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Three Dimensional Geometry: Mathematics | JEE Main

Let the plane

P : 4x - y + z = 10

be rotated by an angle

\frac{\pi}{2}

about its line of intersection with the plane

x + y - z = 4

. If

\alpha

is the distance of the point

(2, 3, -4)

from the new position of the plane P, then

35\alpha

Your answer stays private

What feels right?

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90

85

105

126

Step-by-step solutionView

Correct answer

The rotated plane has equation $x - 4y + 4z + 2 = 0$ , and the perpendicular distance from $(2, 3, -4)$ yields $\alpha = \frac{18}{5}$ , giving $35\alpha = 126$ .

Option analysis

Why each option works or fails

A ·

90

Believing that evaluating the distance formula with reversed signs in the plane's constant term or normal vector gives $|-18|/\sqrt{33}$ or miscomputing the numerator as 18 with a denominator of 7 instead of the true value. Ensure the normal vector is normalized using $\sqrt{A^2 + B^2 + C^2} = \sqrt{1^2 + (-4)^2 + 4^2} = \sqrt{33}$ is avoided; check that the equation is $x - 4y + 4z + 2 = 0$ with magnitude $\sqrt{1 + 16 + 16} = \sqrt{33}$ properly simplified to $\sqrt{33}$ vs $\sqrt{18}$ .

B ·

85

Making an arithmetic error when evaluating the perpendicular distance numerator $|1(2) - 4(3) + 4(-4) + 2|$ . Carefully compute $2 - 12 - 16 + 2 = -24$ , giving an absolute value of $24$ , but check the exact plane parameter $\lambda$ .

C ·

105

Taking $\lambda$ with an arithmetic error in the dot product equation $\vec{n}_1 \cdot \vec{n}_{new} = 0$ , leading to an incorrect normal vector. Check the dot product: $4(4+\lambda) - 1(-1+\lambda) + 1(1-\lambda) = 16 + 4\lambda + 1 - \lambda + 1 - \lambda = 18 + 2\lambda = 0$ , giving $\lambda = -9$ .

D ·

126

None. The student correctly sets up the family of planes, uses orthogonality to determine $\lambda = -9$ , finds the rotated plane equation $-5x - 10y + 10z + 26 = 0$ or equivalent, and computes the distance correctly. This is the correct option.

Reviewed route

Solution

StepWorking

01given

Given planes $P_1: 4x - y + z - 10 = 0$ and $P_2: x + y - z - 4 = 0$ . Point $A = (2, 3, -4)$ . $P_1$ is rotated by $\frac{\pi}{2}$ about the line of intersection of $P_1$ and $P_2$ .

02goal

Find $35\alpha$ , where $\alpha$ is the perpendicular distance from $(2, 3, -4)$ to the rotated plane $P_{new}$ .

03approach

Any plane passing through the line of intersection of $P_1$ and $P_2$ has the equation $P_1 + \lambda P_2 = 0$ . Since $P_{new}$ is rotated by $\frac{\pi}{2}$ from $P_1$ , the normal to $P_{new}$ is perpendicular to the normal to $P_1$ . Use the dot product of normal vectors $\vec{n}_{new} \cdot \vec{n}_1 = 0$ to find $\lambda$ , then calculate distance $\alpha$ .

04execute

The plane is $(4x - y + z - 10) + \lambda(x + y - z - 4) = 0 \implies (4 + \lambda)x + (-1 + \lambda)y + (1 - \lambda)z - (10 + 4\lambda) = 0$ . Its normal vector is $\vec{n}_{new} = (4 + \lambda, -1 + \lambda, 1 - \lambda)$ . The normal to $P_1$ is $\vec{n}_1 = (4, -1, 1)$ . Applying orthogonality $\vec{n}_{new} \cdot \vec{n}_1 = 0$ : $4(4 + \lambda) - 1(-1 + \lambda) + 1(1 - \lambda) = 0$ $16 + 4\lambda + 1 - \lambda + 1 - \lambda = 0 \implies 18 + 2\lambda = 0 \implies \lambda = -9.$

05execute

Substitute $\lambda = -9$ into $P_{new}$ : $(4 - 9)x + (-1 - 9)y + (1 - (-9))z - (10 + 4(-9)) = 0$ $-5x - 10y + 10z - (10 - 36) = 0 \implies -5x - 10y + 10z + 26 = 0$ which can be written as $5x + 10y - 10z - 26 = 0$ . Now find the distance $\alpha$ from $(2, 3, -4)$ : $\alpha = \frac{|5(2) + 10(3) - 10(-4) - 26|}{\sqrt{5^2 + 10^2 + (-10)^2}} = \frac{|10 + 30 + 40 - 26|}{\sqrt{25 + 100 + 100}} = \frac{54}{\sqrt{225}} = \frac{54}{15} = \frac{18}{5}.$

06execute

Calculate $35\alpha$ : $35\alpha = 35 \times \frac{18}{5} = 7 \times 18 = 126.$

✓verify

Check orthogonality of normal vectors: $\vec{n}_{new} = (-5, -10, 10)$ , $\vec{n}_1 = (4, -1, 1)$ . $\vec{n}_{new} \cdot \vec{n}_1 = (-5)(4) + (-10)(-1) + (10)(1) = -20 + 10 + 10 = 0$ . Thus the rotation angle is indeed $\frac{\pi}{2}$ .

Hints that build this answer step by step

Any plane passing through the intersection of $P_1: 4x - y + z - 10 = 0$ and $P_2: x + y - z - 4 = 0$ can be represented using a parameter $\lambda$ . What is its normal vector $\vec{n}_{new}$ ?
$(4 + \lambda)\hat{i} + (-1 + \lambda)\hat{j} + (1 - \lambda)\hat{k}$
Since the new plane is rotated by $\frac{\pi}{2}$ relative to the original plane $P_1$ , their normal vectors must be perpendicular. What is the value of $\lambda$ ?
$\lambda = -9$
Substitute $\lambda = -9$ into $(4+\lambda)x + (-1+\lambda)y + (1-\lambda)z - (10 + 4\lambda) = 0$ and simplify. What is the equation of the rotated plane?
$-5x - 10y + 10z + 26 = 0$
What is the perpendicular distance $\alpha$ from $(2, 3, -4)$ to $-5x - 10y + 10z + 26 = 0$ , and what is $35\alpha$ ?
$\alpha = \frac{54}{15} = \frac{18}{5}$ , so $35\alpha = 126$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 8 September 2026

Quick checks

Students also ask

Why is the normal of the new plane perpendicular to the normal of P1?

Rotating a plane by π/2 about a line in it rotates its normal vector by π/2. Therefore, the new normal vector is perpendicular to the original normal vector.

Answer

The rotated plane has equation $x - 4y + 4z + 2 = 0$ , and the perpendicular distance from $(2, 3, -4)$ yields $\alpha = \frac{18}{5}$ , giving $35\alpha = 126$ .

Why each option works or fails

A: $90$ - Believing that evaluating the distance formula with reversed signs in the plane's constant term or normal vector gives $|-18|/\sqrt{33}$ or miscomputing the numerator as 18 with a denominator of 7 instead of the true value. Ensure the normal vector is normalized using $\sqrt{A^2 + B^2 + C^2} = \sqrt{1^2 + (-4)^2 + 4^2} = \sqrt{33}$ is avoided; check that the equation is $x - 4y + 4z + 2 = 0$ with magnitude $\sqrt{1 + 16 + 16} = \sqrt{33}$ properly simplified to $\sqrt{33}$ vs $\sqrt{18}$ .
B: $85$ - Making an arithmetic error when evaluating the perpendicular distance numerator $|1(2) - 4(3) + 4(-4) + 2|$ . Carefully compute $2 - 12 - 16 + 2 = -24$ , giving an absolute value of $24$ , but check the exact plane parameter $\lambda$ .
C: $105$ - Taking $\lambda$ with an arithmetic error in the dot product equation $\vec{n}_1 \cdot \vec{n}_{new} = 0$ , leading to an incorrect normal vector. Check the dot product: $4(4+\lambda) - 1(-1+\lambda) + 1(1-\lambda) = 16 + 4\lambda + 1 - \lambda + 1 - \lambda = 18 + 2\lambda = 0$ , giving $\lambda = -9$ .
D · correct: $126$ - None. The student correctly sets up the family of planes, uses orthogonality to determine $\lambda = -9$ , finds the rotated plane equation $-5x - 10y + 10z + 26 = 0$ or equivalent, and computes the distance correctly. This is the correct option.

Step-by-step solution

given: Given planes $P_1: 4x - y + z - 10 = 0$ and $P_2: x + y - z - 4 = 0$ . Point $A = (2, 3, -4)$ . $P_1$ is rotated by $\frac{\pi}{2}$ about the line of intersection of $P_1$ and $P_2$ .
goal: Find $|1(2) - 4(3) + 4(-4) + 2|$ , where $\alpha$ is the perpendicular distance from $(2, 3, -4)$ to the rotated plane $P_{new}$ .
approach: Any plane passing through the line of intersection of $P_1$ and $24$ has the equation $P_1 + \lambda P_2 = 0$ . Since $P_{new}$ is rotated by $\frac{\pi}{2}$ from $P_1$ , the normal to $P_{new}$ is perpendicular to the normal to $P_1$ . Use the dot product of normal vectors $\vec{n}_{new} \cdot \vec{n}_1 = 0$ to find $\lambda$ , then calculate distance $\alpha$ .
execute: The plane is $(4x - y + z - 10) + \lambda(x + y - z - 4) = 0 \implies (4 + \lambda)x + (-1 + \lambda)y + (1 - \lambda)z - (10 + 4\lambda) = 0$ . Its normal vector is $\vec{n}_{new} = (4 + \lambda, -1 + \lambda, 1 - \lambda)$ . The normal to $P_1$ is $\vec{n}_1 = (4, -1, 1)$ . Applying orthogonality $\alpha$ : $4(4 + \lambda) - 1(-1 + \lambda) + 1(1 - \lambda) = 0$ $16 + 4\lambda + 1 - \lambda + 1 - \lambda = 0 \implies 18 + 2\lambda = 0 \implies \lambda = -9.$
execute: Substitute $\lambda = -9$ into $P_{new}$ : $(4 - 9)x + (-1 - 9)y + (1 - (-9))z - (10 + 4(-9)) = 0$ $-5x - 10y + 10z - (10 - 36) = 0 \implies -5x - 10y + 10z + 26 = 0$ which can be written as $5x + 10y - 10z - 26 = 0$ . Now find the distance $\alpha$ from $(2, 3, -4)$ : $\alpha = \frac{|5(2) + 10(3) - 10(-4) - 26|}{\sqrt{5^2 + 10^2 + (-10)^2}} = \frac{|10 + 30 + 40 - 26|}{\sqrt{25 + 100 + 100}} = \frac{54}{\sqrt{225}} = \frac{54}{15} = \frac{18}{5}.$
execute: Calculate $35\alpha$ : $35\alpha = 35 \times \frac{18}{5} = 7 \times 18 = 126.$
verify: Check orthogonality of normal vectors: $\vec{n}_{new} = (-5, -10, 10)$ , $\vec{n}_1 = (4, -1, 1)$ . $\vec{n}_{new} \cdot \vec{n}_1 = (-5)(4) + (-10)(-1) + (10)(1) = -20 + 10 + 10 = 0$ . Thus the rotation angle is indeed $\frac{\pi}{2}$ .

Three Dimensional Geometry: Mathematics | JEE Main

What feels right?

The rotated plane has equation $x - 4y + 4z + 2 = 0$ , and the perpendicular distance from $(2, 3, -4)$ yields $\alpha = \frac{18}{5}$ , giving $35\alpha = 126$ .

Why each option works or fails

Solution

01given

02goal

03approach

04execute

05execute

06execute

✓verify

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