Doubtify – JEE Ki Taiyaari, Simple Bhasha Mein

  ❓ Concept 🎬 Shortest Distance in 3D (α-based) in 59 Sec JEE 3D geometry mein jab shortest distance diya ho aur α parameter direction ratios mein aa jaaye — samajh lo cross product ka real test start ho gaya hai 🔥 🖼️ Concept Image ✍️ Short Explanation Shortest distance problems formula-based hote hain, lekin jab parameter (α) involve ho, toh goal hota hai: 👉 SD ko α ke function mein convert karna 👉 Given distance se equation banana 👉 α ke possible values nikaalna (ya unka sum) 🔹 Step 1 — Lines in Symmetric Form Agar line di ho: x − x 1 a 1 = y − y 1 b 1 = z − z 1 c 1 Toh: Direction vector d ⃗ 1 = ( a 1 , b 1 , c 1 ) \vec d_1=(a_1,b_1,c_1) Point on line A ( x 1 , y 1 , z 1 ) A(x_1,y_1,z_1) Similarly, second line ke liye: d ⃗ 2 = ( a 2 , b 2 , c 2 ) \vec d_2=(a_2,b_2,c_2) and point B ( x 2 , y 2 , z 2 ) B(x_2,y_2,z_2) . 📌 Agar α direction ratios mein hai, toh d ⃗ 2 \vec d_2 ​ α pe depend karega . 🔹 Step 2 — Shortest Distance Formula (...

❓ Question If the shortest distance between the lines x − 1 2 = y − 2 3 = z − 3 4 and x 1 = y α = z − 5 1 \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 5 6 , then the sum of all possible values of α \alpha α is equal to ? 🖼️ Question Image ✍️ Short Solution We’ll use the vector formula for the shortest distance between two skew lines : SD = ∣ ( b ⃗ − a ⃗ ) ⋅ ( d 1 ⃗ × d 2 ⃗ ) ∣ ∣ d 1 ⃗ × d 2 ⃗ ∣ where d ⃗ 1 , d ⃗ 2 \vec d_1,\vec d_2 are direction vectors and a ⃗ , b ⃗ \vec a,\vec b  are position vectors of any points on the lines. 🔹 Step 1 — Identify data from line equations Line 1 x − 1 2 = y − 2 3 = z − 3 4 \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4} Point A ( 1 , 2 , 3 ) , d ⃗ 1 = ( 2 , 3 , 4 ) A(1,2,3),\quad \vec d_1=(2,3,4) Line 2 x 1 = y α = z − 5 1 \frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1} Point (when parameter = 0): B ( 0 , 0 , 5 ) , d ⃗ 2 = ( 1 , α , 1 ) B(0,0,5),\quad \vec d_2=(1,\alpha...