A line with direction ratios 1, -1, 2 intersects the lines x/2 = y/3 = (z+1)/3 and (x+1)/-1 = (y-2)/1 = z/4 at the points P and Q, respectively. If the length of the line segment PQ is alpha, then 225alpha^2 is equal to:

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q14 – JEE-MAIN 2026

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Home Maths 3D GeometryQuestion #7

A line with direction ratios

1, - 1, 2

intersects the lines

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

and

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

at the points P and Q, respectively. If the length of the line segment PQ is

α

, then

225 α^{2}

is equal to:

Clue 1 of 3: Parametric Form of Lines

Represent points P and Q on the given lines using parametric forms, say $P (x_{P}, y_{P}, z_{P})$ and $Q (x_{Q}, y_{Q}, z_{Q})$ .

🧠 Full Solution Path

Step 1: Represent Points P and Q Parametrically✦ Active

Let the first line be $L_{1}$ and the second line be $L_{2}$ . We express points P on $L_{1}$ and Q on $L_{2}$ in parametric form:

L_{1} : \frac{x}{2} = \frac{y}{3} = \frac{z + 1}{3} = λ ⟹ P (2 λ, 3 λ, 3 λ - 1)

L_{2} : \frac{x + 1}{- 1} = \frac{y - 2}{1} = \frac{z}{4} = μ ⟹ Q (- μ - 1, μ + 2, 4 μ)

Step 2: Determine Parameters Using Direction Ratios○ Expand

The direction ratios of the line segment PQ are $(x_{Q} - x_{P}, y_{Q} - y_{P}, z_{Q} - z_{P})$ . We are given that these direction ratios are proportional to $(1, - 1, 2)$ . Let the proportionality constant be $k$ .

x_{Q} - x_{P} = - μ - 1 - 2 λ = k (1)

y_{Q} - y_{P} = μ + 2 - 3 λ = - k (2)

z_{Q} - z_{P} = 4 μ - (3 λ - 1) = 2 k (3)

Adding equations (1) and (2): $(- 2 λ - μ - 1) + (- 3 λ + μ + 2) = k + (- k) ⟹ - 5 λ + 1 = 0 ⟹ λ = \frac{1}{5}$ . Substitute $λ = \frac{1}{5}$ into (1) and (3) and solve for $μ$ and $k$ . From (1), $k = - 2 (\frac{1}{5}) - μ - 1 = - \frac{7}{5} - μ$ . From (3), $- 3 (\frac{1}{5}) + 4 μ + 1 = 2 k$ . Substituting $k$ into (3): $- \frac{3}{5} + 4 μ + 1 = 2 (- \frac{7}{5} - μ) ⟹ \frac{2}{5} + 4 μ = - \frac{14}{5} - 2 μ ⟹ 6 μ = - \frac{16}{5} ⟹ μ = - \frac{8}{15}$ . Then, $k = - \frac{7}{5} - (- \frac{8}{15}) = - \frac{21}{15} + \frac{8}{15} = - \frac{13}{15}$ .

💡 Teacher's Secret Hint

Ensure careful algebraic manipulation when solving the system of equations to avoid sign errors.

Step 3: Calculate Length PQ and Final Expression○ Expand

Now substitute the values of $λ$ and $μ$ to find the coordinates of P and Q:

P (2 (\frac{1}{5}), 3 (\frac{1}{5}), 3 (\frac{1}{5}) - 1) = P (\frac{2}{5}, \frac{3}{5}, - \frac{2}{5})

Q (- (- \frac{8}{15}) - 1, - \frac{8}{15} + 2, 4 (- \frac{8}{15})) = Q (\frac{8}{15} - \frac{15}{15}, - \frac{8}{15} + \frac{30}{15}, - \frac{32}{15}) = Q (- \frac{7}{15}, \frac{22}{15}, - \frac{32}{15})

The differences in coordinates are $(x_{Q} - x_{P}, y_{Q} - y_{P}, z_{Q} - z_{P}) = (- \frac{7}{15} - \frac{2}{5}, \frac{22}{15} - \frac{3}{5}, - \frac{32}{15} - (- \frac{2}{5})) = (- \frac{13}{15}, \frac{13}{15}, - \frac{26}{15})$ . The length squared $α^{2} = P Q^{2}$ is:

α^{2} = {(- \frac{13}{15})}^{2} + {(\frac{13}{15})}^{2} + {(- \frac{26}{15})}^{2}

α^{2} = \frac{169}{225} + \frac{169}{225} + \frac{676}{225} = \frac{169 + 169 + 676}{225} = \frac{1014}{225}

The question asks for $225 α^{2}$ :

225 α^{2} = 225 \times \frac{1014}{225} = 1014

💡 Teacher's Secret Hint

Remember that the length squared is the sum of the squares of the differences in coordinates. Double-check calculations for fractions.

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Maths Question 14 – JEE-MAIN 2026

Clue 1 of 3: Parametric Form of Lines

🧠 Full Solution Path