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Shortest Distance Between Lines with Parameter α

Learn how to find parameter α using the shortest distance between two lines formula in 3D geometry. This method uses cross product and helps solve...

❓ Question

If the shortest distance between the lines

x−12=y−23=z−34andx1=yα=z−51\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \quad\text{and}\quad \frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}

is

56,

then the sum of all possible values of α\alpha is equal to ?


🖼️ Question Image

JEE 3D Geometry: Shortest Distance Between Two Lines — Find α Fast! 🔥


✍️ Short Solution

We’ll use the vector formula for the shortest distance between two skew lines:

SD=∣(b⃗−a⃗)⋅(d1⃗×d2⃗)∣∣d1⃗×d2⃗∣

where
d⃗1,d⃗2\vec d_1,\vec d_2are direction vectors
and a⃗,b⃗\vec a,\vec b are position vectors of any points on the lines.

JEE 3D Geometry: Shortest Distance Between Two Lines — Find α Fast! 🔥


🔹 Step 1 — Identify data from line equations

Line 1

x−12=y−23=z−34\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}

Point

A(1,2,3),d⃗1=(2,3,4)A(1,2,3),\quad \vec d_1=(2,3,4)

Line 2

x1=yα=z−51\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}

Point (when parameter = 0):

B(0,0,5),d⃗2=(1,α,1)B(0,0,5),\quad \vec d_2=(1,\alpha,1)

And

B⃗−A⃗=(−1,−2,2)\vec{B}-\vec{A}=(-1,-2,2)

🔹 Step 2 — Find cross product d⃗1×d⃗2\vec d_1\times \vec d_2

d⃗1×d⃗2=(3−4α, 2, 2α−3)\vec d_1\times \vec d_2 = (3-4\alpha,\ 2,\ 2\alpha-3)

Magnitude squared:

(3−4α)2+22+(2α−3)2=20α2−36α+22(3-4\alpha)^2+2^2+(2\alpha-3)^2 = 20\alpha^2-36\alpha+22

🔹 Step 3 — Dot product with B⃗−A⃗

B⃗−A⃗)⋅(d⃗1×d⃗2)=8α−13(\vec{B}-\vec{A})\cdot(\vec d_1\times\vec d_2) = 8\alpha-13

🔹 Step 4 — Apply shortest-distance formula

∣8α−13∣20α2−36α+22=56​

Square both sides:

(8α−13)220α2−36α+22=256\frac{(8\alpha-13)^2}{20\alpha^2-36\alpha+22} = \frac{25}{6}

Cross-multiply and simplify:

29α2+87α−116=029\alpha^2+87\alpha-116=0

🔹 Step 5 — Solve the quadratic

Discriminant:

D=21025=1452D = 21025 = 145^2
α=−87±14558\alpha=\frac{-87\pm145}{58}

So,

α1=1,α2=−4\alpha_1 = 1,\quad \alpha_2 = -4

🔹 Step 6 — Sum of all possible values

α1+α2=1+(−4)=−3\alpha_1+\alpha_2 = 1+(-4)= -3

✅ Final Answer

−3\boxed{-3}

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