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CBSE Sample Paper 2026-27 · Section B · 2 marks

Q23: Parallel lines through given points: find qp\frac{q}{p}⁠; or the sides of a rectangle from its diagonals

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

(a) A line passing through the points (2,6,5)(2, 6, 5)⁠, and (3,p,7)(3, p, 7) is parallel to the line joining the points (−4,3,4)(-4, 3, 4) and (−1,0,q)(-1, 0, q)⁠. Find the value of qp\dfrac{q}{p}⁠.

OR

(b) A rectangle ABCD is such that AC→=u⃗\overrightarrow{AC} = \vec u⁠, and BD→=v⃗\overrightarrow{BD} = \vec v⁠, then find the vectors representing four sides of the rectangle (in a cyclic order starting from AB→\overrightarrow{AB}⁠).

Exam answer

(a)

Let the points be A(2,6,5)A(2, 6, 5)⁠, B(3,p,7)B(3, p, 7)⁠, C(−4,3,4)C(-4, 3, 4) and D(−1,0,q)D(-1, 0, q)⁠.

Direction ratios of AB are 3−2,p−6,7−53 - 2, p - 6, 7 - 5⁠, i.e. 1,p−6,21, p - 6, 2⁠.

Direction ratios of CD are −1+4,0−3,q−4-1 + 4, 0 - 3, q - 4⁠, i.e. 3,−3,q−43, -3, q - 4⁠.

Why?

The direction ratios of the line through (x1,y1,z1)(x_1, y_1, z_1) and (x2,y2,z2)(x_2, y_2, z_2) are the differences of the coordinates:

x2−x1, y2−y1, z2−z1x_2 - x_1,\ y_2 - y_1,\ z_2 - z_1

For CD: −1−(−4)=3-1 - (-4) = 3⁠, 0−3=−30 - 3 = -3 and q−4q - 4⁠.

As the lines are parallel,

13=p−6−3=2q−4\frac13 = \frac{p - 6}{-3} = \frac{2}{q - 4}
Why?

Parallel lines have proportional direction ratios:

a1a2=b1b2=c1c2\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}

Each ratio is (ratio of AB) over (ratio of CD). Here the first ratio, 13\frac13⁠, has no unknown, so it fixes the common value.

⇒p−6=−1, q−4=6⇒p=5, q=10\begin{aligned} &\Rightarrow p - 6 = -1,\ q - 4 = 6 \\ &\Rightarrow p = 5,\ q = 10 \end{aligned}
Why?

From p−6−3=13\frac{p - 6}{-3} = \frac13⁠: p−6=−3×13=−1p - 6 = -3 \times \frac13 = -1⁠.

p−6=−3×13=−33=−1p=−1+6=5\begin{aligned} p - 6 &= -3 \times \frac13 = -\frac{3}{3} = -1 \\ p &= -1 + 6 = 5 \end{aligned}

From 2q−4=13\frac{2}{q - 4} = \frac13⁠: cross-multiply, q−4=2×3=6q - 4 = 2 \times 3 = 6⁠.

1×(q−4)=2×3q−4=6q=6+4=10\begin{aligned} 1 \times (q - 4) &= 2 \times 3 \\ q - 4 &= 6 \\ q &= 6 + 4 = 10 \end{aligned}

∴ qp=105=2\therefore\ \frac{q}{p} = \frac{10}{5} = 2

(b)

Let AB→=a⃗\overrightarrow{AB} = \vec a⁠, BC→=b⃗\overrightarrow{BC} = \vec b⁠, CD→=c⃗\overrightarrow{CD} = \vec c and DA→=d⃗\overrightarrow{DA} = \vec d⁠. In the rectangle, c⃗=−a⃗\vec c = -\vec a and d⃗=−b⃗\vec d = -\vec b⁠.

Why?

Opposite sides of a rectangle are equal and parallel. Going round in the cyclic order A, B, C, D, the side CD runs the opposite way to AB, and DA runs the opposite way to BC. That is why the signs are negative.

u⃗=AC→=a⃗+b⃗…(i)\vec u = \overrightarrow{AC} = \vec a + \vec b \quad\ldots(\text{i}) v⃗=BD→=b⃗+c⃗=b⃗−a⃗…(ii)\begin{aligned} \vec v = \overrightarrow{BD} &= \vec b + \vec c \\ &= \vec b - \vec a \quad\ldots(\text{ii}) \end{aligned}
Why?

Triangle law of addition: going from A to B and then from B to C is the same as going straight from A to C.

AC→=AB→+BC→\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{BC}

In the same way BD→=BC→+CD→\overrightarrow{BD} = \overrightarrow{BC} + \overrightarrow{CD}⁠, and c⃗=−a⃗\vec c = -\vec a⁠.

Solving (i) and (ii),

(i)+(ii):u⃗+v⃗=2b⃗(i)−(ii):u⃗−v⃗=2a⃗\begin{aligned} \text{(i)} + \text{(ii)}: \vec u + \vec v &= 2\vec b \\ \text{(i)} - \text{(ii)}: \vec u - \vec v &= 2\vec a \end{aligned}
Why?

Adding the two equations cancels a⃗\vec a⁠: (a⃗+b⃗)+(b⃗−a⃗)=2b⃗(\vec a + \vec b) + (\vec b - \vec a) = 2\vec b⁠.

Subtracting cancels b⃗\vec b⁠: (a⃗+b⃗)−(b⃗−a⃗)=2a⃗(\vec a + \vec b) - (\vec b - \vec a) = 2\vec a⁠.

a⃗=12(u⃗−v⃗),b⃗=12(u⃗+v⃗)\vec a = \frac12(\vec u - \vec v),\quad \vec b = \frac12(\vec u + \vec v)

Also c⃗=−a⃗=12(v⃗−u⃗)\vec c = -\vec a = \frac12(\vec v - \vec u) and d⃗=−b⃗=−12(u⃗+v⃗)\vec d = -\vec b = -\frac12(\vec u + \vec v)⁠.

Why?

Divide each equation by 2:

2a⃗=u⃗−v⃗⇒a⃗=12(u⃗−v⃗)2b⃗=u⃗+v⃗⇒b⃗=12(u⃗+v⃗)\begin{aligned} 2\vec a = \vec u - \vec v &\Rightarrow \vec a = \tfrac12(\vec u - \vec v) \\ 2\vec b = \vec u + \vec v &\Rightarrow \vec b = \tfrac12(\vec u + \vec v) \end{aligned}

Then put a minus sign in front. For c⃗\vec c⁠, the minus goes inside the bracket:

c⃗=−a⃗=−12(u⃗−v⃗)=12(−u⃗+v⃗)=12(v⃗−u⃗)d⃗=−b⃗=−12(u⃗+v⃗)\begin{aligned} \vec c = -\vec a &= -\tfrac12(\vec u - \vec v) \\ &= \tfrac12(-\vec u + \vec v) \\ &= \tfrac12(\vec v - \vec u) \\[4pt] \vec d = -\vec b &= -\tfrac12(\vec u + \vec v) \end{aligned}

∴ AB→=12(u⃗−v⃗)\therefore\ \overrightarrow{AB} = \dfrac12(\vec u - \vec v)⁠, BC→=12(u⃗+v⃗)\overrightarrow{BC} = \dfrac12(\vec u + \vec v)⁠, CD→=12(v⃗−u⃗)\overrightarrow{CD} = \dfrac12(\vec v - \vec u)⁠, DA→=−12(u⃗+v⃗)\overrightarrow{DA} = -\dfrac12(\vec u + \vec v)

Where this comes from

Chapter 11, Three Dimensional Geometry: Parallel lines; sides of a rectangle as vectors.

Question type: (a) The line through two points + Perpendicular or parallel (combined)

How to spot it: click to see

(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: First, subtract the coordinates of each pair: the first line has direction ratios 1,p−6,21, p - 6, 2 and the second has 3,−3,q−43, -3, q - 4⁠.

(a) 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: The lines are parallel, so these are in proportion: 13=p−6−3=2q−4\frac13 = \frac{p - 6}{-3} = \frac{2}{q - 4}⁠, giving p=5p = 5⁠, q=10q = 10 and qp=2\frac{q}{p} = 2⁠.

Quick lesson · Worked NCERT solutions · Interactive lesson

Also uses Chapter 10, Vector Algebra.

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