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3D Geometry question

2025 · 7 Apr · Shift 1 · Q42
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  5. /2025 · 7 Apr · Shift 1 · Q42

3D Geometry question

2025 · 7 Apr · Shift 1 · Q42

JEE MainMathematics3D GeometryMCQ+4 / −1
If the shortest distance between the lines x−12=y−23=z−34\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}2x−1​=3y−2​=4z−3​ and x1=yα=z−51\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}1x​=αy​=1z−5​ is 56\frac{5}{\sqrt{6}}6​5​, then the sum of all possible values of α\alphaα is
  1. A
    32\frac{3}{2}23​
  2. B
    333
  3. C
    −3-3−3
  4. D
    −32-\frac{3}{2}−23​
View written solutionFree

Correct answer: C

  1. Write the two lines in vector form

The lines are

x−12=y−23=z−34\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}2x−1​=3y−2​=4z−3​

and

x1=yα=z−51.\frac{x}{1}=\frac{y}{\alpha}=\frac{z-5}{1}.1x​=αy​=1z−5​.

So we can take:

  • A point on the first line: A(1,2,3)A(1,2,3)A(1,2,3)
  • Direction vector of the first line: d⃗1=(2,3,4)\vec d_1=(2,3,4)d1​=(2,3,4)

For the second line:

  • A point on the second line: B(0,0,5)B(0,0,5)B(0,0,5)
  • Direction vector of the second line: d⃗2=(1,α,1)\vec d_2=(1,\alpha,1)d2​=(1,α,1)
  1. Formula for shortest distance between two skew lines

If two lines are

r⃗=a⃗+td⃗1,r⃗=b⃗+sd⃗2,\vec r=\vec a+t\vec d_1, \qquad \vec r=\vec b+s\vec d_2,r=a+td1​,r=b+sd2​,

then shortest distance 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,

b⃗−a⃗=(0−1,0−2,5−3)=(−1,−2,2).\vec b-\vec a=(0-1,0-2,5-3)=(-1,-2,2).b−a=(0−1,0−2,5−3)=(−1,−2,2).
  1. Compute the cross product
d⃗1×d⃗2=∣i^j^k^2341α1∣\vec d_1\times \vec d_2= \begin{vmatrix} \hat i & \hat j & \hat k\\ 2&3&4\\ 1&\alpha&1 \end{vmatrix}d1​×d2​=​i^21​j^​3α​k^41​​ =i^(3⋅1−4α)−j^(2⋅1−4⋅1)+k^(2α−3⋅1)=\hat i(3\cdot 1-4\alpha)-\hat j(2\cdot 1-4\cdot 1)+\hat k(2\alpha-3\cdot 1)=i^(3⋅1−4α)−j^​(2⋅1−4⋅1)+k^(2α−3⋅1) =(3−4α,2,2α−3).=(3-4\alpha,2,2\alpha-3).=(3−4α,2,2α−3).
  1. Compute the numerator
(b⃗−a⃗)⋅(d⃗1×d⃗2)=(−1,−2,2)⋅(3−4α,2,2α−3)(\vec b-\vec a)\cdot(\vec d_1\times \vec d_2) =(-1,-2,2)\cdot(3-4\alpha,2,2\alpha-3)(b−a)⋅(d1​×d2​)=(−1,−2,2)⋅(3−4α,2,2α−3) =−(3−4α)−4+2(2α−3)=-(3-4\alpha)-4+2(2\alpha-3)=−(3−4α)−4+2(2α−3) =−3+4α−4+4α−6=8α−13.=-3+4\alpha-4+4\alpha-6=8\alpha-13.=−3+4α−4+4α−6=8α−13.

So numerator is ∣8α−13∣|8\alpha-13|∣8α−13∣.

  1. Compute the denominator
∣d⃗1×d⃗2∣=(3−4α)2+22+(2α−3)2.|\vec d_1\times \vec d_2|=\sqrt{(3-4\alpha)^2+2^2+(2\alpha-3)^2}.∣d1​×d2​∣=(3−4α)2+22+(2α−3)2​.

Expand:

(3−4α)2=16α2−24α+9,(3-4\alpha)^2=16\alpha^2-24\alpha+9,(3−4α)2=16α2−24α+9, (2α−3)2=4α2−12α+9.(2\alpha-3)^2=4\alpha^2-12\alpha+9.(2α−3)2=4α2−12α+9.

Therefore,

∣d⃗1×d⃗2∣=20α2−36α+22.|\vec d_1\times \vec d_2|=\sqrt{20\alpha^2-36\alpha+22}.∣d1​×d2​∣=20α2−36α+22​.
  1. Use the given shortest distance

Given

∣8α−13∣20α2−36α+22=56.\frac{|8\alpha-13|}{\sqrt{20\alpha^2-36\alpha+22}}=\frac{5}{\sqrt6}.20α2−36α+22​∣8α−13∣​=6​5​.

Square both sides:

(8α−13)220α2−36α+22=256.\frac{(8\alpha-13)^2}{20\alpha^2-36\alpha+22}=\frac{25}{6}.20α2−36α+22(8α−13)2​=625​.

So,

6(8α−13)2=25(20α2−36α+22).6(8\alpha-13)^2=25(20\alpha^2-36\alpha+22).6(8α−13)2=25(20α2−36α+22).

Now,

(8α−13)2=64α2−208α+169.(8\alpha-13)^2=64\alpha^2-208\alpha+169.(8α−13)2=64α2−208α+169.

Thus,

6(64α2−208α+169)=25(20α2−36α+22).6(64\alpha^2-208\alpha+169)=25(20\alpha^2-36\alpha+22).6(64α2−208α+169)=25(20α2−36α+22). 384α2−1248α+1014=500α2−900α+550.384\alpha^2-1248\alpha+1014=500\alpha^2-900\alpha+550.384α2−1248α+1014=500α2−900α+550.

Bring all terms to one side:

0=116α2+348α−464.0=116\alpha^2+348\alpha-464.0=116α2+348α−464.

Divide by 116116116:

α2+3α−4=0.\alpha^2+3\alpha-4=0.α2+3α−4=0. (α+4)(α−1)=0.(\alpha+4)(\alpha-1)=0.(α+4)(α−1)=0.

Hence,

α=1orα=−4.\alpha=1 \quad \text{or} \quad \alpha=-4.α=1orα=−4.
  1. Find the sum of all possible values
1+(−4)=−3.1+(-4)=-3.1+(−4)=−3.

So the correct answer is

−3.\boxed{-3}.−3​.

This matches Option C.

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