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MCQMediumJEE 2023Skew Lines & Shortest Distance

JEE Mathematics 2023 Question with Solution

Let SS be the set of all values of λ\lambda, for which the shortest distance between the lines x−λ0=y−34=z+61\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1} and x+λ3=y−4=z−60\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0} is 1313. Then 8∣∑λ∈Sλ∣8\left|\sum_{\lambda\in S}\lambda\right| is equal to

  • A

    302302

  • B

    304304

  • C

    306306

  • D

    308308

Answer

Correct answer:C

Step-by-step solution

Standard Method

Given: The shortest distance between the lines x−λ0=y−34=z+61\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1} and x+λ3=y−4=z−60\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0} is 1313.

Find: 8∣∑λ∈Sλ∣8\left|\sum_{\lambda\in S}\lambda\right|.

Write the lines in vector form.

x=λ,y=3+4t,z=−6+tx=\lambda,\quad y=3+4t,\quad z=-6+t

So, the direction vector of the first line is

d⃗1=(0,4,1)\vec d_1=(0,4,1)

For the second line,

x=−λ+3s,y=−4s,z=6x=-\lambda+3s,\quad y=-4s,\quad z=6

So, the direction vector is

d⃗2=(3,−4,0)\vec d_2=(3,-4,0)

Take points on the two lines:

A(λ,3,−6),B(−λ,0,6)A(\lambda,3,-6),\quad B(-\lambda,0,6)

Hence,

AB⃗=(−λ−λ, −3, 12)=(−2λ,−3,12)\vec{AB}=(-\lambda-\lambda,\,-3,\,12)=(-2\lambda,-3,12)

Use the shortest distance formula between two skew lines:

D=∣AB⃗⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣D=\frac{|\vec{AB}\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}

Compute the cross product:

d⃗1×d⃗2=∣ijk0413−40∣=(4,3,−12)\vec d_1\times \vec d_2= \begin{vmatrix} \mathbf{i}&\mathbf{j}&\mathbf{k}\\ 0&4&1\\ 3&-4&0 \end{vmatrix} =(4,3,-12)

Therefore,

∣d⃗1×d⃗2∣=42+32+(−12)2=13|\vec d_1\times \vec d_2|=\sqrt{4^2+3^2+(-12)^2}=13

Now compute the scalar triple product:

AB⃗⋅(d⃗1×d⃗2)=(−2λ,−3,12)⋅(4,3,−12)\vec{AB}\cdot(\vec d_1\times \vec d_2)=(-2\lambda,-3,12)\cdot(4,3,-12) =−8λ−9−144=−8λ−153=-8\lambda-9-144=-8\lambda-153

Thus,

D=∣8λ+153∣13D=\frac{|8\lambda+153|}{13}

Given that D=13D=13,

∣8λ+153∣13=13\frac{|8\lambda+153|}{13}=13

So,

∣8λ+153∣=169|8\lambda+153|=169

Hence,

8λ+153=±1698\lambda+153=\pm 169

Case I:

8λ+153=1698\lambda+153=169 8λ=168\lambda=16 λ=2\lambda=2

Case II:

8λ+153=−1698\lambda+153=-169 8λ=−3228\lambda=-322 λ=−1614\lambda=-\frac{161}{4}

Therefore,

S={2,−1614}S=\left\{2,-\frac{161}{4}\right\}

Now,

∑λ∈Sλ=2−1614=−1534\sum_{\lambda\in S}\lambda=2-\frac{161}{4}=-\frac{153}{4}

Hence,

8∣∑λ∈Sλ∣=8×1534=3068\left|\sum_{\lambda\in S}\lambda\right|=8\times\frac{153}{4}=306

Therefore, the correct option is C.

Direct Distance Evaluation

Given: The shortest distance between the two lines is 1313.

Find: 8∣∑λ∈Sλ∣8\left|\sum_{\lambda\in S}\lambda\right|.

A quick route is to immediately note the direction vectors

d⃗1=(0,4,1),d⃗2=(3,−4,0)\vec d_1=(0,4,1),\quad \vec d_2=(3,-4,0)

and points

A(λ,3,−6),B(−λ,0,6)A(\lambda,3,-6),\quad B(-\lambda,0,6)

Thus,

AB⃗=(−2λ,−3,12)\vec{AB}=(-2\lambda,-3,12)

Since

d⃗1×d⃗2=(4,3,−12)\vec d_1\times \vec d_2=(4,3,-12)

and

∣(4,3,−12)∣=13,|(4,3,-12)|=13,

the denominator of the skew-line distance formula is already 1313. So only the numerator needs evaluation:

(−2λ,−3,12)⋅(4,3,−12)=−8λ−153(-2\lambda,-3,12)\cdot(4,3,-12)=-8\lambda-153

Hence,

D=∣8λ+153∣13D=\frac{|8\lambda+153|}{13}

Using D=13D=13,

∣8λ+153∣=169|8\lambda+153|=169

which gives

λ=2,−1614\lambda=2,\quad -\frac{161}{4}

Therefore,

8∣2−1614∣=3068\left|2-\frac{161}{4}\right|=306

So, the correct option is C.

Common mistakes

  • Using the distance formula for parallel lines or point-to-line distance is incorrect because these lines are skew. Use the scalar triple product formula with a connecting vector and the cross product of direction vectors instead.

  • Taking the connecting vector incorrectly, such as using BA⃗\vec{BA} inconsistently with the dot product sign, can change the expression. The absolute value removes overall sign, but the vector components must still be formed carefully from chosen points.

  • Errors in the cross product d⃗1×d⃗2\vec d_1\times\vec d_2 are common. A wrong determinant expansion gives an incorrect denominator and numerator. Compute each component systematically before substituting into the distance formula.

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