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

❓ Question

If the shortest distance between the lines

$$\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} \quad\text{and}\quad \frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$$

is

$$\frac{5}{\sqrt{6}},$$

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

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🖼️ Question Image

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✍️ Short Solution

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

$$\text{SD}=\frac{\mathrm{\mid }(\vec{b}-\vec{a})\cdot (\overrightarrow{d_1}\times \overrightarrow{d_2})\mathrm{\mid }}{\mathrm{\mid }\overrightarrow{d_1}\times \overrightarrow{d_2}\mathrm{\mid }}$$

where
$\vec d_1,\vec d_2$are direction vectors
and $\vec a,\vec b$ are position vectors of any points on the lines.

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🔹 Step 1 — Identify data from line equations

Line 1

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

Point

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

Line 2

$$\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}$$

Point (when parameter = 0):

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

And

$$\vec{B}-\vec{A}=(-1,-2,2)$$

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🔹 Step 2 — Find cross product $\vec d_1\times \vec d_2$

$$\vec d_1\times \vec d_2 = (3-4\alpha,\ 2,\ 2\alpha-3)$$

Magnitude squared:

$$(3-4\alpha)^2+2^2+(2\alpha-3)^2 = 20\alpha^2-36\alpha+22$$

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🔹 Step 3 — Dot product with $\vec{B}-\vec{A}$

$$(\vec{B}-\vec{A})\cdot(\vec d_1\times\vec d_2) = 8\alpha-13$$

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🔹 Step 4 — Apply shortest-distance formula

$$\frac{\mathrm{\mid }8\alpha -13\mathrm{\mid }}{\sqrt{20{\alpha }^2-36\alpha +22}}=\frac{5}{\sqrt{\mathrm{6}}}$$

Square both sides:

$$\frac{(8\alpha-13)^2}{20\alpha^2-36\alpha+22} = \frac{25}{6}$$

Cross-multiply and simplify:

$$29\alpha^2+87\alpha-116=0$$

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🔹 Step 5 — Solve the quadratic

Discriminant:

$$D = 21025 = 145^2$$
$$\alpha=\frac{-87\pm145}{58}$$

So,

$$\alpha_1 = 1,\quad \alpha_2 = -4$$

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🔹 Step 6 — Sum of all possible values

$$\alpha_1+\alpha_2 = 1+(-4)= -3$$

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✅ Final Answer

$$\boxed{-3}$$