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3D Geometry question

2023 · 15 Apr · Shift 1 · Q36
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  5. /2023 · 15 Apr · Shift 1 · Q36

3D Geometry question

2023 · 15 Apr · Shift 1 · Q36

JEE MainMathematics3D GeometryMCQ+4 / −1
Let S\mathrm{S}S be the set of all values of λ\lambdaλ, 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}0x−λ​=4y−3​=1z+6​ and x+λ3=y−4=z−60\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}3x+λ​=−4y​=0z−6​ is 13. Then 8∣∑λ∈Sλ∣8\left|\sum\limits_{\lambda \in S} \lambda\right|8​λ∈S∑​λ​ is equal to :
  1. A
    306
  2. B
    304
  3. C
    308
  4. D
    302
View written solutionFree

Correct answer: A

  1. Write the lines in parametric form

The given lines are

x−λ0=y−34=z+61\frac{x-\lambda}{0}=\frac{y-3}{4}=\frac{z+6}{1}0x−λ​=4y−3​=1z+6​ and x+λ3=y−4=z−60.\frac{x+\lambda}{3}=\frac{y}{-4}=\frac{z-6}{0}.3x+λ​=−4y​=0z−6​.

For the first line, since the denominator of x−λx-\lambdax−λ is 000, we get x=λ.x=\lambda.x=λ. Let the common parameter be ttt. Then y=3+4t,z=−6+t.y=3+4t, \qquad z=-6+t.y=3+4t,z=−6+t. So line L1L_1L1​ passes through A=(λ,3,−6)A=(\lambda,3,-6)A=(λ,3,−6) with direction vector d⃗1=(0,4,1).\vec d_1=(0,4,1).d1​=(0,4,1).

For the second line, since the denominator of z−6z-6z−6 is 000, we get z=6.z=6.z=6. Let the common parameter be sss. Then x=−λ+3s,y=−4s.x=-\lambda+3s, \qquad y=-4s.x=−λ+3s,y=−4s. So line L2L_2L2​ passes through B=(−λ,0,6)B=(-\lambda,0,6)B=(−λ,0,6) with direction vector d⃗2=(3,−4,0).\vec d_2=(3,-4,0).d2​=(3,−4,0).

  1. Formula for shortest distance between two skew lines

The shortest distance between lines r⃗=a⃗+td⃗1,r⃗=b⃗+sd⃗2\vec r=\vec a+t\vec d_1, \qquad \vec r=\vec b+s\vec d_2r=a+td1​,r=b+sd2​ is D=∣(b⃗−a⃗)⋅(d⃗1×d⃗2)∣∣d⃗1×d⃗2∣.D=\frac{|(\vec b-\vec a)\cdot(\vec d_1\times \vec d_2)|}{|\vec d_1\times \vec d_2|}.D=∣d1​×d2​∣∣(b−a)⋅(d1​×d2​)∣​.

Here, AB⃗=B−A=(−λ−λ, 0−3, 6−(−6))=(−2λ,−3,12).\vec{AB}=B-A=(-\lambda-\lambda,\,0-3,\,6-(-6))=(-2\lambda,-3,12).AB=B−A=(−λ−λ,0−3,6−(−6))=(−2λ,−3,12).

  1. Compute the cross product
\begin{vmatrix} \hat i & \hat j & \hat k\\ 0 & 4 & 1\\ 3 & -4 & 0 \end{vmatrix}.$$ Expanding, $$\vec d_1\times \vec d_2= \hat i(4\cdot 0-1\cdot(-4)) -\hat j(0\cdot 0-1\cdot 3) +\hat k(0\cdot(-4)-4\cdot 3).$$ So, $$\vec d_1\times \vec d_2=(4,3,-12).$$ Its magnitude is $$|\vec d_1\times \vec d_2|=\sqrt{4^2+3^2+(-12)^2}= \sqrt{16+9+144}= \sqrt{169}=13.$$ --- 4. **Use the distance condition** Given shortest distance is $13$, so $$\frac{|\vec{AB}\cdot(\vec d_1\times \vec d_2)|}{13}=13.$$ Hence, $$|\vec{AB}\cdot(4,3,-12)|=169.$$ Now, $$\vec{AB}\cdot(4,3,-12)=(-2\lambda)(4)+(-3)(3)+(12)(-12).$$ So, $$=-8\lambda-9-144=-8\lambda-153.$$ Thus, $$|-8\lambda-153|=169.$$ --- 5. **Solve for $\lambda$** Case 1: $$-8\lambda-153=169$$ $$-8\lambda=322$$ $$\lambda=-\frac{322}{8}=-\frac{161}{4}.$$ Case 2: $$-8\lambda-153=-169$$ $$-8\lambda=-16$$ $$\lambda=2.$$ Therefore, $$S=\left\{-\frac{161}{4},\,2\right\}.$$ --- 6. **Compute the required value** $$\sum_{\lambda\in S}\lambda=-\frac{161}{4}+2=-\frac{161}{4}+\frac{8}{4}=-\frac{153}{4}.$$ So, $$\left|\sum_{\lambda\in S}\lambda\right|=\frac{153}{4}.$$ Hence, $$8\left|\sum_{\lambda\in S}\lambda\right|=8\cdot \frac{153}{4}=2\cdot 153=306.$$ --- 7. **Compare with options** The value is $$\boxed{306}$$ which corresponds to **Option A**. --- 8. **Comparison with stored answer** Stored correct answer: **A** Our derived answer: **A** So the answer agrees with the stored correct answer.
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