Consider the lines L1 : x - 1 = y - 2 = z and L2 : x - 2 = y = z - 1. Let the feet of the perpendiculars from the point P(5, 1, -3) on the lines L1 and L2 be Q and R respectively. If the area of the triangle PQR is A, then 4A^2 is equal to :

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q17 – JEE-MAIN 2025

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Home Maths 3D GeometryQuestion #29

Consider the lines

L_{1} : x - 1 = y - 2 = z

and

L_{2} : x - 2 = y = z - 1

. Let the feet of the perpendiculars from the point

P (5, 1, - 3)

on the lines

L_{1}

and

L_{2}

Q

and

R

respectively. If the area of the triangle

P Q R

A

, then

4 A^{2}

is equal to :

Clue 1 of 3: Foot of Perpendicular

To find the coordinates of Q and R, determine the general points on lines $L_{1}$ and $L_{2}$ respectively.

🧠 Full Solution Path

Step 1: Find the coordinates of Q and R✦ Active

The line $L_{1}$ is given by $x - 1 = y - 2 = z = λ$ . A general point on $L_{1}$ is $Q (λ + 1, λ + 2, λ)$ . The direction vector of $L_{1}$ is $\vec{d_{1}} = (1, 1, 1)$ . The vector $\vec{P Q}$ from $P (5, 1, - 3)$ to $Q$ is $(λ + 1 - 5, λ + 2 - 1, λ - (- 3)) = (λ - 4, λ + 1, λ + 3)$ . Since $\vec{P Q}$ is perpendicular to $L_{1}$ , their dot product is zero: $(λ - 4) (1) + (λ + 1) (1) + (λ + 3) (1) = 0 ⟹ 3 λ = 0 ⟹ λ = 0$ . Thus, $Q = (1, 2, 0)$ . Similarly, for line $L_{2} : x - 2 = y = z - 1 = μ$ . A general point on $L_{2}$ is $R (μ + 2, μ, μ + 1)$ . The direction vector of $L_{2}$ is $\vec{d_{2}} = (1, 1, 1)$ . The vector $\vec{P R}$ from $P (5, 1, - 3)$ to $R$ is $(μ + 2 - 5, μ - 1, μ + 1 - (- 3)) = (μ - 3, μ - 1, μ + 4)$ . Since $\vec{P R}$ is perpendicular to $L_{2}$ , their dot product is zero: $(μ - 3) (1) + (μ - 1) (1) + (μ + 4) (1) = 0 ⟹ 3 μ = 0 ⟹ μ = 0$ . Thus, $R = (2, 0, 1)$ .

Step 2: Calculate vectors

\vec{P Q}

and

\vec{P R}

○ Expand

Given $P (5, 1, - 3)$ , $Q (1, 2, 0)$ , and $R (2, 0, 1)$ , we calculate the vectors:

\vec{P Q} = Q - P = (1 - 5, 2 - 1, 0 - (- 3)) = (- 4, 1, 3)

\vec{P R} = R - P = (2 - 5, 0 - 1, 1 - (- 3)) = (- 3, - 1, 4)

💡 Teacher's Secret Hint

Ensure correct subtraction of coordinates to form the vectors.

Step 3: Calculate the area of triangle PQR and

4 A^{2}

○ Expand

The area $A$ of triangle $P Q R$ is given by $A = \frac{1}{2} | \vec{P Q} \times \vec{P R} |$ . First, calculate the cross product:

\vec{P Q} \times \vec{P R} = | \begin{matrix} i & j & k \\ - 4 & 1 & 3 \\ - 3 & - 1 & 4 \end{matrix} | = i (1 \cdot 4 - 3 \cdot (- 1)) - j (- 4 \cdot 4 - 3 \cdot (- 3)) + k (- 4 \cdot (- 1) - 1 \cdot (- 3))

= i (4 + 3) - j (- 16 + 9) + k (4 + 3) = 7 i + 7 j + 7 k = (7, 7, 7)

Next, find the magnitude of the cross product:

| \vec{P Q} \times \vec{P R} | = \sqrt{7^{2} + 7^{2} + 7^{2}} = \sqrt{3 \cdot 49} = 7 \sqrt{3}

A = \frac{1}{2} (7 \sqrt{3})

Finally, calculate $4 A^{2}$ :

4 A^{2} = 4 {(\frac{7 \sqrt{3}}{2})}^{2} = 4 (\frac{49 \cdot 3}{4}) = 49 \cdot 3 = 147

💡 Teacher's Secret Hint

Remember that $4 A^{2} = (| \vec{P Q} \times \vec{P R} |)^{2}$ . This simplifies the calculation by avoiding the square root until the final step.

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Maths Question 17 – JEE-MAIN 2025

Clue 1 of 3: Foot of Perpendicular

🧠 Full Solution Path