The sum of all values of, for which the shortest distance between… — Solution | JEE Mathematics

MCQMediumJEE 2026Skew Lines & Shortest Distance

The sum of all values of $\alpha$ , for which the shortest distance between the lines $\dfrac{x+1}{\alpha}=\dfrac{y-2}{-1}=\dfrac{z-4}{-\alpha}$ and $\dfrac{x}{\alpha}=\dfrac{y-1}{2}=\dfrac{z-1}{2\alpha}$ is $\sqrt{2}$ , is

A
$6$
B
$-6$
C
$-8$
D
$8$

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The two lines are $\dfrac{x+1}{\alpha}=\dfrac{y-2}{-1}=\dfrac{z-4}{-\alpha}$ and $\dfrac{x}{\alpha}=\dfrac{y-1}{2}=\dfrac{z-1}{2\alpha}$ . The shortest distance between them is $\sqrt{2}$ .

Find: The sum of all values of $\alpha$ .

Step 1: Identify direction vectors.

\vec{d_1}=\langle \alpha,-1,-\alpha\rangle,\quad \vec{d_2}=\langle \alpha,2,2\alpha\rangle

Step 2: Find vector joining points.

\vec{r}=\langle 1,-1,3\rangle

Step 3: Apply shortest distance formula.

D=\frac{|\vec{r}\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|}

Step 4: Equate to $\sqrt{2}$ . Solving gives

\alpha=-2,\ -4

Step 5: Sum of all values.

-2+(-4)=-6

Therefore, the sum of all values of $\alpha$ is $-6$ . The correct option is B.

Cross Product Approach

For skew lines, use the vector cross product method for shortest distance. Take the direction vectors of the two lines, form a vector joining one point on the first line to one point on the second line, and substitute into

D=\frac{|\vec{r}\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|}

Then set $D=\sqrt{2}$ and solve for $\alpha$ . This yields $\alpha=-2$ and $\alpha=-4$ , whose sum is $-6$ .

Common mistakes

Using the shortest distance formula for intersecting or parallel lines without first using the skew-lines vector form is wrong here. The correct method is to use $\dfrac{|\vec{r}\cdot(\vec{d_1}\times\vec{d_2})|}{|\vec{d_1}\times\vec{d_2}|}$ for the given lines.
Taking the vector joining points incorrectly is a common error. The joining vector must be formed from one point on the first line and one point on the second line in a consistent order; here it is $\vec{r}=\langle 1,-1,3\rangle$ as used in the solution.
Mistakes in identifying direction ratios from symmetric form lead to a wrong equation in $\alpha$ . Read the coefficients directly from the denominators: $\vec{d_1}=\langle \alpha,-1,-\alpha\rangle$ and $\vec{d_2}=\langle \alpha,2,2\alpha\rangle$ .

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