JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Area of an Isosceles Triangle Formed by a Point and a Line of Intersection

If two distinct point Q, R lie on the line of intersection of the planes -x+2 y-z=0 and 3 x-5 y+2 z=0 and P Q=P R=√(18) where the point P is (1,-2,3), then the area of the triangle PQR is equal to

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A2/3 √(38)B4/3 √(38)C8/3 √(38)D√(152/3)

Step-by-step solutionView

Correct answer

The area of triangle $P Q R$ is $\frac{4}{3} \sqrt{38}$ .

Option analysis

Why each option works or fails

A · 2/3 √(38)

The student computed the area of only one half of the isosceles triangle formed by the altitude, forgetting to multiply by $2$ or using half-base instead of base. The total area of isosceles triangle $P Q R$ with altitude $h$ and half-base $b / 2$ is $2 \times (\frac{1}{2} \frac{b}{2} h) = \frac{b}{2} h$ , which evaluates to $\frac{4}{3} \sqrt{38}$ rather than $\frac{2}{3} \sqrt{38}$ .

B · 4/3 √(38)

This is the correct option. The line of intersection has direction vector $\vec{d} = (- 1, - 1, - 1) ∥ (1, 1, 1)$ and passes through $(0, 0, 0)$ . The perpendicular distance from $P (1, - 2, 3)$ to the line is $\sqrt{\frac{38}{3}}$ . Using $P Q = \sqrt{18}$ , the half-base is $\sqrt{18 - \frac{38}{3}} = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}}$ , giving an area of $\frac{4}{\sqrt{3}} \times \sqrt{\frac{38}{3}} = \frac{4}{3} \sqrt{38}$ .

C · 8/3 √(38)

The student multiplied the full base by the altitude without dividing by $2$ , effectively computing the area of a parallelogram. Remember that the area of a triangle is $\frac{1}{2} \times base \times height$ , not $base \times height$ .

D · √(152/3)

The student incorrectly combined radicals or factored inside the square root, writing $\frac{4}{3} \sqrt{38} = \sqrt{\frac{16 \times 38}{9}} = \sqrt{\frac{608}{9}}$ erroneously as $\sqrt{\frac{152}{3}}$ . Carefully check fractional powers and square roots when moving constants under the radical: $\frac{4}{3} \sqrt{38} = \sqrt{\frac{16 \times 38}{9}} = \sqrt{\frac{608}{9}}$ , which does not simplify to $\sqrt{\frac{152}{3}}$ .

Reviewed route

Solution

StepWorking

01Given

Planes: $P_{1} : - x + 2 y - z = 0$ and $P_{2} : 3 x - 5 y + 2 z = 0$ . Point $P (1, - 2, 3)$ . Points $Q, R$ lie on the line of intersection, and $P Q = P R = \sqrt{18}$ .

02Goal

Find the area of the isosceles triangle $P Q R$ .

03Approach

First, determine the line of intersection of the two planes passing through $(0, 0, 0)$ . Second, drop a perpendicular from $P$ to the line at point $T$ (the midpoint of $Q R$ ). Third, calculate the altitude $h = P T$ and the half-base $Q T = \sqrt{P Q^{2} - P T^{2}}$ . Finally, compute the area as $Area = h \times Q T$ .

04Execute

Direction vector of the line is $\vec{d} = {\vec{n}}_{1} \times {\vec{n}}_{2} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 2 & - 1 \\ 3 & - 5 & 2 \end{array} ∣ = (- 1) \hat{i} - (1) \hat{j} + (- 1) \hat{k} = - (\hat{i} + \hat{j} + \hat{k})$ . Since $(0, 0, 0)$ satisfies both plane equations, the line equation is $\frac{x}{1} = \frac{y}{1} = \frac{z}{1}$ . A general point on the line is $T (α, α, α)$ .

05Execute

Vector $\vec{P T} = (α - 1) \hat{i} + (α + 2) \hat{j} + (α - 3) \hat{k}$ . Since $P T ⊥$ line, $\vec{P T} \cdot (1, 1, 1) = 0 ⟹ (α - 1) + (α + 2) + (α - 3) = 0 ⟹ 3 α - 2 = 0 ⟹ α = \frac{2}{3}$ .

06Execute

Calculate the squared altitude: $P T^{2} = {(\frac{2}{3} - 1)}^{2} + {(\frac{2}{3} + 2)}^{2} + {(\frac{2}{3} - 3)}^{2} = {(- \frac{1}{3})}^{2} + {(\frac{8}{3})}^{2} + {(- \frac{7}{3})}^{2} = \frac{1 + 64 + 49}{9} = \frac{114}{9} = \frac{38}{3}$ .

07Execute

In right triangle $P T Q$ , $Q T = \sqrt{P Q^{2} - P T^{2}} = \sqrt{18 - \frac{38}{3}} = \sqrt{\frac{54 - 38}{3}} = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}}$ . Area of $Δ P Q R = \frac{1}{2} \times Q R \times P T = Q T \times P T = \frac{4}{\sqrt{3}} \times \sqrt{\frac{38}{3}} = \frac{4}{3} \sqrt{38}$ .

✓Verify

Double check using the angle approach from the examiner's solution: $\cos θ = \frac{P T}{P Q} = \frac{\sqrt{114} / 3}{3 \sqrt{2}} = \frac{\sqrt{57}}{9} = \frac{\sqrt{19}}{3 \sqrt{3}}$ . $\sin (2 θ) = 2 \sin θ \cos θ = 2 \sqrt{1 - 19 / 27} \cdot \sqrt{19 / 27} = 2 \sqrt{8 / 27} \sqrt{19 / 27} = 2 \frac{2 \sqrt{2} \sqrt{19}}{27} = \frac{4 \sqrt{38}}{27}$ . Area $= \frac{1}{2} (18) \sin (2 θ) = 9 \cdot \frac{4 \sqrt{38}}{27} = \frac{4}{3} \sqrt{38}$ . Both match.

Hints that build this answer step by step

What is the direction vector $\vec{d}$ and a point on the line of intersection of the two planes $- x + 2 y - z = 0$ and $3 x - 5 y + 2 z = 0$ ?
The line passes through $(0, 0, 0)$ with direction vector $\vec{d} = (1, 1, 1)$ .
What is the perpendicular distance $h$ from the point $P (1, - 2, 3)$ to the line $\vec{r} = t (1, 1, 1)$ ?
$h = \sqrt{\frac{38}{3}}$
Since $△ P Q R$ is isosceles with $P Q = P R = \sqrt{18}$ and altitude $h = \sqrt{\frac{38}{3}}$ , what is the area of $△ P Q R$ ?
$Area = \frac{4}{3} \sqrt{38}$

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 22 September 2026

Quick checks

Students also ask

Why does the line pass through the origin?

Both plane equations have no constant term (homogeneous linear equations), so (0,0,0) satisfies both and lies on their intersection.

Why is Area = QT * PT and not (1/2) * QT * PT?

Because QT is only half of the base QR (since T is the midpoint). The base is QR = 2*QT, so Area = (1/2)*(2*QT)*PT = QT*PT.

Answer

The area of triangle $P Q R$ is $\frac{4}{3} \sqrt{38}$ .

Why each option works or fails

A: 2/3 √(38) - The student computed the area of only one half of the isosceles triangle formed by the altitude, forgetting to multiply by $2$ or using half-base instead of base. The total area of isosceles triangle $P Q R$ with altitude $h$ and half-base $b / 2$ is $2 \times (\frac{1}{2} \frac{b}{2} h) = \frac{b}{2} h$ , which evaluates to $\frac{4}{3} \sqrt{38}$ rather than $\frac{2}{3} \sqrt{38}$ .
B · correct: 4/3 √(38) - This is the correct option. The line of intersection has direction vector $\vec{d} = (- 1, - 1, - 1) ∥ (1, 1, 1)$ and passes through $(0, 0, 0)$ . The perpendicular distance from $P (1, - 2, 3)$ to the line is $\sqrt{\frac{38}{3}}$ . Using $P Q = \sqrt{18}$ , the half-base is $\sqrt{18 - \frac{38}{3}} = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}}$ , giving an area of $\frac{4}{\sqrt{3}} \times \sqrt{\frac{38}{3}} = \frac{4}{3} \sqrt{38}$ .
C: 8/3 √(38) - The student multiplied the full base by the altitude without dividing by $2$ , effectively computing the area of a parallelogram. Remember that the area of a triangle is $\frac{1}{2} \times base \times height$ , not $base \times height$ .
D: √(152/3) - The student incorrectly combined radicals or factored inside the square root, writing $\frac{4}{3} \sqrt{38} = \sqrt{\frac{16 \times 38}{9}} = \sqrt{\frac{608}{9}}$ erroneously as $\sqrt{\frac{152}{3}}$ . Carefully check fractional powers and square roots when moving constants under the radical: $\frac{4}{3} \sqrt{38} = \sqrt{\frac{16 \times 38}{9}} = \sqrt{\frac{608}{9}}$ , which does not simplify to $\sqrt{\frac{152}{3}}$ .

Step-by-step solution

Given: Planes: $P_{1} : - x + 2 y - z = 0$ and $P_{2} : 3 x - 5 y + 2 z = 0$ . Point $P (1, - 2, 3)$ . Points $Q, R$ lie on the line of intersection, and $P Q = P R = \sqrt{18}$ .
Goal: Find the area of the isosceles triangle $P Q R$ .
Approach: First, determine the line of intersection of the two planes passing through $(0, 0, 0)$ . Second, drop a perpendicular from $P$ to the line at point $T$ (the midpoint of $Q R$ ). Third, calculate the altitude $h = P T$ and the half-base $Q T = \sqrt{P Q^{2} - P T^{2}}$ . Finally, compute the area as $Area = h \times Q T$ .
Execute: Direction vector of the line is $\vec{d} = {\vec{n}}_{1} \times {\vec{n}}_{2} = ∣ \begin{array}{ccc} \hat{i} & \hat{j} & \hat{k} \\ - 1 & 2 & - 1 \\ 3 & - 5 & 2 \end{array} ∣ = (- 1) \hat{i} - (1) \hat{j} + (- 1) \hat{k} = - (\hat{i} + \hat{j} + \hat{k})$ . Since $(0, 0, 0)$ satisfies both plane equations, the line equation is $\frac{x}{1} = \frac{y}{1} = \frac{z}{1}$ . A general point on the line is $T (α, α, α)$ .
Execute: Vector $\vec{P T} = (α - 1) \hat{i} + (α + 2) \hat{j} + (α - 3) \hat{k}$ . Since $P T ⊥$ line, $\vec{P T} \cdot (1, 1, 1) = 0 ⟹ (α - 1) + (α + 2) + (α - 3) = 0 ⟹ 3 α - 2 = 0 ⟹ α = \frac{2}{3}$ .
Execute: Calculate the squared altitude: $P T^{2} = {(\frac{2}{3} - 1)}^{2} + {(\frac{2}{3} + 2)}^{2} + {(\frac{2}{3} - 3)}^{2} = {(- \frac{1}{3})}^{2} + {(\frac{8}{3})}^{2} + {(- \frac{7}{3})}^{2} = \frac{1 + 64 + 49}{9} = \frac{114}{9} = \frac{38}{3}$ .
Execute: In right triangle $P T Q$ , $Q T = \sqrt{P Q^{2} - P T^{2}} = \sqrt{18 - \frac{38}{3}} = \sqrt{\frac{54 - 38}{3}} = \sqrt{\frac{16}{3}} = \frac{4}{\sqrt{3}}$ . Area of $Δ P Q R = \frac{1}{2} \times Q R \times P T = Q T \times P T = \frac{4}{\sqrt{3}} \times \sqrt{\frac{38}{3}} = \frac{4}{3} \sqrt{38}$ .
Verify: Double check using the angle approach from the examiner's solution: $\cos θ = \frac{P T}{P Q} = \frac{\sqrt{114} / 3}{3 \sqrt{2}} = \frac{\sqrt{57}}{9} = \frac{\sqrt{19}}{3 \sqrt{3}}$ . $\sin (2 θ) = 2 \sin θ \cos θ = 2 \sqrt{1 - 19 / 27} \cdot \sqrt{19 / 27} = 2 \sqrt{8 / 27} \sqrt{19 / 27} = 2 \frac{2 \sqrt{2} \sqrt{19}}{27} = \frac{4 \sqrt{38}}{27}$ . Area $= \frac{1}{2} (18) \sin (2 θ) = 9 \cdot \frac{4 \sqrt{38}}{27} = \frac{4}{3} \sqrt{38}$ . Both match.

Area of an Isosceles Triangle Formed by a Point and a Line of Intersection

What feels right?

The area of triangle $P Q R$ is $\frac{4}{3} \sqrt{38}$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

07Execute

✓Verify

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