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CBSE Board Examination 2022 (Term 2) · Section A · 2 marks

Q6: Values of λ\lambda for which (2,1,λ)(2, 1, \lambda) is 222\sqrt2 units from 3x+5y+4z=113x + 5y + 4z = 11

Board 2022 T2Ch 11 · Three Dimensional Geometry
The question

Find the values of λ\lambda⁠, for which the distance of point (2,1,λ)(2, 1, \lambda) from plane 3x+5y+4z=113x + 5y + 4z = 11 is 222\sqrt2 units.

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

Exam answer

The distance of (x1,y1,z1)(x_1, y_1, z_1) from the plane ax+by+cz=dax + by + cz = d is

∣ax1+by1+cz1−d∣a2+b2+c2\frac{|ax_1 + by_1 + cz_1 - d|}{\sqrt{a^2 + b^2 + c^2}}
Why?

This is the standard formula for the perpendicular distance of a point from a plane. Write the plane with the constant on the right, ax+by+cz=dax + by + cz = d⁠, before reading off the numbers.

Here a=3a = 3⁠, b=5b = 5⁠, c=4c = 4⁠, d=11d = 11⁠, and the point is (2,1,λ)(2, 1, \lambda)⁠.

∴ ∣3(2)+5(1)+4λ−11∣9+25+16=22\begin{aligned} \therefore\ &\frac{|3(2) + 5(1) + 4\lambda - 11|}{\sqrt{9 + 25 + 16}} \\ &= 2\sqrt2 \end{aligned} ⇒∣4λ∣52=22\Rightarrow \frac{|4\lambda|}{5\sqrt2} = 2\sqrt2
Why?

Numerator: 6+5+4λ−11=4λ6 + 5 + 4\lambda - 11 = 4\lambda⁠.

3(2)+5(1)+4λ−11=6+5+4λ−11=11+4λ−11=4λ\begin{aligned} &3(2) + 5(1) + 4\lambda - 11 \\ &= 6 + 5 + 4\lambda - 11 \\ &= 11 + 4\lambda - 11 \\ &= 4\lambda \end{aligned}

Denominator: 50=25×2=52\sqrt{50} = \sqrt{25 \times 2} = 5\sqrt2⁠.

(3)2+(5)2+(4)2=9+25+16=50=25×2=52\begin{aligned} &\sqrt{(3)^2 + (5)^2 + (4)^2} \\ &= \sqrt{9 + 25 + 16} \\ &= \sqrt{50} \\ &= \sqrt{25 \times 2} = 5\sqrt2 \end{aligned}
⇒∣4λ∣=20\Rightarrow |4\lambda| = 20
Why?

Multiply both sides by 525\sqrt2⁠: 22×52=10×2=202\sqrt2 \times 5\sqrt2 = 10 \times 2 = 20⁠.

∣4λ∣=22×52=(2×5)(2×2)=10×2=20\begin{aligned} |4\lambda| &= 2\sqrt2 \times 5\sqrt2 \\ &= (2 \times 5)(\sqrt2 \times \sqrt2) \\ &= 10 \times 2 = 20 \end{aligned}

Watch out: ∣u∣=20|u| = 20 means u=20u = 20 or u=−20u = -20⁠. So 4λ=±204\lambda = \pm 20⁠, and there are two values of λ\lambda⁠. Each gives a point at the same distance, one on each side of the plane.

4λ=20⇒λ=204=54λ=−20⇒λ=−204=−5\begin{aligned} 4\lambda = 20 &\Rightarrow \lambda = \frac{20}{4} = 5 \\ 4\lambda = -20 &\Rightarrow \lambda = \frac{-20}{4} = -5 \end{aligned}

∴ λ=±5\therefore\ \lambda = \pm 5

Where this comes from

Chapter 11, Three Dimensional Geometry: Distance of a point from a plane.

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