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3D Geometry question

2025 · 29 Jan · Shift 1 · Q26
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  5. /2025 · 29 Jan · Shift 1 · Q26

3D Geometry question

2025 · 29 Jan · Shift 1 · Q26

JEE MainMathematics3D GeometryMCQ+4 / −1
Let L1:x−11=y−2−1=z−12\mathrm{L}_1: \frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}L1​:1x−1​=−1y−2​=2z−1​ and L2:x+1−1=y−22=z1\mathrm{L}_2: \frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}L2​:−1x+1​=2y−2​=1z​ be two lines. Let L3L_3L3​ be a line passing through the point (α,β,γ)(\alpha, \beta, \gamma)(α,β,γ) and be perpendicular to both L1L_1L1​ and L2L_2L2​. If L3L_3L3​ intersects L1\mathrm{L}_1L1​, then ∣5α−11β−8γ∣|5 \alpha-11 \beta-8 \gamma|∣5α−11β−8γ∣ equals :
  1. A
    25
  2. B
    20
  3. C
    16
  4. D
    18
View written solutionFree

Correct answer: A

  1. Write the lines in parametric form

For L1L_1L1​:

x−11=y−2−1=z−12=t\frac{x-1}{1}=\frac{y-2}{-1}=\frac{z-1}{2}=t1x−1​=−1y−2​=2z−1​=t

So a general point on L1L_1L1​ is

P=(1+t, 2−t, 1+2t)P=(1+t,\,2-t,\,1+2t)P=(1+t,2−t,1+2t)

and its direction vector is

d⃗1=(1,−1,2).\vec d_1=(1,-1,2).d1​=(1,−1,2).

For L2L_2L2​:

x+1−1=y−22=z1=s\frac{x+1}{-1}=\frac{y-2}{2}=\frac{z}{1}=s−1x+1​=2y−2​=1z​=s

So a general point on L2L_2L2​ is

Q=(−1−s, 2+2s, s)Q=(-1-s,\,2+2s,\,s)Q=(−1−s,2+2s,s)

and its direction vector is

d⃗2=(−1,2,1).\vec d_2=(-1,2,1).d2​=(−1,2,1).
  1. Interpret the condition on L3L_3L3​

L3L_3L3​ is perpendicular to both L1L_1L1​ and L2L_2L2​, and it intersects L1L_1L1​.

Hence, if (α,β,γ)(\alpha,\beta,\gamma)(α,β,γ) is a point on L3L_3L3​, then since L3L_3L3​ meets L1L_1L1​, we may take the intersection point with L1L_1L1​ to be

P=(1+t,2−t,1+2t).P=(1+t,2-t,1+2t).P=(1+t,2−t,1+2t).

Also, because L3L_3L3​ is perpendicular to both lines, its direction must be perpendicular to both d⃗1\vec d_1d1​ and d⃗2\vec d_2d2​. A line perpendicular to both is along the common perpendicular, so the vector joining a point on L1L_1L1​ to a point on L2L_2L2​ must be perpendicular to both d⃗1\vec d_1d1​ and d⃗2\vec d_2d2​.

So let

PQ→=Q−P.\overrightarrow{PQ}=Q-P.PQ​=Q−P.

Then we require

PQ→⋅d⃗1=0,PQ→⋅d⃗2=0.\overrightarrow{PQ}\cdot \vec d_1=0, \qquad \overrightarrow{PQ}\cdot \vec d_2=0.PQ​⋅d1​=0,PQ​⋅d2​=0.
  1. Compute PQ→\overrightarrow{PQ}PQ​
PQ→=(−1−s−(1+t), 2+2s−(2−t), s−(1+2t)).\overrightarrow{PQ}=(-1-s-(1+t),\,2+2s-(2-t),\,s-(1+2t)).PQ​=(−1−s−(1+t),2+2s−(2−t),s−(1+2t)).

So

PQ→=(−2−s−t, 2s+t, s−1−2t).\overrightarrow{PQ}=(-2-s-t,\,2s+t,\,s-1-2t).PQ​=(−2−s−t,2s+t,s−1−2t).
  1. Use perpendicularity with L1L_1L1​
PQ→⋅d⃗1=0\overrightarrow{PQ}\cdot \vec d_1=0PQ​⋅d1​=0 (−2−s−t)(1)+(2s+t)(−1)+(s−1−2t)(2)=0.(-2-s-t)(1)+(2s+t)(-1)+(s-1-2t)(2)=0.(−2−s−t)(1)+(2s+t)(−1)+(s−1−2t)(2)=0.

Expand:

−2−s−t−2s−t+2s−2−4t=0-2-s-t-2s-t+2s-2-4t=0−2−s−t−2s−t+2s−2−4t=0 −4−s−6t=0-4-s-6t=0−4−s−6t=0

So,

s+6t=−4.(1)s+6t=-4. \qquad (1)s+6t=−4.(1)
  1. Use perpendicularity with L2L_2L2​
PQ→⋅d⃗2=0\overrightarrow{PQ}\cdot \vec d_2=0PQ​⋅d2​=0 (−2−s−t)(−1)+(2s+t)(2)+(s−1−2t)(1)=0.(-2-s-t)(-1)+(2s+t)(2)+(s-1-2t)(1)=0.(−2−s−t)(−1)+(2s+t)(2)+(s−1−2t)(1)=0.

Expand:

2+s+t+4s+2t+s−1−2t=02+s+t+4s+2t+s-1-2t=02+s+t+4s+2t+s−1−2t=0 1+6s+t=01+6s+t=01+6s+t=0

So,

6s+t=−1.(2)6s+t=-1. \qquad (2)6s+t=−1.(2)
  1. Solve for ttt and sss

From (1):

s=−4−6t.s=-4-6t.s=−4−6t.

Substitute into (2):

6(−4−6t)+t=−16(-4-6t)+t=-16(−4−6t)+t=−1 −24−36t+t=−1-24-36t+t=-1−24−36t+t=−1 −35t=23-35t=23−35t=23 t=−2335.t=-\frac{23}{35}.t=−3523​.

Then

s=−4−6(−2335)=−14035+13835=−235.s=-4-6\left(-\frac{23}{35}\right) =-\frac{140}{35}+\frac{138}{35} =-\frac{2}{35}.s=−4−6(−3523​)=−35140​+35138​=−352​.
  1. Find the intersection point on L1L_1L1​

Using t=−2335t=-\frac{23}{35}t=−3523​ in P=(1+t,2−t,1+2t)P=(1+t,2-t,1+2t)P=(1+t,2−t,1+2t):

α=1+t=1−2335=1235,\alpha=1+t=1-\frac{23}{35}=\frac{12}{35},α=1+t=1−3523​=3512​, β=2−t=2+2335=9335,\beta=2-t=2+\frac{23}{35}=\frac{93}{35},β=2−t=2+3523​=3593​, γ=1+2t=1−4635=−1135.\gamma=1+2t=1-\frac{46}{35}=-\frac{11}{35}.γ=1+2t=1−3546​=−3511​.

Thus

(α,β,γ)=(1235,9335,−1135).(\alpha,\beta,\gamma)=\left(\frac{12}{35},\frac{93}{35},-\frac{11}{35}\right).(α,β,γ)=(3512​,3593​,−3511​).
  1. Compute the required expression
5α−11β−8γ=5(1235)−11(9335)−8(−1135).5\alpha-11\beta-8\gamma =5\left(\frac{12}{35}\right)-11\left(\frac{93}{35}\right)-8\left(-\frac{11}{35}\right).5α−11β−8γ=5(3512​)−11(3593​)−8(−3511​). =6035−102335+8835=148−102335=−87535=−25.=\frac{60}{35}-\frac{1023}{35}+\frac{88}{35} =\frac{148-1023}{35} =\frac{-875}{35}=-25.=3560​−351023​+3588​=35148−1023​=35−875​=−25.

Therefore,

∣5α−11β−8γ∣=25.|5\alpha-11\beta-8\gamma|=25.∣5α−11β−8γ∣=25.
  1. Check options

The value is 252525, which matches Option A.

  1. Comparison with stored answer

Stored correct answer: A

Derived answer: A

So they agree.

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