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3D Geometry question

2021 · 20 Jul · Shift 2 · Q26
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  5. /2021 · 20 Jul · Shift 2 · Q26

3D Geometry question

2021 · 20 Jul · Shift 2 · Q26

JEE MainMathematics3D GeometryMCQ+4 / −1
The lines x = ay −-− 1 = z −-− 2 and x = 3y −-− 2 = bz −-− 2, (ab eee 0) are coplanar, if :
  1. A
    b = 1, a ∈\in∈ R −-− {0}
  2. B
    a = 1, b ∈\in∈ R −-− {0}
  3. C
    a = 2, b = 2
  4. D
    a = 2, b = 3
View written solutionFree

Correct answer: A

  1. Interpret the symmetric forms correctly

    The lines are x=ay−1=z−2x = ay-1 = z-2x=ay−1=z−2 and x=3y−2=bz−2,x = 3y-2 = bz-2,x=3y−2=bz−2, with ab≠0ab\ne 0ab=0.

    Let the common parameter for the first line be ttt. Then x=t,ay−1=t,z−2=t.x=t,\quad ay-1=t,\quad z-2=t.x=t,ay−1=t,z−2=t. So, x=t,y=t+1a,z=t+2.x=t,\quad y=\frac{t+1}{a},\quad z=t+2.x=t,y=at+1​,z=t+2.

    Hence line L1L_1L1​ passes through P1=(0,1a,2)P_1=(0,\tfrac1a,2)P1​=(0,a1​,2) and has direction vector d⃗1=(1,1a,1).\vec d_1=\left(1,\frac1a,1\right).d1​=(1,a1​,1).

    Now let the common parameter for the second line be sss. Then x=s,3y−2=s,bz−2=s.x=s,\quad 3y-2=s,\quad bz-2=s.x=s,3y−2=s,bz−2=s. So, x=s,y=s+23,z=s+2b.x=s,\quad y=\frac{s+2}{3},\quad z=\frac{s+2}{b}.x=s,y=3s+2​,z=bs+2​.

    Hence line L2L_2L2​ passes through P2=(0,23,2b)P_2=\left(0,\frac23,\frac2b\right)P2​=(0,32​,b2​) and has direction vector d⃗2=(1,13,1b).\vec d_2=\left(1,\frac13,\frac1b\right).d2​=(1,31​,b1​).

  2. Condition for two lines to be coplanar

    Two lines are coplanar iff [P1P2→,d⃗1,d⃗2]=0,[\overrightarrow{P_1P_2},\vec d_1,\vec d_2]=0,[P1​P2​​,d1​,d2​]=0, i.e. the scalar triple product is zero.

    First, P1P2→=P2−P1=(0,23−1a,2b−2).\overrightarrow{P_1P_2}=P_2-P_1=\left(0,\frac23-\frac1a,\frac2b-2\right).P1​P2​​=P2​−P1​=(0,32​−a1​,b2​−2).

  3. Compute the determinant

    We need

    0 & \frac23-\frac1a & \frac2b-2\\[4pt] 1 & \frac1a & 1\\[4pt] 1 & \frac13 & \frac1b \end{pmatrix}=0.$$ Expanding along the first column: $$0-1\cdot\det\begin{pmatrix} \frac23-\frac1a & \frac2b-2\\[4pt] \frac13 & \frac1b \end{pmatrix} +1\cdot\det\begin{pmatrix} \frac23-\frac1a & \frac2b-2\\[4pt] \frac1a & 1 \end{pmatrix}=0.$$ So, $$-\left(\left(\frac23-\frac1a\right)\frac1b-\left(\frac2b-2\right)\frac13\right) +\left(\left(\frac23-\frac1a\right)-\left(\frac2b-2\right)\frac1a\right)=0.$$

  4. Simplify

    First bracket:

    =\frac{2}{3b}-\frac{1}{ab}-\frac{2}{3b}+\frac23 =\frac23-\frac1{ab}.$$ Second bracket: $$\left(\frac23-\frac1a\right)-\left(\frac2b-2\right)\frac1a =\frac23-\frac1a-\frac{2}{ab}+\frac2a =\frac23+\frac1a-\frac{2}{ab}.$$ Therefore, $$-\left(\frac23-\frac1{ab}\right)+\left(\frac23+\frac1a-\frac{2}{ab}\right)=0.$$ $$-\frac23+\frac1{ab}+\frac23+\frac1a-\frac{2}{ab}=0$$ $$\frac1a-\frac1{ab}=0$$ $$\frac1a\left(1-\frac1b\right)=0.$$ Since $a\ne 0$, we get $$1-\frac1b=0 \implies b=1.$$

  5. Check options

    • A: b=1, a∈R−{0}b=1,\ a\in \mathbb R-\{0\}b=1, a∈R−{0} ✅ Correct
    • B: a=1, b∈R−{0}a=1,\ b\in \mathbb R-\{0\}a=1, b∈R−{0} ❌ Not sufficient; coplanarity requires b=1b=1b=1
    • C: a=2, b=2a=2,\ b=2a=2, b=2 ❌ Here b≠1b\ne 1b=1
    • D: a=2, b=3a=2,\ b=3a=2, b=3 ❌ Here b≠1b\ne 1b=1

  6. Final answer

    The lines are coplanar when b=1, a≠0\boxed{b=1,\ a\ne 0}b=1, a=0​ so the correct option is A.

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