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CBSE Sample Paper 2021-22 (Term 2) · Section C · 4 marks

Q13: Foot of the perpendicular from (1,2,0)(1, 2, 0) to x−3y+2z=9x - 3y + 2z = 9⁠, and the distance

Sample 2021-22 T2Ch 11 · Three Dimensional Geometry
The question

Find the foot of the perpendicular from the point (1,2,0)(1, 2, 0) upon the plane x−3y+2z=9x - 3y + 2z = 9⁠. Hence, find the distance of the point (1,2,0)(1, 2, 0) from the given plane.

This topic is no longer in the CBSE syllabus (2025-26).

Exam answer

The normal to the plane has direction ratios 1,−3,21, -3, 2⁠. So the line through (1,2,0)(1, 2, 0) perpendicular to the plane is

Why?

In a plane ax+by+cz=dax + by + cz = d⁠, the coefficients a,b,ca, b, c are direction ratios of its normal. The perpendicular from the point runs along the normal.

A 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}
x−11=y−2−3=z2\frac{x - 1}{1} = \frac{y - 2}{-3} = \frac{z}{2}

The coordinates of the foot of the perpendicular are (μ+1,−3μ+2,2μ)(\mu + 1, -3\mu + 2, 2\mu) for some μ\mu⁠.

Why?

Set each ratio equal to μ\mu⁠: x=μ+1x = \mu + 1⁠, y=−3μ+2y = -3\mu + 2⁠, z=2μz = 2\mu⁠. Every point of the line has this form, the foot included.

These coordinates satisfy the equation of the plane:

(μ+1)−3(−3μ+2)+2(2μ)=9⇒14μ−5=9⇒μ=1\begin{aligned} &(\mu + 1) - 3(-3\mu + 2) + 2(2\mu) = 9 \\ &\Rightarrow 14\mu - 5 = 9 \Rightarrow \mu = 1 \end{aligned}
Why?

The foot lies on the line and on the plane. Expanding the left side:

μ+1+9μ−6+4μ=14μ−5\begin{aligned} &\mu + 1 + 9\mu - 6 + 4\mu \\ &= 14\mu - 5 \end{aligned}

Each bracket first, then collect like terms:

−3(−3μ+2)=9μ−62(2μ)=4μμ+9μ+4μ=14μ1−6=−5\begin{aligned} -3(-3\mu + 2) &= 9\mu - 6 \\ 2(2\mu) &= 4\mu \\[4pt] \mu + 9\mu + 4\mu &= 14\mu \\ 1 - 6 &= -5 \end{aligned}

Then solve:

14μ−5=914μ=9+5=14μ=1414=1\begin{aligned} 14\mu - 5 &= 9 \\ 14\mu &= 9 + 5 = 14 \\ \mu &= \frac{14}{14} = 1 \end{aligned}

∴\therefore The foot of the perpendicular is (2,−1,2)(2, -1, 2)⁠.

Why?

Put μ=1\mu = 1⁠: (1+1,−3+2,2)=(2,−1,2)(1 + 1, -3 + 2, 2) = (2, -1, 2)⁠.

x=μ+1=1+1=2y=−3μ+2=−3+2=−1z=2μ=2(1)=2\begin{aligned} x &= \mu + 1 = 1 + 1 = 2 \\ y &= -3\mu + 2 = -3 + 2 = -1 \\ z &= 2\mu = 2(1) = 2 \end{aligned}

Check: 2−3(−1)+2(2)=92 - 3(-1) + 2(2) = 9⁠, so the point is on the plane.

2−3(−1)+2(2)=2+3+4=9 ✓\begin{aligned} &2 - 3(-1) + 2(2) \\ &= 2 + 3 + 4 = 9\ \checkmark \end{aligned}

The required distance is the distance between (1,2,0)(1, 2, 0) and (2,−1,2)(2, -1, 2)⁠:

d2=(1−2)2+(2+1)2+(0−2)2=1+9+4=14\begin{aligned} d^2 &= (1 - 2)^2 + (2 + 1)^2 \\ &\quad + (0 - 2)^2 \\ &= 1 + 9 + 4 = 14 \end{aligned}
Why?

Distance formula:

d2=(x2−x1)2+(y2−y1)2+(z2−z1)2\begin{aligned} d^2 &= (x_2 - x_1)^2 \\ &\quad + (y_2 - y_1)^2 \\ &\quad + (z_2 - z_1)^2 \end{aligned}

Each square:

(1−2)2=(−1)2=1(2+1)2=(3)2=9(0−2)2=(−2)2=41+9+4=14,d=14\begin{aligned} (1 - 2)^2 &= (-1)^2 = 1 \\ (2 + 1)^2 &= (3)^2 = 9 \\ (0 - 2)^2 &= (-2)^2 = 4 \\[4pt] 1 + 9 + 4 &= 14,\quad d = \sqrt{14} \end{aligned}

Check with the distance of a point from a plane:

∣1−3(2)+2(0)−9∣1+9+4=1414=14\begin{aligned} &\frac{|1 - 3(2) + 2(0) - 9|}{\sqrt{1 + 9 + 4}} \\ &= \frac{14}{\sqrt{14}} = \sqrt{14} \end{aligned}

Top: 1−6+0−9=−141 - 6 + 0 - 9 = -14⁠, and ∣−14∣=14|-14| = 14⁠. Bottom: 12+(−3)2+22=14\sqrt{1^2 + (-3)^2 + 2^2} = \sqrt{14}⁠. Since 14=14×1414 = \sqrt{14} \times \sqrt{14}⁠, 1414=14\frac{14}{\sqrt{14}} = \sqrt{14}⁠.

∴\therefore Required distance =14= \sqrt{14} units.

Where this comes from

Chapter 11, Three Dimensional Geometry: Foot of the perpendicular to a plane; distance.

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