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CBSE Sample Paper 2024-25 · Section D · 5 marks

Q35: Shortest distance between two lines, or the image of (1,2,1)(1, 2, 1) in a line

Sample 2024-25Ch 11 · Three Dimensional Geometry
The question

(a) Find the shortest distance between the lines l1l_1 and l2l_2 whose vector equations are

r⃗=(−i^−j^−k^)+λ(7i^−6j^+k^)\begin{aligned} \vec r &= (-\hat i - \hat j - \hat k) \\ &\quad + \lambda(7\hat i - 6\hat j + \hat k) \end{aligned}

and

r⃗=(3i^+5j^+7k^)+μ(i^−2j^+k^)\begin{aligned} \vec r &= (3\hat i + 5\hat j + 7\hat k) \\ &\quad + \mu(\hat i - 2\hat j + \hat k) \end{aligned}

where λ\lambda and μ\mu are parameters.

OR

(b) Find the image of the point (1,2,1)(1, 2, 1) with respect to the line x−31=y+12=z−13\dfrac{x - 3}{1} = \dfrac{y + 1}{2} = \dfrac{z - 1}{3}⁠. Also find the equation of the line joining the given point and its image.

Exam answer

(a)

The lines are not parallel, as 7i^−6j^+k^7\hat i - 6\hat j + \hat k and i^−2j^+k^\hat i - 2\hat j + \hat k are not parallel. Let PQ, with P on l1l_1 and Q on l2l_2⁠, be the line segment perpendicular to both lines. Then PQ is the shortest distance.

Why?

The direction ratios 7, −6-6⁠, 1 and 1, −2-2⁠, 1 are not proportional (71≠−6−2\frac71 \ne \frac{-6}{-2}⁠), so the directions differ.

For two lines that are not parallel, the shortest distance is along their common perpendicular. So we look for the points P and Q where the segment PQ is perpendicular to both lines.

The position vector of P, for some λ\lambda⁠, is

(7λ−1)i^−(6λ+1)j^+(λ−1)k^(7\lambda - 1)\hat i - (6\lambda + 1)\hat j + (\lambda - 1)\hat k

and of Q, for some μ\mu⁠, is

(μ+3)i^+(−2μ+5)j^+(μ+7)k^(\mu + 3)\hat i + (-2\mu + 5)\hat j + (\mu + 7)\hat k
Why?

Every point of the line r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b has this form for some value of the parameter. Collect the components of l1l_1⁠:

i^:−1+7λj^:−1−6λ=−(6λ+1)k^:−1+λ\begin{aligned} \hat i&: -1 + 7\lambda \\ \hat j&: -1 - 6\lambda = -(6\lambda + 1) \\ \hat k&: -1 + \lambda \end{aligned}

In the same way for l2l_2⁠, with μ\mu⁠.

PQ→=(μ−7λ+4)i^+(−2μ+6λ+6)j^+(μ−λ+8)k^\begin{aligned} \overrightarrow{PQ} &= (\mu - 7\lambda + 4)\hat i \\ &\quad + (-2\mu + 6\lambda + 6)\hat j \\ &\quad + (\mu - \lambda + 8)\hat k \end{aligned}
Why?

PQ→\overrightarrow{PQ} is the position vector of Q minus that of P, component by component:

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

Open the brackets (a minus in front changes every sign inside):

i^:μ+3−7λ+1=μ−7λ+4j^:−2μ+5+6λ+1=−2μ+6λ+6k^:μ+7−λ+1=μ−λ+8\begin{aligned} \hat i&: \mu + 3 - 7\lambda + 1 \\ &= \mu - 7\lambda + 4 \\ \hat j&: -2\mu + 5 + 6\lambda + 1 \\ &= -2\mu + 6\lambda + 6 \\ \hat k&: \mu + 7 - \lambda + 1 \\ &= \mu - \lambda + 8 \end{aligned}

PQ→\overrightarrow{PQ} is perpendicular to 7i^−6j^+k^7\hat i - 6\hat j + \hat k and to i^−2j^+k^\hat i - 2\hat j + \hat k⁠:

7(μ−7λ+4)−6(−2μ+6λ+6)+(μ−λ+8)=0⇒7μ−49λ+28+12μ−36λ−36+μ−λ+8=0⇒20μ−86λ=0⇒10μ−43λ=0\begin{aligned} &7(\mu - 7\lambda + 4) - 6(-2\mu + 6\lambda + 6) \\ &\qquad + (\mu - \lambda + 8) = 0 \\ &\Rightarrow 7\mu - 49\lambda + 28 + 12\mu - 36\lambda \\ &\qquad - 36 + \mu - \lambda + 8 = 0 \\ &\Rightarrow 20\mu - 86\lambda = 0 \\ &\Rightarrow 10\mu - 43\lambda = 0 \end{aligned}
Why?

Perpendicular vectors have dot product 0. Dot PQ→\overrightarrow{PQ} with the direction 7, −6-6⁠, 1 of l1l_1⁠.

Multiply out each bracket:

7(μ−7λ+4)=7μ−49λ+28−6(−2μ+6λ+6)=12μ−36λ−361(μ−λ+8)=μ−λ+8\begin{aligned} &7(\mu - 7\lambda + 4) \\ &= 7\mu - 49\lambda + 28 \\[4pt] &-6(-2\mu + 6\lambda + 6) \\ &= 12\mu - 36\lambda - 36 \\[4pt] &1(\mu - \lambda + 8) \\ &= \mu - \lambda + 8 \end{aligned}

Collect like terms:

7+12+1=20−49−36−1=−8628−36+8=0\begin{aligned} 7 + 12 + 1 &= 20 \\ -49 - 36 - 1 &= -86 \\ 28 - 36 + 8 &= 0 \end{aligned}

Then divide by 2.

20μ2−86λ2=0210μ−43λ=0\begin{aligned} \frac{20\mu}{2} - \frac{86\lambda}{2} &= \frac02 \\ 10\mu - 43\lambda &= 0 \end{aligned}
(μ−7λ+4)−2(−2μ+6λ+6)+(μ−λ+8)=0⇒μ−7λ+4+4μ−12λ−12+μ−λ+8=0⇒6μ−20λ=0⇒3μ−10λ=0\begin{aligned} &(\mu - 7\lambda + 4) - 2(-2\mu + 6\lambda + 6) \\ &\qquad + (\mu - \lambda + 8) = 0 \\ &\Rightarrow \mu - 7\lambda + 4 + 4\mu - 12\lambda \\ &\qquad - 12 + \mu - \lambda + 8 = 0 \\ &\Rightarrow 6\mu - 20\lambda = 0 \\ &\Rightarrow 3\mu - 10\lambda = 0 \end{aligned}
Why?

The same with the direction 1, −2-2⁠, 1 of l2l_2⁠:

1(μ−7λ+4)=μ−7λ+4−2(−2μ+6λ+6)=4μ−12λ−121(μ−λ+8)=μ−λ+8\begin{aligned} &1(\mu - 7\lambda + 4) \\ &= \mu - 7\lambda + 4 \\[4pt] &-2(-2\mu + 6\lambda + 6) \\ &= 4\mu - 12\lambda - 12 \\[4pt] &1(\mu - \lambda + 8) \\ &= \mu - \lambda + 8 \end{aligned}

Collect like terms:

1+4+1=6−7−12−1=−204−12+8=0\begin{aligned} 1 + 4 + 1 &= 6 \\ -7 - 12 - 1 &= -20 \\ 4 - 12 + 8 &= 0 \end{aligned}

Then divide by 2.

6μ2−20λ2=023μ−10λ=0\begin{aligned} \frac{6\mu}{2} - \frac{20\lambda}{2} &= \frac02 \\ 3\mu - 10\lambda &= 0 \end{aligned}
3μ=10λ⇒μ=10λ3⇒100λ3=43λ⇒100λ=129λ⇒λ=0\begin{aligned} &3\mu = 10\lambda \Rightarrow \mu = \frac{10\lambda}{3} \\ &\Rightarrow \frac{100\lambda}{3} = 43\lambda \\ &\Rightarrow 100\lambda = 129\lambda \Rightarrow \lambda = 0 \end{aligned}
Why?

Put μ\mu from the second equation into 10μ−43λ=010\mu - 43\lambda = 0⁠. Then 29λ=029\lambda = 0⁠, so λ=0\lambda = 0⁠, and μ=10(0)3=0\mu = \frac{10(0)}{3} = 0⁠.

10(10λ3)−43λ=0100λ3=43λ100λ=129λ(× 3)0=129λ−100λ29λ=0λ=0(÷ 29)\begin{aligned} 10\left(\frac{10\lambda}{3}\right) - 43\lambda &= 0 \\ \frac{100\lambda}{3} &= 43\lambda \\ 100\lambda &= 129\lambda \quad (\times\,3) \\ 0 &= 129\lambda - 100\lambda \\ 29\lambda &= 0 \\ \lambda &= 0 \quad (\div\,29) \end{aligned}

On solving these equations, μ=λ=0\mu = \lambda = 0⁠.

So, the position vectors of P and Q are −i^−j^−k^-\hat i - \hat j - \hat k and 3i^+5j^+7k^3\hat i + 5\hat j + 7\hat k⁠.

Why?

With λ=0\lambda = 0 and μ=0\mu = 0⁠, P and Q are the given points of the two lines, (−1,−1,−1)(-1, -1, -1) and (3,5,7)(3, 5, 7)⁠.

Then PQ→\overrightarrow{PQ} is Q minus P:

i^:3−(−1)=3+1=4j^:5−(−1)=5+1=6k^:7−(−1)=7+1=8\begin{aligned} \hat i&: 3 - (-1) = 3 + 1 = 4 \\ \hat j&: 5 - (-1) = 5 + 1 = 6 \\ \hat k&: 7 - (-1) = 7 + 1 = 8 \end{aligned}
PQ→=4i^+6j^+8k^\overrightarrow{PQ} = 4\hat i + 6\hat j + 8\hat k ∣PQ→∣=42+62+82=116\lvert\overrightarrow{PQ}\rvert = \sqrt{4^2 + 6^2 + 8^2} = \sqrt{116}
Why?

16+36+64=11616 + 36 + 64 = 116⁠, and 116=4×29116 = 4 \times 29⁠, so

42=16, 62=36, 82=6416+36=5252+64=116\begin{aligned} 4^2 = 16,\ 6^2 &= 36,\ 8^2 = 64 \\ 16 + 36 &= 52 \\ 52 + 64 &= 116 \end{aligned} 116=4×29=229\sqrt{116} = \sqrt{4 \times 29} = 2\sqrt{29}

Check with the shortest distance formula for skew lines. Here b1⃗×b2⃗=−4i^−6j^−8k^\vec{b_1} \times \vec{b_2} = -4\hat i - 6\hat j - 8\hat k and a2⃗−a1⃗=4i^+6j^+8k^\vec{a_2} - \vec{a_1} = 4\hat i + 6\hat j + 8\hat k⁠. Their dot product is −116-116⁠, so the distance is

b1⃗×b2⃗=∣i^j^k^7−611−21∣i^:(−6)(1)−(1)(−2)=−6+2=−4j^:−((7)(1)−(1)(1))=−(7−1)=−6k^:(7)(−2)−(−6)(1)=−14+6=−8(−4)(4)+(−6)(6)+(−8)(8)=−16−36−64=−116∣b1⃗×b2⃗∣=16+36+64=116\begin{aligned} \vec{b_1} \times \vec{b_2} &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 7 & -6 & 1 \\ 1 & -2 & 1 \end{vmatrix} \\[4pt] \hat i&: (-6)(1) - (1)(-2) \\ &= -6 + 2 = -4 \\ \hat j&: -\big((7)(1) - (1)(1)\big) \\ &= -(7 - 1) = -6 \\ \hat k&: (7)(-2) - (-6)(1) \\ &= -14 + 6 = -8 \\[4pt] &(-4)(4) + (-6)(6) + (-8)(8) \\ &= -16 - 36 - 64 = -116 \\[4pt] \lvert\vec{b_1} \times \vec{b_2}\rvert &= \sqrt{16 + 36 + 64} = \sqrt{116} \end{aligned} ∣−116∣116=116\frac{\lvert -116\rvert}{\sqrt{116}} = \sqrt{116}

∴\therefore Shortest distance =229= 2\sqrt{29} units.

(b)

Let P(1,2,1)P(1, 2, 1) be the given point and L the foot of the perpendicular from P to the given line.

Put x−31=y+12=z−13=λ\text{Put } \frac{x - 3}{1} = \frac{y + 1}{2} = \frac{z - 1}{3} = \lambda

Then x=λ+3x = \lambda + 3⁠, y=2λ−1y = 2\lambda - 1⁠, z=3λ+1z = 3\lambda + 1⁠, so L is (λ+3,2λ−1,3λ+1)(\lambda + 3, 2\lambda - 1, 3\lambda + 1)⁠.

Why?

Setting the equal fractions to a parameter λ\lambda gives every point of the line in one form. L lies on the line, so it has this form for one particular λ\lambda⁠.

Direction ratios of PL are

λ+3−1, 2λ−1−2, 3λ+1−1i.e. λ+2, 2λ−3, 3λ\begin{aligned} &\lambda + 3 - 1,\ 2\lambda - 1 - 2,\ 3\lambda + 1 - 1 \\ &\text{i.e. } \lambda + 2,\ 2\lambda - 3,\ 3\lambda \end{aligned}
Why?

The direction ratios of the line joining two points are the differences of their coordinates, here L minus P with P=(1,2,1)P = (1, 2, 1)⁠.

Direction ratios of the given line are 1, 2, 3, and it is perpendicular to PL:

(λ+2)1+(2λ−3)2+3λ⋅3=0⇒λ+2+4λ−6+9λ=0⇒14λ=4⇒λ=27\begin{aligned} &(\lambda + 2)1 + (2\lambda - 3)2 + 3\lambda\cdot 3 = 0 \\ &\Rightarrow \lambda + 2 + 4\lambda - 6 + 9\lambda = 0 \\ &\Rightarrow 14\lambda = 4 \Rightarrow \lambda = \frac27 \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 (λ+2)(1)=λ+2(2λ−3)(2)=4λ−6(3λ)(3)=9λ\begin{aligned} (\lambda + 2)(1) &= \lambda + 2 \\ (2\lambda - 3)(2) &= 4\lambda - 6 \\ (3\lambda)(3) &= 9\lambda \end{aligned}

Collect: 1+4+9=141 + 4 + 9 = 14 and 2−6=−42 - 6 = -4⁠, so 14λ−4=014\lambda - 4 = 0⁠.

14λ−4=014λ=4λ=414=4÷214÷2=27\begin{aligned} 14\lambda - 4 &= 0 \\ 14\lambda &= 4 \\ \lambda &= \frac{4}{14} = \frac{4 \div 2}{14 \div 2} = \frac27 \end{aligned}

Then λ+3=237\lambda + 3 = \frac{23}{7}⁠, 2λ−1=−372\lambda - 1 = -\frac37⁠, 3λ+1=1373\lambda + 1 = \frac{13}{7}⁠, so L is (237,−37,137)\left(\frac{23}{7}, -\frac37, \frac{13}{7}\right)⁠.

Why?

With λ=27\lambda = \frac27⁠:

27+3=2+21747−1=4−7767+1=6+77\begin{aligned} \frac27 + 3 &= \frac{2 + 21}{7} \\ \frac47 - 1 &= \frac{4 - 7}{7} \\ \frac67 + 1 &= \frac{6 + 7}{7} \end{aligned}

(2λ=2×27=472\lambda = 2 \times \frac27 = \frac47⁠, 3λ=3×27=673\lambda = 3 \times \frac27 = \frac67⁠, and 3=2173 = \frac{21}{7}⁠, 1=771 = \frac77⁠.) So

2+217=2374−77=−376+77=137\begin{aligned} \frac{2 + 21}{7} &= \frac{23}{7} \\ \frac{4 - 7}{7} &= -\frac37 \\ \frac{6 + 7}{7} &= \frac{13}{7} \end{aligned}

Let Q(x1,y1,z1)Q(x_1, y_1, z_1) be the image of P. Then L is the mid-point of PQ:

1+x12=237, 2+y12=−37,\frac{1 + x_1}{2} = \frac{23}{7},\ \frac{2 + y_1}{2} = -\frac37, 1+z12=137\frac{1 + z_1}{2} = \frac{13}{7}
Why?

The line acts like a mirror. The image Q is as far beyond the line as P is in front of it, along the same perpendicular. So the foot L is the mid-point of PQ.

Mid-point of (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2)⁠:

(x1+x22,y1+y22,z1+z22)\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}, \frac{z_1 + z_2}{2}\right)
⇒x1=397, y1=−207, z1=197\Rightarrow x_1 = \frac{39}{7},\ y_1 = -\frac{20}{7},\ z_1 = \frac{19}{7}
Why?

Multiply each equation by 2, then subtract:

1+x1=2×237=4672+y1=2×(−37)=−671+z1=2×137=267\begin{aligned} 1 + x_1 &= 2 \times \frac{23}{7} = \frac{46}{7} \\ 2 + y_1 &= 2 \times \left(-\frac37\right) = -\frac67 \\ 1 + z_1 &= 2 \times \frac{13}{7} = \frac{26}{7} \end{aligned}

Subtract 1, 2 and 1 (that is 77\frac77⁠, 147\frac{14}{7} and 77\frac77⁠):

x1=467−1=397y1=−67−2=−207z1=267−1=197\begin{aligned} x_1 &= \frac{46}{7} - 1 = \frac{39}{7} \\ y_1 &= -\frac67 - 2 = -\frac{20}{7} \\ z_1 &= \frac{26}{7} - 1 = \frac{19}{7} \end{aligned}

Hence, the image of P(1,2,1)P(1, 2, 1) is Q(397,−207,197)Q\left(\frac{39}{7}, -\frac{20}{7}, \frac{19}{7}\right)⁠.

The line joining P and Q is

x−1327=y−2−347=z−1127\frac{x - 1}{\dfrac{32}{7}} = \frac{y - 2}{-\dfrac{34}{7}} = \frac{z - 1}{\dfrac{12}{7}}
Why?

Use the point P and the direction ratios Q minus P:

397−1=327−207−2=−347197−1=127\begin{aligned} \frac{39}{7} - 1 &= \frac{32}{7} \\ -\frac{20}{7} - 2 &= -\frac{34}{7} \\ \frac{19}{7} - 1 &= \frac{12}{7} \end{aligned}

(1=771 = \frac77 and 2=1472 = \frac{14}{7}⁠, so −207−147=−347-\frac{20}{7} - \frac{14}{7} = -\frac{34}{7}⁠.)

Multiplying all three by 72\frac72 gives the simpler direction ratios 16, −17-17⁠, 6. Direction ratios can be scaled freely.

327×72=322=16−347×72=−342=−17127×72=122=6\begin{aligned} \frac{32}{7} \times \frac72 &= \frac{32}{2} = 16 \\ -\frac{34}{7} \times \frac72 &= -\frac{34}{2} = -17 \\ \frac{12}{7} \times \frac72 &= \frac{12}{2} = 6 \end{aligned}

⇒x−116=y−2−17=z−16\Rightarrow \dfrac{x - 1}{16} = \dfrac{y - 2}{-17} = \dfrac{z - 1}{6}⁠.

Where this comes from

Chapter 11, Three Dimensional Geometry: Shortest distance; image of a point in a line.

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

How to spot it: click to see

(a) Type 8. The shortest distance between two lines

Spot it: "Find the shortest distance between the lines". First look at the two directions.

Here: 7i^−6j^+k^7\hat i - 6\hat j + \hat k and i^−2j^+k^\hat i - 2\hat j + \hat k are not parallel. Write P with λ\lambda and Q with μ\mu⁠, and make PQ→\overrightarrow{PQ} perpendicular to both directions: λ=μ=0\lambda = \mu = 0⁠, so PQ=4i^+6j^+8k^PQ = 4\hat i + 6\hat j + 8\hat k and the distance is 2292\sqrt{29}⁠.

(b) 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 P(1,2,1)P(1, 2, 1)⁠: L(λ+3,2λ−1,3λ+1)L(\lambda + 3, 2\lambda - 1, 3\lambda + 1) with PL perpendicular to 1,2,31, 2, 3 gives λ=27\lambda = \frac27⁠. L is the midpoint of P and its image, so the image is (397,−207,197)\left(\frac{39}{7}, -\frac{20}{7}, \frac{19}{7}\right)⁠.

(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: Then the line through P and its image: subtracting gives 327,−347,127\frac{32}{7}, -\frac{34}{7}, \frac{12}{7}⁠, or 16,−17,616, -17, 6⁠, as the direction ratios.

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