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MCQMediumJEE 2023Skew Lines & Shortest Distance

JEE Mathematics 2023 Question with Solution

The distance of the point (−1,2,3)(-1, 2, 3) from the plane r⋅(i^−2j^+3k^)=10\mathbf{r} \cdot (\hat{i} - 2\hat{j} + 3\hat{k}) = 10 parallel to the line of the shortest distance between the lines \mathbf{r = (\hat{i - \hat{j) + \lambda(2\hat{i + \hat{k) and \mathbf{r = (2\hat{i - \hat{j) + \mu (\hat{i - \hat{j + \hat{k) is:

  • A

    252\sqrt{5}

  • B

    353\sqrt{5}

  • C

    363\sqrt{6}

  • D

    262\sqrt{6}

Answer

Correct answer:B

Step-by-step solution

Standard Method

Given: The point is (−1,2,3)(-1,2,3) and the plane is r⋅(i^−2j^+3k^)=10\mathbf{r}\cdot(\hat{i}-2\hat{j}+3\hat{k})=10. The required direction is parallel to the line of shortest distance between the two given lines.

Find: The distance from the point to the plane measured along that direction.

For the two lines, the direction vectors are

a1=⟨2,0,1⟩,a2=⟨1,−1,1⟩\mathbf{a}_1=\langle 2,0,1\rangle, \qquad \mathbf{a}_2=\langle 1,-1,1\rangle

The line of shortest distance between two skew lines is parallel to a1×a2\mathbf{a}_1 \times \mathbf{a}_2.

Compute the cross product:

n=a1×a2=∣i^j^k^2011−11∣\mathbf{n}=\mathbf{a}_1\times\mathbf{a}_2=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 2 & 0 & 1 \\ 1 & -1 & 1 \end{vmatrix} n=i^(0⋅1−1⋅(−1))−j^(2⋅1−1⋅1)+k^(2⋅(−1)−0⋅1)\mathbf{n}=\hat{i}(0\cdot 1-1\cdot(-1)) - \hat{j}(2\cdot 1-1\cdot 1) + \hat{k}(2\cdot(-1)-0\cdot 1) n=i^−j^−2k^=⟨1,−1,−2⟩\mathbf{n}=\hat{i}-\hat{j}-2\hat{k}=\langle 1,-1,-2\rangle

Now take the line through the point (−1,2,3)(-1,2,3) in this direction:

R(t)=⟨−1,2,3⟩+t⟨1,−1,−2⟩\mathbf{R}(t)=\langle -1,2,3\rangle + t\langle 1,-1,-2\rangle

A point on this line lies on the plane when

R(t)⋅⟨1,−2,3⟩=10\mathbf{R}(t)\cdot \langle 1,-2,3\rangle = 10

Substitute:

(−1+t,  2−t,  3−2t)⋅⟨1,−2,3⟩=10(-1+t,\;2-t,\;3-2t)\cdot \langle 1,-2,3\rangle = 10 (−1+t)(1)+(2−t)(−2)+(3−2t)(3)=10(-1+t)(1)+(2-t)(-2)+(3-2t)(3)=10 (−1+t)−(4−2t)+(9−6t)=10(-1+t)-(4-2t)+(9-6t)=10 4−3t=104-3t=10 t=−2t=-2

Hence the displacement vector from the point to the plane along this direction is

(−2)⟨1,−1,−2⟩=⟨−2,2,4⟩(-2)\langle 1,-1,-2\rangle = \langle -2,2,4\rangle

Its magnitude is

(−2)2+22+42=4+4+16=24=26\sqrt{(-2)^2+2^2+4^2}=\sqrt{4+4+16}=\sqrt{24}=2\sqrt{6}

Therefore, the required distance is 262\sqrt{6}. This matches option D. The solution states B, but the worked result is 262\sqrt{6}, so the derived answer from the working is D.

Using direction of shortest distance explicitly

Given: The required path must be parallel to the shortest-distance line between the two skew lines.

Find: The corresponding distance from the given point to the plane.

The shortest-distance line between two skew lines is perpendicular to both lines. Therefore its direction vector is the cross product of their direction vectors:

⟨2,0,1⟩×⟨1,−1,1⟩=⟨1,−1,−2⟩\langle 2,0,1\rangle \times \langle 1,-1,1\rangle = \langle 1,-1,-2\rangle

So the required directed line from the point is

(x,y,z)=(−1,2,3)+t(1,−1,−2)(x,y,z)=(-1,2,3)+t(1,-1,-2)

which gives

x=−1+t,y=2−t,z=3−2tx=-1+t, \qquad y=2-t, \qquad z=3-2t

The plane equation is

x−2y+3z=10x-2y+3z=10

Substitute the parametric coordinates:

(−1+t)−2(2−t)+3(3−2t)=10(-1+t)-2(2-t)+3(3-2t)=10 −1+t−4+2t+9−6t=10-1+t-4+2t+9-6t=10 4−3t=104-3t=10 t=−2t=-2

Distance along this line equals ∣t∣|t| times the magnitude of the direction vector:

∣t∣ ∥⟨1,−1,−2⟩∥=212+(−1)2+(−2)2|t|\,\|\langle 1,-1,-2\rangle\| = 2\sqrt{1^2+(-1)^2+(-2)^2} =26=2\sqrt{6}

Hence the correct option is D.

Common mistakes

  • Using the plane’s normal vector directly as the direction of motion. That is wrong because the question asks for distance along a line parallel to the shortest distance between the two given lines, not perpendicular distance to the plane. First find the cross product of the two line direction vectors.

  • Trusting the listed statement “The Correct Option is B” without checking the algebra. The worked solution gives the final distance as 262\sqrt{6}, which corresponds to option D. Always verify the final value against the options.

  • Computing a1×a2\mathbf{a}_1 \times \mathbf{a}_2 incorrectly by sign error in the determinant expansion. A wrong cross product changes the entire direction line. Expand the determinant carefully, especially the negative sign in the j^\hat{j} term.

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