Let the foot of perpendicular from the point (λ, 2, 3) on the line (x - 4)/1 = (y - 9)/2 = (z - 5)/1 be the point (1, μ, 2). Then the distance between the lines (x - 1)/2 = (y - 2)/3 = (z + 4)/6 and (x - λ)/2 = (y - μ)/3 = (z + 5)/6 is equal to :

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

Maths Q15 – JEE-MAIN 2026

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Home Maths 3D GeometryQuestion #15

Let the foot of perpendicular from the point

(λ, 2, 3)

on the line

\frac{x - 4}{1} = \frac{y - 9}{2} = \frac{z - 5}{1}

be the point

(1, μ, 2)

. Then the distance between the lines

\frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}

and

\frac{x - λ}{2} = \frac{y - μ}{3} = \frac{z + 5}{6}

is equal to :

Clue 1 of 3: Foot of Perpendicular

The foot of the perpendicular from a point to a line is the point on the line such that the vector connecting the given point to the foot of the perpendicular is orthogonal to the direction vector of the line.

🧠 Full Solution Path

Step 1: Determine the values of

λ

and

μ

✦ Active

The foot of the perpendicular from $A (λ, 2, 3)$ to the line $L_{1} : \frac{x - 4}{1} = \frac{y - 9}{2} = \frac{z - 5}{1}$ is $F (1, μ, 2)$ . Since $F$ lies on $L_{1}$ , substitute its coordinates into the line equation:

\frac{1 - 4}{1} = \frac{μ - 9}{2} = \frac{2 - 5}{1} ⟹ - 3 = \frac{μ - 9}{2} = - 3

From $- 3 = \frac{μ - 9}{2}$ , we get $μ - 9 = - 6 ⟹ μ = 3$ . The vector $\vec{A F} = (1 - λ, 3 - 2, 2 - 3) = (1 - λ, 1, - 1)$ . The direction vector of $L_{1}$ is $\vec{d_{1}} = (1, 2, 1)$ . Since $\vec{A F} ⊥ \vec{d_{1}}$ , their dot product is zero:

\vec{A F} \cdot \vec{d_{1}} = (1 - λ) (1) + (1) (2) + (- 1) (1) = 0 ⟹ 1 - λ + 2 - 1 = 0 ⟹ 2 - λ = 0 ⟹ λ = 2

Thus, $λ = 2$ and $μ = 3$ .

Step 2: Identify the two lines and their properties○ Expand

The first line is $L_{2} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z + 4}{6}$ . It passes through $A_{1} (1, 2, - 4)$ and has direction vector $\vec{d_{2}} = (2, 3, 6)$ . The second line, using $λ = 2$ and $μ = 3$ , is $L_{3} : \frac{x - 2}{2} = \frac{y - 3}{3} = \frac{z + 5}{6}$ . It passes through $A_{2} (2, 3, - 5)$ and has direction vector $\vec{d_{3}} = (2, 3, 6)$ . Since $\vec{d_{2}} = \vec{d_{3}} = (2, 3, 6)$ , the two lines are parallel.

💡 Teacher's Secret Hint

Recognizing parallel lines simplifies the distance calculation significantly.

Step 3: Calculate the shortest distance between the parallel lines○ Expand

The vector connecting a point on $L_{2}$ to a point on $L_{3}$ is $\vec{A_{1} A_{2}} = A_{2} - A_{1} = (2 - 1, 3 - 2, - 5 - (- 4)) = (1, 1, - 1)$ . The common direction vector is $\vec{d} = (2, 3, 6)$ . The shortest distance $D$ between two parallel lines is given by $D = \frac{| \vec{A_{1} A_{2}} \times \vec{d} |}{| \vec{d} |}$ . First, calculate the cross product:

\vec{A_{1} A_{2}} \times \vec{d} = | \begin{matrix} i & j & k \\ 1 & 1 & - 1 \\ 2 & 3 & 6 \end{matrix} | = i (6 - (- 3)) - j (6 - (- 2)) + k (3 - 2) = 9 i - 8 j + 1 k

Next, find the magnitudes:

| \vec{A_{1} A_{2}} \times \vec{d} | = \sqrt{9^{2} + (- 8)^{2} + 1^{2}} = \sqrt{81 + 64 + 1} = \sqrt{146}

| \vec{d} | = \sqrt{2^{2} + 3^{2} + 6^{2}} = \sqrt{4 + 9 + 36} = \sqrt{49} = 7

Finally, the distance is:

D = \frac{\sqrt{146}}{7}

💡 Teacher's Secret Hint

Ensure correct calculation of cross product and magnitudes to avoid errors.

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Maths Question 15 – JEE-MAIN 2026

Clue 1 of 3: Foot of Perpendicular

🧠 Full Solution Path