Let the vertices Q and R of the triangle PQR lie on the line (x + 3)/5 = (y - 1)/2 = (z + 4)/3, QR=5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is m/n then :

The correct answer is Option 4. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q13 – JEE-MAIN 2025

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Home Maths 3D GeometryQuestion #19

Let the vertices Q and R of the triangle PQR lie on the line

\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}

, QR=5 and the coordinates of the point P be (0, 2, 3). If the area of the triangle PQR is

\frac{m}{n}

then :

Clue 1 of 3: Geometric Interpretation

The area of a triangle can be calculated using its base and corresponding height.

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● Video Walkthrough

Step 1: Identify components for height calculation✦ Active

The line $L$ is given by $\frac{x + 3}{5} = \frac{y - 1}{2} = \frac{z + 4}{3}$ . A point on the line is $A (- 3, 1, - 4)$ and its direction vector is $\vec{d} = (5, 2, 3)$ . The point $P$ is $(0, 2, 3)$ . The vector $\vec{A P}$ is $P - A = (0 - (- 3), 2 - 1, 3 - (- 4)) = (3, 1, 7)$ . The magnitude of the direction vector is $| \vec{d} | = \sqrt{5^{2} + 2^{2} + 3^{2}} = \sqrt{25 + 4 + 9} = \sqrt{38}$ .

Step 2: Calculate the perpendicular distance (height) from P to line L○ Expand

The cross product $\vec{A P} \times \vec{d}$ is calculated as:

\vec{A P} \times \vec{d} = | \begin{matrix} i & j & k \\ 3 & 1 & 7 \\ 5 & 2 & 3 \end{matrix} | = i (3 - 14) - j (9 - 35) + k (6 - 5) = - 11 i + 26 j + k

The magnitude of the cross product is $| \vec{A P} \times \vec{d} | = \sqrt{(- 11)^{2} + 26^{2} + 1^{2}} = \sqrt{121 + 676 + 1} = \sqrt{798}$ . The height $h$ is given by $h = \frac{| \vec{A P} \times \vec{d} |}{| \vec{d} |} = \frac{\sqrt{798}}{\sqrt{38}} = \sqrt{\frac{798}{38}} = \sqrt{21}$ .

Step 3: Calculate the area of the triangle and match with options○ Expand

The base $Q R = 5$ . The area of triangle PQR is $\frac{1}{2} \times Q R \times h = \frac{1}{2} \times 5 \times \sqrt{21} = \frac{5 \sqrt{21}}{2}$ . Given that the area is $\frac{m}{n}$ , we have $\frac{m}{n} = \frac{5 \sqrt{21}}{2}$ . This implies $2 m = 5 \sqrt{21} n$ , which can be written as $2 m - 5 \sqrt{21} n = 0$ . Comparing this with the given options, option 4 is the correct choice.

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Maths Question 13 – JEE-MAIN 2025

Clue 1 of 3: Geometric Interpretation

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