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MCQMediumJEE 2025Skew Lines & Shortest Distance

JEE Mathematics 2025 Question with Solution

The line L1L_1 is parallel to the vector a=−3i^+2j^+4k^\mathbf{a} = -3\hat{i} + 2\hat{j} + 4\hat{k} and passes through the point (7,6,2)(7, 6, 2), and the line L2L_2 is parallel to the vector b=2i^+j^+3k^\mathbf{b} = 2\hat{i} + \hat{j} + 3\hat{k} and passes through the point (5,3,4)(5, 3, 4). The shortest distance between the lines L1L_1 and L2L_2 is:

  • A

    2338\frac{23}{\sqrt{38}}

  • B

    2157\frac{21}{\sqrt{57}}

  • C

    2357\frac{23}{\sqrt{57}}

  • D

    2138\frac{21}{\sqrt{38}}

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: Line L1L_1 passes through (7,6,2)(7,6,2) and has direction vector d1=−3i^+2j^+4k^\mathbf{d}_1=-3\hat{i}+2\hat{j}+4\hat{k}. Line L2L_2 passes through (5,3,4)(5,3,4) and has direction vector d2=2i^+j^+3k^\mathbf{d}_2=2\hat{i}+\hat{j}+3\hat{k}.

Find: The shortest distance between the two skew lines.

For two skew lines, the shortest distance is

d=∣(p2−p1)⋅(d1×d2)∣∣d1×d2∣d=\frac{|(\mathbf{p}_2-\mathbf{p}_1)\cdot(\mathbf{d}_1\times \mathbf{d}_2)|}{|\mathbf{d}_1\times \mathbf{d}_2|}

Take

p1=(7,6,2),p2=(5,3,4)\mathbf{p}_1=(7,6,2),\qquad \mathbf{p}_2=(5,3,4)

So,

p2−p1=(−2,−3,2)\mathbf{p}_2-\mathbf{p}_1=(-2,-3,2)

Now compute the cross product:

d1×d2=∣i^j^k^−324213∣\mathbf{d}_1\times \mathbf{d}_2= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 4 \\ 2 & 1 & 3 \end{vmatrix} =i^(2⋅3−4⋅1)−j^((−3)⋅3−4⋅2)+k^((−3)⋅1−2⋅2)=\hat{i}(2\cdot 3-4\cdot 1)-\hat{j}((-3)\cdot 3-4\cdot 2)+\hat{k}((-3)\cdot 1-2\cdot 2) =2i^+17j^−7k^=2\hat{i}+17\hat{j}-7\hat{k}

Then the scalar triple product is

(−2,−3,2)⋅(2,17,−7)=(−2)(2)+(−3)(17)+(2)(−7)=−69(-2,-3,2)\cdot(2,17,-7)=(-2)(2)+(-3)(17)+(2)(-7)=-69

Hence,

∣(p2−p1)⋅(d1×d2)∣=69|(\mathbf{p}_2-\mathbf{p}_1)\cdot(\mathbf{d}_1\times \mathbf{d}_2)|=69

Also,

∣d1×d2∣=22+172+(−7)2=342=338|\mathbf{d}_1\times \mathbf{d}_2|=\sqrt{2^2+17^2+(-7)^2}=\sqrt{342}=3\sqrt{38}

Therefore,

d=69342=69338=2338d=\frac{69}{\sqrt{342}}=\frac{69}{3\sqrt{38}}=\frac{23}{\sqrt{38}}

So, the shortest distance is 2338\frac{23}{\sqrt{38}} and the correct option is A.

Step-by-step vector setup

Given:

  • A point on L1L_1 is (7,6,2)(7,6,2) and its direction vector is (−3,2,4)(-3,2,4).
  • A point on L2L_2 is (5,3,4)(5,3,4) and its direction vector is (2,1,3)(2,1,3).

Find: The shortest distance between L1L_1 and L2L_2.

First write the parametric forms mentioned in the solution:

L1:(7−3λ, 6+2λ, 2+4λ)L_1:(7-3\lambda,\,6+2\lambda,\,2+4\lambda) L2:(5+2μ, 3+μ, 4+3μ)L_2:(5+2\mu,\,3+\mu,\,4+3\mu)

Now use the shortest-distance formula for skew lines:

d=∣(p2−p1)⋅(d1×d2)∣∣d1×d2∣d=\frac{|(\mathbf{p}_2-\mathbf{p}_1)\cdot(\mathbf{d}_1\times \mathbf{d}_2)|}{|\mathbf{d}_1\times \mathbf{d}_2|}

Compute the joining vector:

p2−p1=(5−7, 3−6, 4−2)=(−2,−3,2)\mathbf{p}_2-\mathbf{p}_1=(5-7,\,3-6,\,4-2)=(-2,-3,2)

Compute

d1×d2=∣i^j^k^−324213∣=2i^+17j^−7k^\mathbf{d}_1\times \mathbf{d}_2= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -3 & 2 & 4 \\ 2 & 1 & 3 \end{vmatrix} =2\hat{i}+17\hat{j}-7\hat{k}

Then,

(−2,−3,2)⋅(2,17,−7)=−4−51−14=−69(-2,-3,2)\cdot(2,17,-7)=-4-51-14=-69

Taking modulus gives 6969.

Magnitude of the cross product:

22+172+(−7)2=4+289+49=342\sqrt{2^2+17^2+(-7)^2}=\sqrt{4+289+49}=\sqrt{342}

So,

d=69342d=\frac{69}{\sqrt{342}}

Since 342=9×38342=9\times 38,

342=338\sqrt{342}=3\sqrt{38}

Therefore,

d=69338=2338d=\frac{69}{3\sqrt{38}}=\frac{23}{\sqrt{38}}

Thus, the shortest distance between the two lines is 2338\frac{23}{\sqrt{38}}.

Common mistakes

  • Using the distance formula between two points instead of the skew-lines formula is incorrect because the lines do not intersect and are not parallel in the same plane. Use the scalar triple product with d1×d2\mathbf{d}_1\times\mathbf{d}_2 instead.

  • Computing d1×d2\mathbf{d}_1\times\mathbf{d}_2 with a sign error in the j^\hat{j} term is common. In a determinant expansion, the middle term carries a minus sign, so evaluate that component carefully.

  • Taking the connecting vector in the wrong order without modulus can change the sign of the numerator. Either use p2−p1\mathbf{p}_2-\mathbf{p}_1 or p1−p2\mathbf{p}_1-\mathbf{p}_2, but always apply absolute value in the formula.

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