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3D Geometry question

2025 · 3 Apr · Shift 1 · Q35
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3D Geometry question

2025 · 3 Apr · Shift 1 · Q35

JEE MainMathematics3D GeometryMCQ+4 / −1
Let a line passing through the point (4,1,0)(4,1,0)(4,1,0) intersect the line L1:x−12=y−23=z−34\mathrm{L}_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}L1​:2x−1​=3y−2​=4z−3​ at the point A(α,β,γ)A(\alpha, \beta, \gamma)A(α,β,γ) and the line L2:x−6=y=−z+4\mathrm{L}_2: x-6=y=-z+4L2​:x−6=y=−z+4 at the point B(a,b,c)B(a, b, c)B(a,b,c). Then ∣101αβγabc∣\left|\begin{array}{lll}1 & 0 & 1 \\ \alpha & \beta & \gamma \\ a & b & c\end{array}\right|​1αa​0βb​1γc​​ is equal to
  1. A
    16
  2. B
    6
  3. C
    8
  4. D
    12
View written solutionFree

Correct answer: C

  1. Parametrize the given lines

For L1L_1L1​:

x−12=y−23=z−34=t\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}=t2x−1​=3y−2​=4z−3​=t

So a general point on L1L_1L1​ is

A(α,β,γ)=(1+2t, 2+3t, 3+4t).A(\alpha,\beta,\gamma)=(1+2t,\,2+3t,\,3+4t).A(α,β,γ)=(1+2t,2+3t,3+4t).

For L2L_2L2​:

x−6=y=−z+4=sx-6=y=-z+4=sx−6=y=−z+4=s

So

x=6+s,y=s,z=4−s.x=6+s,\quad y=s,\quad z=4-s.x=6+s,y=s,z=4−s.

Hence a general point on L2L_2L2​ is

B(a,b,c)=(6+s, s, 4−s).B(a,b,c)=(6+s,\,s,\,4-s).B(a,b,c)=(6+s,s,4−s).

The line through AAA and BBB passes through (4,1,0)(4,1,0)(4,1,0). Therefore points A,B,A,B,A,B, and P=(4,1,0)P=(4,1,0)P=(4,1,0) are collinear.

  1. Use collinearity condition

Since PPP lies on line ABABAB, vectors AP→\overrightarrow{AP}AP and AB→\overrightarrow{AB}AB are parallel.

First compute:

AB→=B−A=(5+s−2t, s−2−3t, 1−s−4t)\overrightarrow{AB}=B-A=(5+s-2t,\,s-2-3t,\,1-s-4t)AB=B−A=(5+s−2t,s−2−3t,1−s−4t)

and

AP→=P−A=(3−2t, −1−3t, −3−4t).\overrightarrow{AP}=P-A=(3-2t,\,-1-3t,\,-3-4t).AP=P−A=(3−2t,−1−3t,−3−4t).

So there exists λ\lambdaλ such that

(3−2t, −1−3t, −3−4t)=λ(5+s−2t, s−2−3t, 1−s−4t).(3-2t,\,-1-3t,\,-3-4t)=\lambda(5+s-2t,\,s-2-3t,\,1-s-4t).(3−2t,−1−3t,−3−4t)=λ(5+s−2t,s−2−3t,1−s−4t).

But a cleaner way is to write PPP as a section point of AAA and BBB:

P=A+μ(B−A).P=A+\mu(B-A).P=A+μ(B−A).

Thus,

(4,1,0)=(1+2t,2+3t,3+4t)+μ((5+s−2t),(s−2−3t),(1−s−4t)).(4,1,0)=(1+2t,2+3t,3+4t)+\mu\big((5+s-2t),(s-2-3t),(1-s-4t)\big).(4,1,0)=(1+2t,2+3t,3+4t)+μ((5+s−2t),(s−2−3t),(1−s−4t)).

This gives the system:

3−2t=μ(5+s−2t)(1)3-2t=\mu(5+s-2t) \qquad (1)3−2t=μ(5+s−2t)(1) −1−3t=μ(s−2−3t)(2)-1-3t=\mu(s-2-3t) \qquad (2)−1−3t=μ(s−2−3t)(2) −3−4t=μ(1−s−4t)(3)-3-4t=\mu(1-s-4t) \qquad (3)−3−4t=μ(1−s−4t)(3)

We solve for t,s,μt,s,\mut,s,μ.

  1. Solve the system

From (1) and (2), subtract suitably. Observe that if we try μ=12\mu=\frac12μ=21​, then

From (1):

3−2t=12(5+s−2t)3-2t=\frac12(5+s-2t)3−2t=21​(5+s−2t) 6−4t=5+s−2t6-4t=5+s-2t6−4t=5+s−2t 1−2t=s1-2t=s 1−2t=s

From (2):

−1−3t=12(s−2−3t)-1-3t=\frac12(s-2-3t)−1−3t=21​(s−2−3t) −2−6t=s−2−3t-2-6t=s-2-3t−2−6t=s−2−3t s=−3ts=-3ts=−3t

Equating the two expressions for sss:

1−2t=−3t  ⟹  t=−1.1-2t=-3t \implies t=-1.1−2t=−3t⟹t=−1.

Then

s=−3(−1)=3.s=-3(-1)=3.s=−3(−1)=3.

Check in (3):

−3−4(−1)=1,-3-4(-1)=1,−3−4(−1)=1,

and

12(1−3−4(−1))=12(2)=1,\frac12(1-3-4(-1))=\frac12(2)=1,21​(1−3−4(−1))=21​(2)=1,

so it works.

Thus,

t=−1,s=3.t=-1,\quad s=3.t=−1,s=3.

Therefore,

A=(1+2(−1), 2+3(−1), 3+4(−1))=(−1,−1,−1),A=(1+2(-1),\,2+3(-1),\,3+4(-1))=(-1,-1,-1),A=(1+2(−1),2+3(−1),3+4(−1))=(−1,−1,−1), B=(6+3, 3, 4−3)=(9,3,1).B=(6+3,\,3,\,4-3)=(9,3,1).B=(6+3,3,4−3)=(9,3,1).

So

α=−1, β=−1, γ=−1,\alpha=-1,\ \beta=-1,\ \gamma=-1,α=−1, β=−1, γ=−1, a=9, b=3, c=1.a=9,\ b=3,\ c=1.a=9, b=3, c=1.
  1. Evaluate the determinant

We need

∣101αβγabc∣=∣101−1−1−1931∣.\left|\begin{array}{ccc} 1 & 0 & 1\\ \alpha & \beta & \gamma\\ a & b & c \end{array}\right| = \left|\begin{array}{ccc} 1 & 0 & 1\\ -1 & -1 & -1\\ 9 & 3 & 1 \end{array}\right|.​1αa​0βb​1γc​​=​1−19​0−13​1−11​​.

Expand along the first row:

D=1∣−1−131∣+1∣−1−193∣D=1\begin{vmatrix}-1 & -1\\ 3 & 1\end{vmatrix}+1\begin{vmatrix}-1 & -1\\ 9 & 3\end{vmatrix}D=1​−13​−11​​+1​−19​−13​​

(since the middle term is zero, and sign of third term is positive).

Now,

∣−1−131∣=(−1)(1)−(−1)(3)=−1+3=2,\begin{vmatrix}-1 & -1\\ 3 & 1\end{vmatrix}=(-1)(1)-(-1)(3)=-1+3=2,​−13​−11​​=(−1)(1)−(−1)(3)=−1+3=2, ∣−1−193∣=(−1)(3)−(−1)(9)=−3+9=6.\begin{vmatrix}-1 & -1\\ 9 & 3\end{vmatrix}=(-1)(3)-(-1)(9)=-3+9=6.​−19​−13​​=(−1)(3)−(−1)(9)=−3+9=6.

So,

D=2+6=8.D=2+6=8.D=2+6=8.
  1. Compare with options

The determinant equals

8.8.8.

So the correct option is C.

  1. Comparison with stored answer

Stored correct answer: C

Our derived answer: C

They agree.

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