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CBSE Sample Paper 2026-27 · Section C · 3 marks

Q30: Foot of the perpendicular from (2,1,4)(2, 1, 4) to a line; or the angle between two lines and a common perpendicular

Sample 2026-27Ch 11 · Three Dimensional Geometry
The question

(a) Find the foot of the perpendicular drawn from the point (2,1,4)(2, 1, 4) on the line x−2=2y+34=2−zx - 2 = \dfrac{2y + 3}{4} = 2 - z⁠. Also find the equation of the line passing through this point and the point (2,1,4)(2, 1, 4)⁠.

OR

(b) Find the angle between the lines:

x−12=y−12=z−21\frac{x - 1}{2} = \frac{y - 1}{2} = \frac{z - 2}{1}

and

x−52=y−12=3−z2.\frac{x - 5}{2} = y - \frac12 = \frac{3 - z}{2}.

Also find the cartesian equation of a line passing through the point (1,1,1)(1, 1, 1) and perpendicular to both lines.

Exam answer

(a)

x−2=2y+34=2−z=kx - 2 = \frac{2y + 3}{4} = 2 - z = k

In standard form, the line is

x−21=y+322=z−2−1\frac{x - 2}{1} = \frac{y + \frac32}{2} = \frac{z - 2}{-1}
Why?

Direction ratios can be read only when each of xx⁠, yy⁠, zz has coefficient 1 on top:

2y+34=y+3222−z=z−2−1\begin{aligned} \frac{2y + 3}{4} &= \frac{y + \frac32}{2} \\ 2 - z &= \frac{z - 2}{-1} \end{aligned}

In detail: take 2 out of the top, then cancel it with the 4. For zz⁠, take out −1-1⁠:

2y+34=2(y+32)4=y+3222−z=−(z−2)=z−2−1\begin{aligned} \frac{2y + 3}{4} &= \frac{2\left(y + \frac32\right)}{4} \\ &= \frac{y + \frac32}{2} \\[4pt] 2 - z &= -(z - 2) = \frac{z - 2}{-1} \end{aligned}

So the direction ratios are 1,2,−11, 2, -1⁠. Watch out: reading 44 for yy⁠, or 11 for zz⁠, is a common slip.

A point on the line is P(k+2,4k−32,2−k)P\left(k + 2, \frac{4k - 3}{2}, 2 - k\right)⁠, and the direction ratios of the line are 1,2,−11, 2, -1⁠.

Why?

Set each fraction equal to kk and solve for the coordinates: x−2=kx - 2 = k⁠, 2y+3=4k2y + 3 = 4k and 2−z=k2 - z = k⁠. Every point of the line has this form for some kk⁠.

x−2=k⇒x=k+22y+3=4k2y=4k−3y=4k−322−z=k⇒z=2−k\begin{aligned} x - 2 = k &\Rightarrow x = k + 2 \\[4pt] 2y + 3 &= 4k \\ 2y &= 4k - 3 \\ y &= \frac{4k - 3}{2} \\[4pt] 2 - z = k &\Rightarrow z = 2 - k \end{aligned}

Direction ratios of AP, where A(2,1,4)A(2, 1, 4)⁠, are

k+2−2, 4k−32−1, 2−k−4=k, 4k−52, −k−2\begin{aligned} &k + 2 - 2,\ \frac{4k - 3}{2} - 1,\ 2 - k - 4 \\ &= k,\ \frac{4k - 5}{2},\ -k - 2 \end{aligned}
Why?

Subtract the coordinates of A from those of P, one at a time:

(k+2)−2=k4k−32−1=4k−32−22=4k−3−22=4k−52(2−k)−4=−k−2\begin{aligned} (k + 2) - 2 &= k \\[4pt] \frac{4k - 3}{2} - 1 &= \frac{4k - 3}{2} - \frac{2}{2} \\ &= \frac{4k - 3 - 2}{2} = \frac{4k - 5}{2} \\[4pt] (2 - k) - 4 &= -k - 2 \end{aligned}

As AP is perpendicular to the given line,

1×k+2×4k−52+(−1)(−k−2)=0\begin{aligned} &1 \times k + 2 \times \frac{4k - 5}{2} \\ &\quad + (-1)(-k - 2) = 0 \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

Here one set is 1,2,−11, 2, -1 (the line) and the other is the direction ratios of AP.

⇒k+4k−5+k+2=0\Rightarrow k + 4k - 5 + k + 2 = 0 ⇒6k−3=0⇒k=12\Rightarrow 6k - 3 = 0 \Rightarrow k = \frac12
Why?

2×4k−52=4k−52 \times \frac{4k - 5}{2} = 4k - 5 and (−1)(−k−2)=k+2(-1)(-k - 2) = k + 2⁠. Adding: 6k−36k - 3⁠.

k+4k+k=6k−5+2=−36k−3=06k=3k=36=12\begin{aligned} k + 4k + k &= 6k \\ -5 + 2 &= -3 \\[4pt] 6k - 3 &= 0 \\ 6k &= 3 \\ k &= \frac{3}{6} = \frac12 \end{aligned}

So the foot of the perpendicular is P(52,−12,32)P\left(\frac52, -\frac12, \frac32\right)⁠.

Why?

Put k=12k = \frac12 in P:

k+2=12+42=524k=4×12=24k−32=2−32=−122−k=42−12=32\begin{aligned} k + 2 &= \frac12 + \frac42 = \frac52 \\[4pt] 4k &= 4 \times \frac12 = 2 \\ \frac{4k - 3}{2} &= \frac{2 - 3}{2} = -\frac12 \\[4pt] 2 - k &= \frac42 - \frac12 = \frac32 \end{aligned}

Direction ratios of AP are 12,−32,−52\frac12, -\frac32, -\frac52⁠, or 1,−3,−51, -3, -5⁠.

Why?

With k=12k = \frac12⁠: k=12k = \frac12⁠, 4k−52=−32\frac{4k - 5}{2} = -\frac32 and −k−2=−52-k - 2 = -\frac52⁠. Multiplying all three by 2 gives the same direction, with whole numbers.

4k−52=2−52=−32−k−2=−12−42=−522×12=12×(−32)=−32×(−52)=−5\begin{aligned} \frac{4k - 5}{2} &= \frac{2 - 5}{2} = -\frac32 \\ -k - 2 &= -\frac12 - \frac42 = -\frac52 \\[4pt] 2 \times \tfrac12 &= 1 \\ 2 \times \left(-\tfrac32\right) &= -3 \\ 2 \times \left(-\tfrac52\right) &= -5 \end{aligned}

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}

Use the point A(2,1,4)A(2, 1, 4)⁠.

∴\therefore Equation of AP: x−21=y−1−3=z−4−5\frac{x - 2}{1} = \frac{y - 1}{-3} = \frac{z - 4}{-5}

(b)

The second line is x−52=y−121=z−3−2\frac{x - 5}{2} = \frac{y - \frac12}{1} = \frac{z - 3}{-2}⁠.

Why?

Rewrite it so that xx⁠, yy⁠, zz each have coefficient 1 on top: y−12=y−121y - \frac12 = \frac{y - \frac12}{1} and 3−z2=z−3−2\frac{3 - z}{2} = \frac{z - 3}{-2}⁠.

Watch out: the direction ratio for zz is −2-2⁠, not 22⁠.

Direction ratios of the lines are 2,2,12, 2, 1 and 2,1,−22, 1, -2⁠. Let θ\theta be the angle between them.

cos⁡θ=2×2+2×1+1×(−2)9 9=49\begin{aligned} \cos\theta &= \frac{2 \times 2 + 2 \times 1 + 1 \times (-2)}{\sqrt9\,\sqrt9} \\ &= \frac49 \end{aligned}
Why?

Angle between two lines with direction vectors m⃗\vec m and n⃗\vec n⁠:

cos⁡θ=∣a1a2+b1b2+c1c2∣∣m⃗∣ ∣n⃗∣\cos\theta = \frac{|a_1a_2 + b_1b_2 + c_1c_2|}{|\vec m|\,|\vec n|}

Here ∣m⃗∣=22+22+12=3|\vec m| = \sqrt{2^2 + 2^2 + 1^2} = 3 and ∣n⃗∣=22+12+(−2)2=3|\vec n| = \sqrt{2^2 + 1^2 + (-2)^2} = 3⁠. The top is 4+2−2=44 + 2 - 2 = 4⁠.

2×2=4, 2×1=2,1×(−2)=−24+2=6,6−2=4∣m⃗∣=4+4+1=9=3∣n⃗∣=4+1+4=9=3cos⁡θ=43×3=49\begin{aligned} &2 \times 2 = 4,\ 2 \times 1 = 2, \\ &1 \times (-2) = -2 \\[4pt] &4 + 2 = 6,\quad 6 - 2 = 4 \\[4pt] &|\vec m| = \sqrt{4 + 4 + 1} = \sqrt9 = 3 \\ &|\vec n| = \sqrt{4 + 1 + 4} = \sqrt9 = 3 \\[4pt] &\cos\theta = \frac{4}{3 \times 3} = \frac49 \end{aligned}

∴ θ=cos⁡−1(49)\therefore\ \theta = \cos^{-1}\left(\frac49\right)

The required line is along

∣i^j^k^22121−2∣=i^(−4−1)−j^(−4−2)+k^(2−4)=−5i^+6j^−2k^\begin{aligned} &\begin{vmatrix} \hat i & \hat j & \hat k \\ 2 & 2 & 1 \\ 2 & 1 & -2 \end{vmatrix} \\ &= \hat i(-4 - 1) - \hat j(-4 - 2) \\ &\quad + \hat k(2 - 4) \\ &= -5\hat i + 6\hat j - 2\hat k \end{aligned}
Why?

A line perpendicular to both lines is along the cross product of their direction vectors, because a cross product is perpendicular to each of its two factors.

Expand along the first row. Watch the minus sign in front of j^\hat j⁠.

i^:(2)(−2)−(1)(1)=−4−1=−5j^:(2)(−2)−(1)(2)=−4−2=−6with its minus: −(−6)=6k^:(2)(1)−(2)(2)=2−4=−2\begin{aligned} \hat i&: (2)(-2) - (1)(1) \\ &= -4 - 1 = -5 \\[4pt] \hat j&: (2)(-2) - (1)(2) \\ &= -4 - 2 = -6 \\ &\text{with its minus: } -(-6) = 6 \\[4pt] \hat k&: (2)(1) - (2)(2) \\ &= 2 - 4 = -2 \end{aligned}

Then write the line through (1,1,1)(1, 1, 1) with direction ratios −5,6,−2-5, 6, -2⁠, in the form x−x1a=y−y1b=z−z1c\frac{x - x_1}{a} = \frac{y - y_1}{b} = \frac{z - z_1}{c}⁠.

∴\therefore The required line is x−1−5=y−16=z−1−2\dfrac{x - 1}{-5} = \dfrac{y - 1}{6} = \dfrac{z - 1}{-2}

Where this comes from

Chapter 11, Three Dimensional Geometry: Foot of a perpendicular; angle between two lines.

Question type: (a) From Cartesian to vector form + the foot of a perpendicular + The line through two points (combined); (b) The angle between two lines + A line perpendicular to two lines (combined)

How to spot it: click to see

(a) Type 5. From Cartesian to vector form: read off the point and the direction

Spot it: a line is given as three equal fractions, and you need its vector form, its point or its direction.

Here: First, x−2=2y+34=2−zx - 2 = \frac{2y + 3}{4} = 2 - z is not in standard form: rewrite it as x−21=y+322=z−2−1\frac{x - 2}{1} = \frac{y + \frac32}{2} = \frac{z - 2}{-1}⁠, so the direction ratios are 1,2,−11, 2, -1⁠.

(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: Then take the foot as P(k+2,4k−32,2−k)P\left(k + 2, \frac{4k - 3}{2}, 2 - k\right)⁠. AP perpendicular to 1,2,−11, 2, -1 gives 6k−3=06k - 3 = 0⁠, so k=12k = \frac12 and the foot is (52,−12,32)\left(\frac52, -\frac12, \frac32\right)⁠.

(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: Last, the line through A and the foot: subtracting gives 12,−32,−52\frac12, -\frac32, -\frac52⁠, or 1,−3,−51, -3, -5⁠, as the direction ratios.

(b) Type 6. The angle between two lines: only the directions matter

Spot it: "Find the angle between the lines". The points on the lines play no part.

Here: Rewrite y−12y - \frac12 as y−121\frac{y - \frac12}{1} and 3−z2\frac{3 - z}{2} as z−3−2\frac{z - 3}{-2}⁠, so the directions are 2,2,12, 2, 1 and 2,1,−22, 1, -2⁠. Then cos⁡θ=43×3=49\cos\theta = \frac{4}{3 \times 3} = \frac49⁠.

(b) Type 9. A line perpendicular to two lines: use the cross product

Spot it: "Find the line through the point … perpendicular to the two lines …". The CBSE sample papers for 2024-25, 2025-26 and 2026-27 all ask it.

Here: The new line is perpendicular to both, so its direction is (2i^+2j^+k^)×(2i^+j^−2k^)=−5i^+6j^−2k^(2\hat i + 2\hat j + \hat k)\times(2\hat i + \hat j - 2\hat k) = -5\hat i + 6\hat j - 2\hat k⁠, through (1,1,1)(1, 1, 1)⁠.

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