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CBSE Sample Paper 2026-27 · Section E · 4 marks

Q36: Case study: laser beams in a mall, parallel and skew lines

Sample 2026-27Ch 11 · Three Dimensional GeometryCh 10 · Vector Algebra
The question

Two security sensors in a multi-storey mall emit laser beams for movement detection. Three of these beams travel along the following lines:

l1:r⃗=(λ+1)i^+(3λ+2)j^+(3−λ)k^ and\begin{aligned} l_1 : \vec r &= (\lambda + 1)\hat i + (3\lambda + 2)\hat j \\ &\quad + (3 - \lambda)\hat k \text{ and} \end{aligned} l2:r⃗=2(μ+1)i^+(3μ−1)j^+(μ+1)k^\begin{aligned} l_2 : \vec r &= 2(\mu + 1)\hat i + (3\mu - 1)\hat j \\ &\quad + (\mu + 1)\hat k \end{aligned} l3:r⃗=(γ+3)i^+(3γ−1)j^+(1−γ)k^\begin{aligned} l_3 : \vec r &= (\gamma + 3)\hat i + (3\gamma - 1)\hat j \\ &\quad + (1 - \gamma)\hat k \end{aligned}

Based on the given information, answer the following questions:

(i) Show that the lines l1l_1 and l2l_2 are not parallel to each other. (1 mark)

(ii) Show that the lines l1l_1 and l3l_3 are parallel to each other. (1 mark)

(iii) (a) Find the shortest distance between the lines l1l_1 and l2l_2⁠. (2 marks)

OR

(iii) (b) Find the shortest distance between the lines l1l_1 and l3l_3⁠. (2 marks)

Exam answer

(i) The given lines are

l1:r⃗=i^+2j^+3k^+λ(i^+3j^−k^)≡a⃗+λb⃗\begin{aligned} l_1 : \vec r &= \hat i + 2\hat j + 3\hat k \\ &\quad + \lambda(\hat i + 3\hat j - \hat k) \equiv \vec a + \lambda\vec b \end{aligned} l2:r⃗=2i^−j^+k^+μ(2i^+3j^+k^)≡c⃗+μd⃗\begin{aligned} l_2 : \vec r &= 2\hat i - \hat j + \hat k \\ &\quad + \mu(2\hat i + 3\hat j + \hat k) \equiv \vec c + \mu\vec d \end{aligned}
Why?

Split each component into a constant part and a part with the parameter. For l1l_1⁠:

(λ+1)i^=i^+λi^(3λ+2)j^=2j^+3λj^(3−λ)k^=3k^−λk^\begin{aligned} (\lambda + 1)\hat i &= \hat i + \lambda\hat i \\ (3\lambda + 2)\hat j &= 2\hat j + 3\lambda\hat j \\ (3 - \lambda)\hat k &= 3\hat k - \lambda\hat k \end{aligned}

This gives the form r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b⁠: a point a⃗\vec a on the line and the direction b⃗\vec b⁠. For l2l_2⁠, note 2(μ+1)=2+2μ2(\mu + 1) = 2 + 2\mu⁠.

2(μ+1)i^=2i^+2μi^(3μ−1)j^=−j^+3μj^(μ+1)k^=k^+μk^\begin{aligned} 2(\mu + 1)\hat i &= 2\hat i + 2\mu\hat i \\ (3\mu - 1)\hat j &= -\hat j + 3\mu\hat j \\ (\mu + 1)\hat k &= \hat k + \mu\hat k \end{aligned}

Here b⃗=i^+3j^−k^\vec b = \hat i + 3\hat j - \hat k and d⃗=2i^+3j^+k^\vec d = 2\hat i + 3\hat j + \hat k⁠. Since 12≠33\frac12 \ne \frac33⁠, b⃗≠λd⃗\vec b \ne \lambda\vec d for any λ∈R\lambda \in R⁠, so b⃗\vec b is not parallel to d⃗\vec d⁠.

Why?

Two lines are parallel exactly when their direction vectors are multiples of each other, so the ratios of the components must all be equal.

Here the i^\hat i ratio is 12\frac12 but the j^\hat j ratio is 33=1\frac33 = 1⁠. They differ, so no single number λ\lambda works.

So, the lines l1l_1 and l2l_2 are not parallel.

(ii)

l3:r⃗=3i^−j^+k^+γ(i^+3j^−k^)≡e⃗+γf⃗\begin{aligned} l_3 : \vec r &= 3\hat i - \hat j + \hat k \\ &\quad + \gamma(\hat i + 3\hat j - \hat k) \equiv \vec e + \gamma\vec f \end{aligned}
Why?

Same splitting: (γ+3)i^=3i^+γi^(\gamma + 3)\hat i = 3\hat i + \gamma\hat i⁠, (3γ−1)j^=−j^+3γj^(3\gamma - 1)\hat j = -\hat j + 3\gamma\hat j and (1−γ)k^=k^−γk^(1 - \gamma)\hat k = \hat k - \gamma\hat k⁠.

Since b⃗=f⃗=i^+3j^−k^\vec b = \vec f = \hat i + 3\hat j - \hat k⁠, the lines l1l_1 and l3l_3 are parallel.

Why?

Equal direction vectors mean parallel lines. They are two different lines: the point (3,−1,1)(3, -1, 1) of l3l_3 is not on l1l_1⁠, because λ+1=3\lambda + 1 = 3 needs λ=2\lambda = 2⁠, and then 3λ+2=8≠−13\lambda + 2 = 8 \ne -1⁠.

(iii) (a)

c⃗−a⃗=(2i^−j^+k^)−(i^+2j^+3k^)=i^−3j^−2k^\begin{aligned} \vec c - \vec a &= (2\hat i - \hat j + \hat k) \\ &\quad - (\hat i + 2\hat j + 3\hat k) \\ &= \hat i - 3\hat j - 2\hat k \end{aligned}
Why?

Subtract matching components:

i^:2−1=1j^:−1−2=−3k^:1−3=−2\begin{aligned} \hat i&: 2 - 1 = 1 \\ \hat j&: -1 - 2 = -3 \\ \hat k&: 1 - 3 = -2 \end{aligned}
b⃗×d⃗=∣i^j^k^13−1231∣=i^(3+3)−j^(1+2)+k^(3−6)=6i^−3j^−3k^=3(2i^−j^−k^)\begin{aligned} \vec b \times \vec d &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 3 & -1 \\ 2 & 3 & 1 \end{vmatrix} \\ &= \hat i(3 + 3) - \hat j(1 + 2) \\ &\quad + \hat k(3 - 6) \\ &= 6\hat i - 3\hat j - 3\hat k \\ &= 3(2\hat i - \hat j - \hat k) \end{aligned}
Why?

Expand the determinant along the first row:

i^:3×1−(−1)×3=6j^:−(1×1−(−1)×2)=−3k^:1×3−3×2=−3\begin{aligned} \hat i&: 3 \times 1 - (-1) \times 3 = 6 \\ \hat j&: -(1 \times 1 - (-1) \times 2) \\ &\quad = -3 \\ \hat k&: 1 \times 3 - 3 \times 2 = -3 \end{aligned}

Product by product:

i^:3−(−3)=3+3=6j^:1−(−2)=1+2=3  with its minus: −3k^:3−6=−3\begin{aligned} \hat i&: 3 - (-3) = 3 + 3 = 6 \\ \hat j&: 1 - (-2) = 1 + 2 = 3 \\ &\ \ \text{with its minus: } -3 \\ \hat k&: 3 - 6 = -3 \end{aligned}

Then take out the common factor 3: 6=3×26 = 3 \times 2 and −3=3×(−1)-3 = 3 \times (-1)⁠.

Watch the minus sign in front of the j^\hat j term.

(c⃗−a⃗)⋅(b⃗×d⃗)=3(2+3+2)=21≠0\begin{aligned} (\vec c - \vec a)\cdot(\vec b \times \vec d) &= 3(2 + 3 + 2) \\ &= 21 \ne 0 \end{aligned}
Why?

Dot product: multiply matching components and add.

(1)(2)+(−3)(−1)+(−2)(−1)=2+3+2=7\begin{aligned} &(1)(2) + (-3)(-1) + (-2)(-1) \\ &= 2 + 3 + 2 = 7 \end{aligned}

The factor 3 in front gives 21.

3×7=213 \times 7 = 21

So the lines are skew lines.

Why?

The lines are not parallel, by (i). If they met, c⃗−a⃗\vec c - \vec a would lie in the same plane as b⃗\vec b and d⃗\vec d⁠, and the product above would be 0. It is 21, so they do not meet: they are skew, and the shortest distance is not 0.

∣b⃗×d⃗∣=322+(−1)2+(−1)2=36\begin{aligned} |\vec b \times \vec d| &= 3\sqrt{2^2 + (-1)^2 + (-1)^2} \\ &= 3\sqrt6 \end{aligned} D=∣(c⃗−a⃗)⋅(b⃗×d⃗)∣b⃗×d⃗∣∣=∣2136∣=76\begin{aligned} D &= \left|\frac{(\vec c - \vec a)\cdot(\vec b \times \vec d)}{|\vec b \times \vec d|}\right| \\ &= \left|\frac{21}{3\sqrt6}\right| = \frac{7}{\sqrt6} \end{aligned}
Why?

This is the shortest distance formula for the skew lines r⃗=a⃗+λb⃗\vec r = \vec a + \lambda\vec b and r⃗=c⃗+μd⃗\vec r = \vec c + \mu\vec d⁠.

The length of b⃗×d⃗\vec b \times \vec d⁠, with the 3 kept outside:

22+(−1)2+(−1)2=4+1+1=6∣b⃗×d⃗∣=36\begin{aligned} 2^2 + (-1)^2 + (-1)^2 &= 4 + 1 + 1 = 6 \\ |\vec b \times \vec d| &= 3\sqrt6 \end{aligned}

Simplify 2136=76\frac{21}{3\sqrt6} = \frac{7}{\sqrt6}⁠, then rationalise: multiply the top and the bottom by 6\sqrt6 to get 766\frac{7\sqrt6}{6}⁠.

2136=21÷36=76=7×66×6=766\begin{aligned} \frac{21}{3\sqrt6} &= \frac{21 \div 3}{\sqrt6} = \frac{7}{\sqrt6} \\ &= \frac{7 \times \sqrt6}{\sqrt6 \times \sqrt6} = \frac{7\sqrt6}{6} \end{aligned}

∴\therefore Shortest distance =766= \frac{7\sqrt6}{6} units

(iii) (b) The lines l1l_1 and l3l_3 are parallel, with b⃗=i^+3j^−k^\vec b = \hat i + 3\hat j - \hat k and

e⃗−a⃗=2i^−3j^−2k^\vec e - \vec a = 2\hat i - 3\hat j - 2\hat k
Why?

e⃗−a⃗\vec e - \vec a joins the point (1,2,3)(1, 2, 3) of l1l_1 to the point (3,−1,1)(3, -1, 1) of l3l_3⁠: 3−1=23 - 1 = 2⁠, −1−2=−3-1 - 2 = -3 and 1−3=−21 - 3 = -2⁠.

Shortest distance D=∣b⃗×(e⃗−a⃗)∣∣b⃗∣D = \frac{|\vec b \times (\vec e - \vec a)|}{|\vec b|}

Why?

For parallel lines the distance is the same everywhere. The cross product with b⃗\vec b keeps only the part of e⃗−a⃗\vec e - \vec a that is perpendicular to the lines, and dividing by ∣b⃗∣|\vec b| removes the length of b⃗\vec b⁠.

b⃗×(e⃗−a⃗)=∣i^j^k^13−12−3−2∣=i^(−6−3)−j^(−2+2)+k^(−3−6)=−9i^−9k^\begin{aligned} \vec b \times (\vec e - \vec a) &= \begin{vmatrix} \hat i & \hat j & \hat k \\ 1 & 3 & -1 \\ 2 & -3 & -2 \end{vmatrix} \\ &= \hat i(-6 - 3) \\ &\quad - \hat j(-2 + 2) \\ &\quad + \hat k(-3 - 6) \\ &= -9\hat i - 9\hat k \end{aligned}
Why?

Expand along the first row, product by product:

i^:(3)(−2)−(−1)(−3)=−6−3=−9j^:(1)(−2)−(−1)(2)=−2+2=0k^:(1)(−3)−(3)(2)=−3−6=−9\begin{aligned} \hat i&: (3)(-2) - (-1)(-3) \\ &= -6 - 3 = -9 \\[4pt] \hat j&: (1)(-2) - (-1)(2) \\ &= -2 + 2 = 0 \\[4pt] \hat k&: (1)(-3) - (3)(2) \\ &= -3 - 6 = -9 \end{aligned}

The j^\hat j part is −0=0-0 = 0⁠, so it drops out.

D=∣−9i^−9k^∣11=9211\begin{aligned} D &= \frac{\lvert -9\hat i - 9\hat k\rvert}{\sqrt{11}} \\ &= \frac{9\sqrt2}{\sqrt{11}} \end{aligned}
Why?

∣−9i^−9k^∣=81+81=92\lvert -9\hat i - 9\hat k\rvert = \sqrt{81 + 81} = 9\sqrt2 and ∣b⃗∣=1+9+1=11|\vec b| = \sqrt{1 + 9 + 1} = \sqrt{11}⁠.

(−9)2+02+(−9)2=81+0+81=162=81×2=92∣b⃗∣=12+32+(−1)2=1+9+1=11\begin{aligned} &\sqrt{(-9)^2 + 0^2 + (-9)^2} \\ &= \sqrt{81 + 0 + 81} = \sqrt{162} \\ &= \sqrt{81 \times 2} = 9\sqrt2 \\[4pt] &|\vec b| = \sqrt{1^2 + 3^2 + (-1)^2} \\ &= \sqrt{1 + 9 + 1} = \sqrt{11} \end{aligned}

Rationalise: multiply the top and the bottom by 11\sqrt{11}⁠. Then 2⋅11=22\sqrt2 \cdot \sqrt{11} = \sqrt{22}⁠.

9211=92×1111×11=92211\begin{aligned} \frac{9\sqrt2}{\sqrt{11}} &= \frac{9\sqrt2 \times \sqrt{11}}{\sqrt{11} \times \sqrt{11}} \\ &= \frac{9\sqrt{22}}{11} \end{aligned}

∴\therefore Shortest distance =92211= \dfrac{9\sqrt{22}}{11} units

Where this comes from

Chapter 11, Three Dimensional Geometry: Lines in space: parallel lines, shortest distance.

Question type: (i) Perpendicular or parallel; (ii) Perpendicular or parallel; (iii)(a) The shortest distance between two lines; (iii)(b) The shortest distance between two lines

How to spot it: click to see

(i) 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: Read b⃗=i^+3j^−k^\vec b = \hat i + 3\hat j - \hat k from l1l_1 and d⃗=2i^+3j^+k^\vec d = 2\hat i + 3\hat j + \hat k from l2l_2⁠: 12≠33\frac12 \ne \frac33⁠, so they are not in proportion and the lines are not parallel.

(ii) 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: Collect the γ\gamma terms of l3l_3⁠: its direction is i^+3j^−k^\hat i + 3\hat j - \hat k⁠, exactly the direction of l1l_1⁠, so the lines are parallel.

(iii)(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: l1l_1 and l2l_2 are not parallel, so use the skew formula: b⃗×d⃗=3(2i^−j^−k^)\vec b\times\vec d = 3(2\hat i - \hat j - \hat k) and the top is 21, giving 76\frac{7}{\sqrt6}⁠.

(iii)(b) Type 8. The shortest distance between two lines

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

Here: l1l_1 and l3l_3 are parallel, so use the parallel formula with e⃗−a⃗=2i^−3j^−2k^\vec e - \vec a = 2\hat i - 3\hat j - 2\hat k⁠: b⃗×(e⃗−a⃗)=−9i^−9k^\vec b\times(\vec e - \vec a) = -9\hat i - 9\hat k⁠, over ∣b⃗∣=11|\vec b| = \sqrt{11}⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 10, Vector Algebra.

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