JEE Challenger
More from Three Dimensional Geometry
MathMedAdv 2013 • Paper 2

Coplanarity Condition for Two Straight Lines in 3D Space

Given two lines L1:x=5,y3−α=z−2L_1 : x = 5, \frac{y}{3 - \alpha} = \frac{z}{-2} and L2:x=α,y−1=z2−αL_2 : x = \alpha, \frac{y}{-1} = \frac{z}{2 - \alpha} which are coplanar. Then the parameter α\alpha can take the value(s)

Options

A

11

Correct
B

22

C

33

D

44

Correct

Topics & Concepts

Three Dimensional GeometryLine in 3D

Step-by-Step Solution

To determine the values of the parameter α\alpha for which the two lines L1L_1 and L2L_2 are coplanar, we first express both lines in standard symmetric form.

For line L1L_1: x=5,y3−α=z−2x = 5, \quad \frac{y}{3 - \alpha} = \frac{z}{-2} This can be written in Cartesian form as: x−50=y−03−α=z−0−2\frac{x - 5}{0} = \frac{y - 0}{3 - \alpha} = \frac{z - 0}{-2} Thus, line L1L_1 passes through the point A(5,0,0)A(5, 0, 0) and has direction vector: d⃗1=0i^+(3−α)j^−2k^\vec{d}_1 = 0\hat{i} + (3 - \alpha)\hat{j} - 2\hat{k}

For line L2L_2: x=α,y−1=z2−αx = \alpha, \quad \frac{y}{-1} = \frac{z}{2 - \alpha} This can be written in Cartesian form as: x−α0=y−0−1=z−02−α\frac{x - \alpha}{0} = \frac{y - 0}{-1} = \frac{z - 0}{2 - \alpha} Thus, line L2L_2 passes through the point B(α,0,0)B(\alpha, 0, 0) and has direction vector: d⃗2=0i^−j^+(2−α)k^\vec{d}_2 = 0\hat{i} - \hat{j} + (2 - \alpha)\hat{k}

The vector connecting points AA and BB is: AB⃗=(α−5)i^+0j^+0k^\vec{AB} = (\alpha - 5)\hat{i} + 0\hat{j} + 0\hat{k}

Coplanarity Condition

Two straight lines are coplanar if and only if the scalar triple product of AB⃗\vec{AB}, d⃗1\vec{d}_1, and d⃗2\vec{d}_2 vanishes, which is represented by the determinant:

∣x2−x1y2−y1z2−z1a1b1c1a2b2c2∣=0\begin{vmatrix} x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ a_1 & b_1 & c_1 \\ a_2 & b_2 & c_2 \end{vmatrix} = 0

Substituting the values into the determinant:

∣α−50003−α−20−12−α∣=0\begin{vmatrix} \alpha - 5 & 0 & 0 \\ 0 & 3 - \alpha & -2 \\ 0 & -1 & 2 - \alpha \end{vmatrix} = 0

Expanding along the first row: (α−5)[(3−α)(2−α)−(−2)(−1)]=0(\alpha - 5) \left[ (3 - \alpha)(2 - \alpha) - (-2)(-1) \right] = 0 (α−5)[(6−5α+α2)−2]=0(\alpha - 5) \left[ (6 - 5\alpha + \alpha^2) - 2 \right] = 0 (α−5)(α2−5α+4)=0(\alpha - 5)(\alpha^2 - 5\alpha + 4) = 0 (α−5)(α−1)(α−4)=0(\alpha - 5)(\alpha - 1)(\alpha - 4) = 0

This yields the solutions: α=1,α=4,α=5\alpha = 1, \quad \alpha = 4, \quad \alpha = 5

  • For α=1\alpha = 1, the direction vectors d⃗1=(0,2,−2)\vec{d}_1 = (0, 2, -2) and d⃗2=(0,−1,1)\vec{d}_2 = (0, -1, 1) are parallel, making the lines parallel and therefore coplanar.
  • For α=4\alpha = 4, the direction vectors d⃗1=(0,−1,−2)\vec{d}_1 = (0, -1, -2) and d⃗2=(0,−1,−2)\vec{d}_2 = (0, -1, -2) are parallel, making the lines parallel and therefore coplanar.

Matching with the given options, the correct values are α=1\alpha = 1 and α=4\alpha = 4.

Correct Options:

  • A (1)
  • D (4)
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