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MCQMediumJEE 2023Skew Lines & Shortest Distance

JEE Mathematics 2023 Question with Solution

One vertex of a rectangular parallelepiped is at the origin OO and the lengths of its edges along xx, yy and zz axes are 33, 44 and 55 units respectively. Let PP be the vertex (3,4,5)(3, 4, 5). Then the shortest distance between the diagonal OPOP and an edge parallel to the zz axis, not passing through OO or PP is:

  • A

    1255\frac{12}{5\sqrt{5}}

  • B

    1255\frac{12\sqrt{5}}{5}

  • C

    125\frac{12}{5}

  • D

    1212

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: One diagonal is OPOP of the rectangular parallelepiped with direction vector 3i^+4j^+5k^3\hat{i}+4\hat{j}+5\hat{k}. We need the shortest distance from this diagonal to an edge parallel to the zz-axis not passing through OO or PP.

Find: The shortest distance between the two skew lines.

The equation of diagonal OEOE is

r⃗=λ(3i^+4j^+5k^)\vec{r}=\lambda(3\hat{i}+4\hat{j}+5\hat{k})

and the equation of edge GDGD is

r⃗=4j^+μk^\vec{r}=4\hat{j}+\mu\hat{k}
Rectangular parallelepiped with origin O, vertex E at (3,4,5), G at (0,4,0), D at (0,4,5), C at (0,0,5), and diagonal from O to E shown.

The shortest distance is the magnitude of the projection of 4i^4\hat{i} on 3j^−4i^3\hat{j}-4\hat{i}.

Shortest distance=129+16=125\text{Shortest distance}=\frac{12}{\sqrt{9+16}}=\frac{12}{5}

Therefore, the shortest distance is 125\frac{12}{5}, so the correct option is C.

Skew Lines Formula

Given: The diagonal OPOP has direction vector (3,4,5)(3,4,5). Take the edge parallel to the zz-axis through (3,0,0)(3,0,0), whose direction vector is (0,0,1)(0,0,1).

Find: The shortest distance between these two skew lines.

Use the formula for distance between two skew lines:

S.D.=∣(a2⃗−a1⃗)⋅(b1⃗×b2⃗)∣∣b1⃗×b2⃗∣\text{S.D.}=\frac{|(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})|}{|\vec{b_1}\times\vec{b_2}|}

Here,

a1⃗=(0,0,0),b1⃗=(3,4,5),a2⃗=(3,0,0),b2⃗=(0,0,1)\vec{a_1}=(0,0,0),\quad \vec{b_1}=(3,4,5),\quad \vec{a_2}=(3,0,0),\quad \vec{b_2}=(0,0,1)

Now,

b1⃗×b2⃗=∣i^j^k^345001∣=4i^−3j^\vec{b_1}\times\vec{b_2}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\\3&4&5\\0&0&1\end{vmatrix}=4\hat{i}-3\hat{j}

So,

∣b1⃗×b2⃗∣=42+(−3)2=16+9=5|\vec{b_1}\times\vec{b_2}|=\sqrt{4^2+(-3)^2}=\sqrt{16+9}=5

Also,

a2⃗−a1⃗=(3,0,0)\vec{a_2}-\vec{a_1}=(3,0,0)

Hence,

(a2⃗−a1⃗)⋅(b1⃗×b2⃗)=(3,0,0)⋅(4,−3,0)=12(\vec{a_2}-\vec{a_1})\cdot(\vec{b_1}\times\vec{b_2})=(3,0,0)\cdot(4,-3,0)=12

Therefore,

S.D.=125\text{S.D.}=\frac{12}{5}

Thus, the shortest distance is 125\frac{12}{5}.

Common mistakes

  • Choosing the edge through PP or through OO is incorrect because the question explicitly excludes edges passing through those points. First identify a valid edge parallel to the zz-axis, such as the one through (3,0,0)(3,0,0) or (0,4,0)(0,4,0).

  • Using the distance formula for parallel lines or point-to-line distance directly is wrong because the diagonal OPOP and the selected edge are skew lines. Use the skew-lines distance formula or an equivalent projection method instead.

  • Computing the cross product incorrectly can change the final answer. For direction vectors (3,4,5)(3,4,5) and (0,0,1)(0,0,1), the correct cross product is 4i^−3j^4\hat{i}-3\hat{j}, whose magnitude is 55, not 41\sqrt{41} or any other value.

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