Case Study - 2 An engineer is designing a new metro rai — CBSE Class 12 Mathematics 2025 PYQ

Question

l_1 : \frac{x-2}{3} = \frac{y+1}{-2} = \frac{z-3}{4}

(i) Find whether the two metro tracks are parallel.

(ii) Solar panels are to be installed on the rooftop of the metro stations. Determine the equation of the line representing the placement of solar panels on the rooftop of Line A's stations, given that panels are to be positioned parallel to Line A's track (

l_1

) and pass through the point

(1, -2, -3)

(iii)

(a) To connect the stations, a pedestrian pathway perpendicular to the two metro lines is to be constructed which passes through point

(3, 2, 1)

. Determine the equation of the pedestrian walkway.

(OR)

(iii)

(b) Find the shortest distance between Line A and Line B.

CBSECBSE Class XII Board 2025Subjective· 4mImportance★★★★★

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Part (a)Concept understanding — Line Perpendicular To Two Lines

Line Perpendicular to Two Lines

In 3D geometry a very common task is this: two lines are given, and you must find the direction of a third line that is perpendicular to both of them. This shows up when finding a common perpendicular, the shortest distance between two lines, or the normal to a plane containing two directions.

The Core Idea

A line in space is fixed by two things: a point it passes through, and its direction vector. So "find a line perpendicular to two lines" really means "find a vector perpendicular to both of the two given direction vectors."

If the two given lines have direction vectors

\vec{b_1} = a_1\hat{i} + b_1\hat{j} + c_1\hat{k}, \qquad \vec{b_2} = a_2\hat{i} + b_2\hat{j} + c_2\hat{k},

then a vector perpendicular to both is their cross product:

\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix}

Why the Cross Product?

The defining property of $\vec{b_1}\times\vec{b_2}$ is that it is perpendicular to each factor:

(\vec{b_1}\times\vec{b_2})\cdot\vec{b_1} = 0, \qquad (\vec{b_1}\times\vec{b_2})\cdot\vec{b_2} = 0.

So it points in exactly the direction we need — along both perpendicularity conditions at once. This is why one cross product replaces solving a pair of dot-product equations by hand.

Note

The cross product only gives the direction of the perpendicular line. To pin down the actual line you still need a point it must pass through, given by the problem.

Using It

Suppose a line must be perpendicular to $\vec{b_1} = \hat{i}+2\hat{j}+3\hat{k}$ and $\vec{b_2}=\hat{i}-\hat{j}+\hat{k}$ .

\vec{b_1}\times\vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 1 & 2 & 3 \\ 1 & -1 & 1 \end{vmatrix} = 5\hat{i} + 2\hat{j} - 3\hat{k}.

So the required line has direction ratios $\langle 5, 2, -3\rangle$ . Through a point $A(x_0,y_0,z_0)$ its equation is …

Part (b)Concept understanding — Skew Lines

Skew Lines

In a plane, two straight lines have only two possibilities: they meet, or they are parallel. In three dimensions a third possibility appears — lines that neither meet nor run parallel. These are skew lines.

What Makes Lines Skew

Two lines in space are skew if they are not parallel and do not intersect. The deeper reason is that skew lines do not lie in the same plane — they are non-coplanar. Parallel lines and intersecting lines always share a plane; skew lines never do.

Note

A classic picture: one edge along the top of a room and a different edge along the floor, running in a different direction. Extend them forever and they still never touch, yet they are clearly not parallel.

The Three Cases in Space

Lines	Directions	Do they meet?	Coplanar?
Intersecting	different	yes, at one point	yes
Parallel	same (proportional)	no	yes
Skew	different	no	no

How to Test for Skew Lines

Take two lines $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$ .

Not parallel: $\vec{b_1}$ and $\vec{b_2}$ are not proportional (so $\vec{b_1}\times\vec{b_2}\neq\vec{0}$ ).
Do not intersect: no values of $\lambda,\mu$ make the points coincide.

Both conditions are captured by one scalar triple product. The lines are skew exactly when

(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2}) \neq 0.

If this value is zero, the lines are coplanar (they intersect or are parallel); if it is non-zero, they are skew.

Shortest Distance Between Skew Lines

Because skew lines miss each other, there is a well-defined shortest distance between them, measured along their common perpendicular:

$d = \frac{\left|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})\right|}{\left|\vec{b_1}\times\vec{b_2}\right|}.$ …

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Part (a): the tracks are not parallel; the solar-panel line is $\frac{x-1}{3}=\frac{y+2}{-2}=\frac{z+3}{4}$ ; the pedestrian walkway is $\frac{x-3}{2}=\frac{y-2}{17}=\frac{z-1}{7}$ . Part (b): the shortest distance between Lines A and B is $\frac{31}{\sqrt{342}}=\frac{31}{3\sqrt{38}}$ .

Line A: $l_1:\frac{x-2}{3}=\frac{y+1}{-2}=\frac{z-3}{4}$ passes through $A(2,-1,3)$ with direction $\vec d_1=(3,-2,4)$ .

Line B: $l_2:\frac{x-1}{2}=\frac{y-3}{1}=\frac{z+2}{-3}$ passes through $B(1,3,-2)$ with direction $\vec d_2=(2,1,-3)$ .

Part (a)

Are the tracks parallel? Two lines are parallel only if their direction ratios are proportional. Here $\frac{3}{2}=1.5,\quad \frac{-2}{1}=-2,\quad \frac{4}{-3}\approx-1.33,$ which are not all equal, so the tracks are not parallel.
Watch out

Check all three ratios — a single matching ratio does not make lines parallel.
Solar-panel line. Panels are parallel to $l_1$ , so the line has direction $(3,-2,4)$ and passes through $(1,-2,-3)$ : $\frac{x-1}{3}=\frac{y+2}{-2}=\frac{z+3}{4}.$ (iii)(a) Pedestrian walkway. A line perpendicular to both $l_1$ and $l_2$ has direction $\vec d_1\times\vec d_2$ : $\vec d_1\times\vec d_2=\begin{vmatrix}\mathbf i&\mathbf j&\mathbf k\\3&-2&4\\2&1&-3\end{vmatrix} =\mathbf i\big[(-2)(-3)-(4)(1)\big]-\mathbf j\big[(3)(-3)-(4)(2)\big]+\mathbf k\big[(3)(1)-(-2)(2)\big].$ $=\mathbf i(6-4)-\mathbf j(-9-8)+\mathbf k(3+4)=(2,\,17,\,7).$ Through $(3,2,1)$ : …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

CBSE 2020Set 65/1/11 markMCQ
$(-1, 5, 4)$
›Reveal solutionSolution
A line perpendicular to the plane $z=0$ must be parallel to the $z$ -axis, so its direction vector is $\hat{k}$ . The line passes through $(-1,5,4)$ , giving $\vec{r} = -\hat{i} + 5\hat{j} + 4\hat{k} + \lambda\hat{k}$ . This matches option (B).

The plane $z = 0$ is the $xy$ -plane — a flat horizontal surface. Any line perpendicular to it must point straight up or down, i.e., parallel to the $z$ -axis. That’s the core geometric insight.

A line’s vector equation is $\vec{r} = \vec{a} + \lambda \vec{d}$ , where $\vec{a}$ is a point on the line and $\vec{d}$ is the direction vector. Here, the direction vector must be along $\hat{k}$ (or any scalar multiple of it). The given point is $(-1, 5, 4)$ , so $\vec{a} = -\hat{i} + 5\hat{j} + 4\hat{k}$ .

Now check each option:

Option (A): $\vec{r} = -\hat{i} + 5\hat{j} + 4\hat{k} + \lambda(\hat{i} + \hat{j})$

Direction is $\hat{i} + \hat{j}$ , which lies in the $xy$ -plane — parallel to $z=0$ , not perpendicular. So this is wrong.

Option (B): $\vec{r} = -\hat{i} + 5\hat{j} + (4 + \lambda)\hat{k}$

Rewrite as $-\hat{i} + 5\hat{j} + 4\hat{k} + \lambda\hat{k}$ . Direction is $\hat{k}$ — exactly what we need. This is correct.

Option (C): $\vec{r} = \hat{i} - 5\hat{j} - 4\hat{k} + \lambda\hat{k}$ …
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CBSE 2019Set ANNUAL1 markMCQ
$Q. The direction cosines of two perpendicular lines are l₁, m₁, n₁ and l₂, m₂, n₂ respectively. Then the direction cosines of a line which is perpendicular to both these lines are:$
(a) l₁ + l₂, m₁ + m₂, n₁ + n₂
(b) l₁ − l₂, m₁ − m₂, n₁ − n₂
(c) m₁n₂ − n₁m₂, n₁l₂ − l₁n₂, l₁m₂ − l₂m₁
(d) m₁n₁ − m₂n₂, l₁n₁ − l₂n₂, l₁m₂ − l₂m₁
›Reveal solutionSolution

A line perpendicular to two given lines is along their cross product.

If two lines have direction ratios (l₁,m₁,n₁) and (l₂,m₂,n₂), a vector perpendicular to both is given by the cross product of the two direction vectors:

(l₁î+m₁ĵ+n₁k̂) × (l₂î+m₂ĵ+n₂k̂) = (m₁n₂−n₁m₂) î + (n₁l₂−l₁n₂) ĵ + (l₁m₂−l₂m₁) k̂

…

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CBSE 2017Set ANNUAL1 mark
$d$
›Reveal solutionSolution

definition recall: shortest distance between two skew (or parallel) lines in vector form

For two lines $\vec r=\vec a_1+\lambda\vec b_1$ and $\vec r=\vec a_2+\mu\vec b_2$ , the shortest distance between them is

…

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Part (a)

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