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MCQMediumJEE 2025Skew Lines & Shortest Distance

JEE Mathematics 2025 Question with Solution

If the shortest distance between the lines x−12=y−23=z−34\frac{x-1}{2} = \frac{y-2}{3} = \frac{z-3}{4} and x1=yα=z−51\frac{x}{1} = \frac{y}{\alpha} = \frac{z-5}{1} is 56\frac{5}{\sqrt{6}}, then the sum of all possible values of α\alpha is:

  • A

    32\frac{3}{2}

  • B

    −32\frac{-3}{2}

  • C

    33

  • D

    −3-3

Answer

Correct answer:D

Step-by-step solution

Standard Method

Given: The lines are

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

and

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

with shortest distance

D=56.D=\frac{5}{\sqrt{6}}.

Find: The sum of all possible values of α\alpha.

Take point A(1,2,3)A(1,2,3) and direction vector d⃗1=⟨2,3,4⟩\vec d_1=\langle 2,3,4\rangle for the first line. For the second line, take point B(0,0,5)B(0,0,5) and direction vector d⃗2=⟨1,α,1⟩\vec d_2=\langle 1,\alpha,1\rangle. Then

AB→=⟨−1,−2,2⟩.\overrightarrow{AB}=\langle -1,-2,2\rangle.

Use the shortest distance formula for skew lines:

D=∣AB→⋅(d⃗1×d⃗2)∣∥d⃗1×d⃗2∥.D=\frac{\left|\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2)\right|}{\left\|\vec d_1\times \vec d_2\right\|}.

Now,

d⃗1×d⃗2=∣ijk2341α1∣=⟨3−4α, 2, 2α−3⟩.\vec d_1\times \vec d_2= \begin{vmatrix} \mathbf i & \mathbf j & \mathbf k\\ 2 & 3 & 4\\ 1 & \alpha & 1 \end{vmatrix} =\langle 3-4\alpha,\,2,\,2\alpha-3\rangle.

Hence,

∥d⃗1×d⃗2∥=(3−4α)2+4+(2α−3)2=20α2−36α+22.\left\|\vec d_1\times \vec d_2\right\|=\sqrt{(3-4\alpha)^2+4+(2\alpha-3)^2}=\sqrt{20\alpha^2-36\alpha+22}.

Also,

AB→⋅(d⃗1×d⃗2)=(−1)(3−4α)+(−2)(2)+2(2α−3)=8α−13.\overrightarrow{AB}\cdot(\vec d_1\times \vec d_2)=(-1)(3-4\alpha)+(-2)(2)+2(2\alpha-3)=8\alpha-13.

Therefore,

∣8α−13∣20α2−36α+22=56.\frac{|8\alpha-13|}{\sqrt{20\alpha^2-36\alpha+22}}=\frac{5}{\sqrt{6}}.

Squaring both sides,

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

So,

29α2+87α−116=0.29\alpha^2+87\alpha-116=0.

Solving,

α=−87±872+4⋅29⋅11658=−87±14558∈{1,−4}.\alpha=\frac{-87\pm\sqrt{87^2+4\cdot29\cdot116}}{58}=\frac{-87\pm145}{58}\in\{1,-4\}.

Hence the sum of all possible values is

1+(−4)=−3.1+(-4)=-3.

Therefore, the correct option is D.

Vector formula shortcut

Given: Two skew lines with direction vectors d⃗1=⟨2,3,4⟩\vec d_1=\langle 2,3,4\rangle and d⃗2=⟨1,α,1⟩\vec d_2=\langle 1,\alpha,1\rangle.

Find: The sum of all possible values of α\alpha.

The quickest route is to directly use one point from each line and the vector formula for distance between skew lines. With A(1,2,3)A(1,2,3) and B(0,0,5)B(0,0,5),

AB→=⟨−1,−2,2⟩.\overrightarrow{AB}=\langle -1,-2,2\rangle.

Then compute only the two quantities needed in the formula:

d⃗1×d⃗2=⟨3−4α,2,2α−3⟩,\vec d_1\times\vec d_2=\langle 3-4\alpha,2,2\alpha-3\rangle, AB→⋅(d⃗1×d⃗2)=8α−13.\overrightarrow{AB}\cdot(\vec d_1\times\vec d_2)=8\alpha-13.

Substitute into

∣AB→⋅(d⃗1×d⃗2)∣∥d⃗1×d⃗2∥=56\frac{|\overrightarrow{AB}\cdot(\vec d_1\times\vec d_2)|}{\|\vec d_1\times\vec d_2\|}=\frac{5}{\sqrt6}

to get

∣8α−13∣20α2−36α+22=56.\frac{|8\alpha-13|}{\sqrt{20\alpha^2-36\alpha+22}}=\frac{5}{\sqrt6}.

This gives

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

so

α=1,−4.\alpha=1,-4.

Therefore, their sum is −3-3, so the correct option is D.

Common mistakes

  • Using the distance formula for parallel lines or intersecting lines is incorrect because these lines are skew. For skew lines, use D=∣AB→⋅(d⃗1×d⃗2)∣∥d⃗1×d⃗2∥D=\frac{|\overrightarrow{AB}\cdot(\vec d_1\times\vec d_2)|}{\|\vec d_1\times\vec d_2\|} instead.

  • Taking the wrong connecting vector between points on the two lines leads to an incorrect dot product. Choose one fixed point from each line carefully, then form AB→\overrightarrow{AB} consistently.

  • Errors in the cross product d⃗1×d⃗2\vec d_1\times\vec d_2 are common. A sign mistake in the determinant changes both numerator and denominator, so expand the determinant carefully.

  • Stopping after finding one value of α\alpha is incomplete because the equation is quadratic. Solve for all possible values and then add them, since the question asks for the sum of all possible values.

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