If the shortest distance between the lines x-1/2=y-2/3=z-3/4 and x/1=y/alpha=z-5/1 is 5/sqrt(6), then the sum of all possible values of alpha is

The correct answer is Option 2. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q14 – JEE-MAIN 2025

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Home Maths 3D GeometryQuestion #27

If the shortest distance between the lines

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

and

\frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}

\frac{5}{\sqrt{6}}

, then the sum of all possible values of

α

Clue 1 of 3: Identify Line Parameters

Extract the position vectors and direction vectors for both lines from their given Cartesian equations.

🥷

Ninja Strategy•Vieta's Formulas for Sum of Roots

After setting up the shortest distance equation and simplifying it to a quadratic equation of the form $a α^{2} + b α + c = 0$ , directly use Vieta's formula for the sum of roots ( $- b / a$ ) without solving for individual $α$ values.

🧠 Full Solution Path

Step 1: Extract Line Parameters✦ Active

The first line $L_{1} : \frac{x - 1}{2} = \frac{y - 2}{3} = \frac{z - 3}{4}$ passes through point ${\vec{a}}_{1} = \hat{i} + 2 \hat{j} + 3 \hat{k}$ and has direction vector ${\vec{b}}_{1} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ . The second line $L_{2} : \frac{x}{1} = \frac{y}{α} = \frac{z - 5}{1}$ passes through point ${\vec{a}}_{2} = 0 \hat{i} + 0 \hat{j} + 5 \hat{k}$ and has direction vector ${\vec{b}}_{2} = \hat{i} + α \hat{j} + \hat{k}$ .

{\vec{a}}_{2} - {\vec{a}}_{1} = (0 - 1) \hat{i} + (0 - 2) \hat{j} + (5 - 3) \hat{k} = - \hat{i} - 2 \hat{j} + 2 \hat{k}

Step 2: Apply Shortest Distance Formula○ Expand

Calculate the cross product of the direction vectors and the scalar triple product. The shortest distance $D$ is given by $D = \frac{| ({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) |}{| {\vec{b}}_{1} \times {\vec{b}}_{2} |}$ . Given $D = \frac{5}{\sqrt{6}}$ .

{\vec{b}}_{1} \times {\vec{b}}_{2} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 3 & 4 \\ 1 & α & 1 \end{matrix} | = (3 - 4 α) \hat{i} + 2 \hat{j} + (2 α - 3) \hat{k}

({\vec{a}}_{2} - {\vec{a}}_{1}) \cdot ({\vec{b}}_{1} \times {\vec{b}}_{2}) = (- 1) (3 - 4 α) + (- 2) (2) + (2) (2 α - 3) = - 3 + 4 α - 4 + 4 α - 6 = 8 α - 13

| {\vec{b}}_{1} \times {\vec{b}}_{2} | = \sqrt{(3 - 4 α)^{2} + 2^{2} + (2 α - 3)^{2}} = \sqrt{9 - 24 α + 16 α^{2} + 4 + 4 α^{2} - 12 α + 9} = \sqrt{20 α^{2} - 36 α + 22}

Substitute these into the distance formula:

\frac{| 8 α - 13 |}{\sqrt{20 α^{2} - 36 α + 22}} = \frac{5}{\sqrt{6}}

💡 Teacher's Secret Hint

Be careful with algebraic signs during the cross product and dot product calculations.

Step 3: Solve for

α

and Sum the Values○ Expand

Square both sides of the equation to eliminate the absolute value and square root:

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

6 (64 α^{2} - 208 α + 169) = 25 (20 α^{2} - 36 α + 22)

384 α^{2} - 1248 α + 1014 = 500 α^{2} - 900 α + 550

116 α^{2} + 348 α - 464 = 0

Divide the quadratic equation by 116:

α^{2} + 3 α - 4 = 0

This quadratic equation can be factored as $(α + 4) (α - 1) = 0$ . The possible values for $α$ are $α = 1$ and $α = - 4$ . The sum of all possible values of $α$ is $1 + (- 4) = - 3$ .

💡 Teacher's Secret Hint

Remember that squaring both sides can introduce extraneous solutions, but in this case, both solutions are valid. For a quadratic equation $a x^{2} + b x + c = 0$ , the sum of roots is $- b / a$ .

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Maths Question 14 – JEE-MAIN 2025

Clue 1 of 3: Identify Line Parameters

🧠 Full Solution Path