &= 16\lambda^2 - 56\lambda + 49 \\[4pt] (-9\lambda + 7)^2 &= (-9\lambda)^2 + 2(-9\lambda)(7) + 7^2 \\ &= 81\lambda^2 - 126\lambda + 49 \end{aligned}

Subtracting 4949 leaves 98λ2−196λ+9898\lambda^2 - 196\lambda + 98⁠, and λ2−2λ+1=(λ−1)2\lambda^2 - 2\lambda + 1 = (\lambda - 1)^2⁠.

147−49=9898λ2−196λ+98=098(λ2−2λ+1)=0(196=2×98)λ2−2λ+1=0(÷ 98)(λ−1)2=0λ−1=0⇒λ=1\begin{aligned} 147 - 49 &= 98 \\ 98\lambda^2 - 196\lambda + 98 &= 0 \\ 98(\lambda^2 - 2\lambda + 1) &= 0 \quad (196 = 2 \times 98) \\ \lambda^2 - 2\lambda + 1 &= 0 \quad (\div\, 98) \\ (\lambda - 1)^2 &= 0 \\ \lambda - 1 &= 0 \Rightarrow \lambda = 1 \end{aligned}

Hence the required point is P(−4,1,−3)P(-4, 1, -3)⁠.

Why?

Put λ=1\lambda = 1⁠: (1−5,4−3,−9+6)=(−4,1,−3)(1 - 5, 4 - 3, -9 + 6) = (-4, 1, -3)⁠.

Check: PQ2=62+32+22=49PQ^2 = 6^2 + 3^2 + 2^2 = 49⁠, so PQ=7PQ = 7⁠.

2−(−4)=64−1=3−1−(−3)=262+32+22=36+9+4=4949=7\begin{aligned} 2 - (-4) &= 6 \\ 4 - 1 &= 3 \\ -1 - (-3) &= 2 \\[4pt] 6^2 + 3^2 + 2^2 &= 36 + 9 + 4 = 49 \\ \sqrt{49} &= 7 \end{aligned}

The double root λ=1\lambda = 1 means there is only one such point on the line.

Direction ratios of PQ are 2−(−4),4−1,−1−(−3)2 - (-4), 4 - 1, -1 - (-3)⁠, that is 6,3,26, 3, 2⁠. The line through P(−4,1,−3)P(-4, 1, -3) is

x+46=y−13=z+32\frac{x + 4}{6} = \frac{y - 1}{3} = \frac{z + 3}{2}
Why?

Same form as in (a): x−(−4)=x+4x - (-4) = x + 4 and z−(−3)=z+3z - (-3) = z + 3⁠.

Using Q instead of P as the point is also correct. This gives the same line:

x−26=y−43=z+12\frac{x - 2}{6} = \frac{y - 4}{3} = \frac{z + 1}{2}

∴ P(−4,1,−3)\therefore\ P(-4, 1, -3)⁠, and line PQ: x+46=y−13=z+32\dfrac{x + 4}{6} = \dfrac{y - 1}{3} = \dfrac{z + 3}{2}⁠.

Where this comes from

Chapter 11, Three Dimensional Geometry: Image of a point in a line; a point at distance 7.

Question type: (a) the foot of a perpendicular + The line through two points (combined); (b) The line through two points

How to spot it: click to see

(a) 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: First find the foot from A(1,6,3)A(1, 6, 3)⁠: M(λ,2λ+1,3λ+2)M(\lambda, 2\lambda + 1, 3\lambda + 2) with AM perpendicular to 1,2,31, 2, 3 gives M(1,3,5)M(1, 3, 5)⁠. M is the midpoint of AA′AA'⁠, so A′=(1,0,7)A' = (1, 0, 7)⁠.

(a) Type 2. The line through two points: subtract, then divide

Spot it: two points are given, and you need the direction of the line through them.

Here: Then the line through A and A′A'⁠: subtracting gives 0,−6,40, -6, 4⁠, or 0,−3,20, -3, 2⁠, so AA' is x−10=y−6−3=z−32\frac{x - 1}{0} = \frac{y - 6}{-3} = \frac{z - 3}{2}⁠.

(b) Type 2. The line through two points: subtract, then divide

Spot it: two points are given, and you need the direction of the line through them.

Here: First write P as (λ−5,4λ−3,−9λ+6)(\lambda - 5, 4\lambda - 3, -9\lambda + 6) and set PQ2=49PQ^2 = 49⁠, which gives P(−4,1,−3)P(-4, 1, -3)⁠. For the line PQ, Q minus P gives the direction ratios 6,3,26, 3, 2⁠.

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