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UPSC 2015 Maths Optional Paper 1 Q3c-ii — Step-by-Step Solution

7 marks · Section A

Straight lines in 3D · Analytic Geometry · asked 5× in 13 yrs · Read the full method →

Question

Verify if the lines

L1:  x−a+dα−δ=y−aα=z−a−dα+δandL2:  x−b+cβ−γ=y−bβ=z−b−cβ+γL_1:\;\dfrac{x-a+d}{\alpha-\delta}=\dfrac{y-a}{\alpha}=\dfrac{z-a-d}{\alpha+\delta}\quad\text{and}\quad L_2:\;\dfrac{x-b+c}{\beta-\gamma}=\dfrac{y-b}{\beta}=\dfrac{z-b-c}{\beta+\gamma}

are coplanar. If yes, then find the equation of the plane in which they lie.

Technique

Recognise the AP structure of each line’s point coordinates and direction ratios; this immediately gives the determinant a column-of-zeros under C1+C3−2C2C_1+C_3-2C_2, proving coplanarity. The plane normal (1,−2,1)(1,-2,1) also follows from the AP structure (a−2b+c=0a-2b+c=0 for any AP).

Solution

Read structure. Both lines have an arithmetic-progression flavour:

Coplanarity test. Two lines through P1,P2P_1,P_2 with directions d⃗1,d⃗2\vec d_1,\vec d_2 are coplanar iff

[P1P2⃗, d⃗1, d⃗2]=det⁡(b−a−c+db−ab−a+c−dα−δαα+δβ−γββ+γ)=0.[\vec{P_1P_2},\,\vec d_1,\,\vec d_2]=\det\begin{pmatrix}b-a-c+d & b-a & b-a+c-d\\ \alpha-\delta & \alpha & \alpha+\delta\\ \beta-\gamma & \beta & \beta+\gamma\end{pmatrix}=0.

Step 1 — Simplify via column operations

For each of the three rows, the entries are in AP — i.e. column 1 + column 3 = 2×2\times column 2. Apply C1→C1+C3−2C2C_1\to C_1+C_3-2C_2 (this makes column 1 become zero for AP rows):

For row 1: (b−a−c+d)+(b−a+c−d)−2(b−a)=2(b−a)−2(b−a)=0(b-a-c+d)+(b-a+c-d)-2(b-a)=2(b-a)-2(b-a)=0. ✓ For row 2: (α−δ)+(α+δ)−2α=0(\alpha-\delta)+(\alpha+\delta)-2\alpha=0. ✓ For row 3: (β−γ)+(β+γ)−2β=0(\beta-\gamma)+(\beta+\gamma)-2\beta=0. ✓

So C1→C1+C3−2C2C_1\to C_1+C_3-2C_2 produces a column of zeros, hence

det⁡=0.\det=0.

Conclusion: the lines are coplanar.

Step 2 — Equation of the common plane

The plane contains P1P_1 and is spanned (as a translated 2-plane) by d⃗1\vec d_1 and d⃗2\vec d_2. Normal n⃗=d⃗1×d⃗2\vec n=\vec d_1\times\vec d_2.

Compute d⃗1×d⃗2\vec d_1\times\vec d_2 with d⃗1=(α−δ,α,α+δ)\vec d_1=(\alpha-\delta,\alpha,\alpha+\delta), d⃗2=(β−γ,β,β+γ)\vec d_2=(\beta-\gamma,\beta,\beta+\gamma):

n⃗=(α(β+γ)−(α+δ)β,    (α+δ)(β−γ)−(α−δ)(β+γ),    (α−δ)β−α(β−γ)).\vec n=\bigl(\alpha(\beta+\gamma)-(\alpha+\delta)\beta,\;\;(\alpha+\delta)(\beta-\gamma)-(\alpha-\delta)(\beta+\gamma),\;\;(\alpha-\delta)\beta-\alpha(\beta-\gamma)\bigr).

Component-by-component:

nxn_x: αβ+αγ−αβ−δβ=αγ−βδ\alpha\beta+\alpha\gamma-\alpha\beta-\delta\beta=\alpha\gamma-\beta\delta.

nzn_z: αβ−βδ−αβ+αγ=αγ−βδ\alpha\beta-\beta\delta-\alpha\beta+\alpha\gamma=\alpha\gamma-\beta\delta.

nyn_y: (α+δ)(β−γ)−(α−δ)(β+γ)(\alpha+\delta)(\beta-\gamma)-(\alpha-\delta)(\beta+\gamma). Expand: (αβ−αγ+δβ−δγ)−(αβ+αγ−δβ−δγ)(\alpha\beta-\alpha\gamma+\delta\beta-\delta\gamma)-(\alpha\beta+\alpha\gamma-\delta\beta-\delta\gamma) =−2αγ+2δβ=−2(αγ−βδ)=-2\alpha\gamma+2\delta\beta=-2(\alpha\gamma-\beta\delta).

So

n⃗=(αγ−βδ)(1, −2, 1).\vec n=(\alpha\gamma-\beta\delta)\bigl(1,\,-2,\,1\bigr).

A scalar multiple, so we can use n⃗=(1,−2,1)\vec n=(1,-2,1) (assuming αγ≠βδ\alpha\gamma\ne\beta\delta; else the two lines have parallel directions and the normal degenerates — see note below).

Step 3 — Plane equation through P1=(a−d,a,a+d)P_1=(a-d,a,a+d) with normal (1,−2,1)(1,-2,1)

(x−(a−d))−2(y−a)+(z−(a+d))=0(x-(a-d))-2(y-a)+(z-(a+d))=0 x−a+d−2y+2a+z−a−d=0x-a+d-2y+2a+z-a-d=0 x−2y+z=0.x-2y+z=0.

Answer

  x−2y+z=0.  \boxed{\;x-2y+z=0.\;}
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