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3D Geometry question

2020 · Shift 1 · Q29
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  5. /2020 · Shift 1 · Q29

3D Geometry question

2020 · Shift 1 · Q29

JEE AdvancedMathematics3D GeometryMultiple correct+4 / −2
Let L1 and L2 be the following straight lines. L1:x−11=y−1=z−13{L_1}:{{x - 1} \over 1} = {y \over { - 1}} = {{z - 1} \over 3}L1​:1x−1​=−1y​=3z−1​ and L2:x−1−3=y−1=z−11{L_2}:{{x - 1} \over { - 3}} = {y \over { - 1}} = {{z - 1} \over 1}L2​:−3x−1​=−1y​=1z−1​. Suppose the straight line L:x−αl=y−1m=z−γ−2L:{{x - \alpha } \over l} = {{y - 1} \over m} = {{z - \gamma } \over { - 2}}L:lx−α​=my−1​=−2z−γ​ lies in the plane containing L1 and L2 and passes through the point of intersection of L1 and L2. If the line L bisects the acute angle between the lines L1 and L2, then which of the following statements is/are TRUE?
  1. A
    α−γ\alpha -\gammaα−γ = 3
  2. B
    l + m = 2
  3. C
    α−γ\alpha -\gammaα−γ = 1
  4. D
    l + m = 0
View written solutionFree

Correct answer: A, B

  1. Write the lines in parametric form

For L1L_1L1​:

x−11=y−1=z−13=t\frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{3}=t1x−1​=−1y​=3z−1​=t

So,

x=1+t,y=−t,z=1+3tx=1+t,\quad y=-t,\quad z=1+3tx=1+t,y=−t,z=1+3t

Hence, a point on L1L_1L1​ is P1=(1,0,1)P_1=(1,0,1)P1​=(1,0,1) and its direction vector is

d⃗1=(1,−1,3).\vec d_1=(1,-1,3).d1​=(1,−1,3).

For L2L_2L2​:

x−1−3=y−1=z−11=s\frac{x-1}{-3}=\frac{y}{-1}=\frac{z-1}{1}=s−3x−1​=−1y​=1z−1​=s

So,

x=1−3s,y=−s,z=1+sx=1-3s,\quad y=-s,\quad z=1+sx=1−3s,y=−s,z=1+s

Hence, a point on L2L_2L2​ is also P2=(1,0,1)P_2=(1,0,1)P2​=(1,0,1) and its direction vector is

d⃗2=(−3,−1,1).\vec d_2=(-3,-1,1).d2​=(−3,−1,1).

Thus, the two lines intersect at

P=(1,0,1).P=(1,0,1).P=(1,0,1).
  1. Find the direction of the angle bisectors

The direction vectors of the angle bisectors between two intersecting lines are proportional to

d⃗1∣d⃗1∣±d⃗2∣d⃗2∣.\frac{\vec d_1}{|\vec d_1|} \pm \frac{\vec d_2}{|\vec d_2|}.∣d1​∣d1​​±∣d2​∣d2​​.

Now,

∣d⃗1∣=12+(−1)2+32=11,|\vec d_1|=\sqrt{1^2+(-1)^2+3^2}=\sqrt{11},∣d1​∣=12+(−1)2+32​=11​, ∣d⃗2∣=(−3)2+(−1)2+12=11.|\vec d_2|=\sqrt{(-3)^2+(-1)^2+1^2}=\sqrt{11}.∣d2​∣=(−3)2+(−1)2+12​=11​.

So the bisector directions are proportional to

d⃗1+d⃗2=(1,−1,3)+(−3,−1,1)=(−2,−2,4)\vec d_1+\vec d_2=(1,-1,3)+(-3,-1,1)=(-2,-2,4)d1​+d2​=(1,−1,3)+(−3,−1,1)=(−2,−2,4)

which simplifies to

(−1,−1,2),(-1,-1,2),(−1,−1,2),

and

d⃗1−d⃗2=(1,−1,3)−(−3,−1,1)=(4,0,2)\vec d_1-\vec d_2=(1,-1,3)-(-3,-1,1)=(4,0,2)d1​−d2​=(1,−1,3)−(−3,−1,1)=(4,0,2)

which simplifies to

(2,0,1).(2,0,1).(2,0,1).

To determine which one is the acute-angle bisector, check the angle each makes with d⃗1\vec d_1d1​ and d⃗2\vec d_2d2​.

Take

b⃗1=(−1,−1,2).\vec b_1=(-1,-1,2).b1​=(−1,−1,2).

Then

b⃗1⋅d⃗1=−1(1)+(−1)(−1)+2(3)=6>0,\vec b_1\cdot \vec d_1=-1(1)+(-1)(-1)+2(3)=6>0,b1​⋅d1​=−1(1)+(−1)(−1)+2(3)=6>0, b⃗1⋅d⃗2=(−1)(−3)+(−1)(−1)+2(1)=6>0.\vec b_1\cdot \vec d_2=(-1)(-3)+(-1)(-1)+2(1)=6>0.b1​⋅d2​=(−1)(−3)+(−1)(−1)+2(1)=6>0.

So b⃗1\vec b_1b1​ makes acute angles with both lines, hence it bisects the acute angle.

Thus, line LLL has direction ratios proportional to

(l,m,−2)∥(−1,−1,2).(l,m,-2) \parallel (-1,-1,2).(l,m,−2)∥(−1,−1,2).

Let

(l,m,−2)=k(−1,−1,2).(l,m,-2)=k(-1,-1,2).(l,m,−2)=k(−1,−1,2).

From the third component,

−2=2k  ⟹  k=−1.-2=2k \implies k=-1.−2=2k⟹k=−1.

Hence,

l=1,m=1.l=1,\quad m=1.l=1,m=1.

Therefore,

l+m=2.l+m=2.l+m=2.

So Option B is true and Option D is false.

  1. Use the fact that line LLL passes through the intersection point

Given

L:x−αl=y−1m=z−γ−2.L:\frac{x-\alpha}{l}=\frac{y-1}{m}=\frac{z-\gamma}{-2}.L:lx−α​=my−1​=−2z−γ​.

Since LLL passes through P=(1,0,1)P=(1,0,1)P=(1,0,1), there exists some parameter rrr such that

1−αl=0−1m=1−γ−2=r.\frac{1-\alpha}{l}=\frac{0-1}{m}=\frac{1-\gamma}{-2}=r.l1−α​=m0−1​=−21−γ​=r.

Using l=1,m=1l=1, m=1l=1,m=1,

1−α=−1  ⟹  α=2,1-\alpha=-1 \implies \alpha=2,1−α=−1⟹α=2, 1−γ−2=−1  ⟹  1−γ=2  ⟹  γ=−1.\frac{1-\gamma}{-2}=-1 \implies 1-\gamma=2 \implies \gamma=-1.−21−γ​=−1⟹1−γ=2⟹γ=−1.

Thus,

α−γ=2−(−1)=3.\alpha-\gamma=2-(-1)=3.α−γ=2−(−1)=3.

So Option A is true and Option C is false.

  1. Final evaluation of options
  • A: α−γ=3\alpha-\gamma=3α−γ=3 ✅ True
  • B: l+m=2l+m=2l+m=2 ✅ True
  • C: α−γ=1\alpha-\gamma=1α−γ=1 ❌ False
  • D: l+m=0l+m=0l+m=0 ❌ False

Therefore, the correct options are

A, B\boxed{A,\ B}A, B​
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