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3D Geometry question

2004 · Shift 0 · Q114
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3D Geometry question

2004 · Shift 0 · Q114

JEE MainMathematics3D GeometryMCQ+4 / −1
If the straight lines x=1+s,y=−3−λs,z=1+λsx=1+s,y=-3 - \lambda s,z = 1 + \lambda sx=1+s,y=−3−λs,z=1+λs and x=t2,y=1+t,z=2−t,x = {t \over 2},y = 1 + t,z = 2 - t,x=2t​,y=1+t,z=2−t, with parameters sss and ttt respectively, are co-planar, then λ\lambdaλ equals :
  1. A
    000
  2. B
    −1-1−1
  3. C
    −12- {1 \over 2}−21​
  4. D
    −2-2−2
View written solutionFree

Correct answer: D

  1. Write the lines in vector form

For the first line, x=1+s,y=−3−λs,z=1+λsx=1+s,\quad y=-3-\lambda s,\quad z=1+\lambda sx=1+s,y=−3−λs,z=1+λs So a point on it is A(1,−3,1)A(1,-3,1)A(1,−3,1) and its direction vector is d⃗1=(1,−λ,λ).\vec d_1=(1,-\lambda,\lambda).d1​=(1,−λ,λ).

For the second line, x=t2,y=1+t,z=2−tx=\frac t2,\quad y=1+t,\quad z=2-tx=2t​,y=1+t,z=2−t Taking t=0t=0t=0, a point on it is B(0,1,2)B(0,1,2)B(0,1,2) and its direction vector is d⃗2=(12,1,−1).\vec d_2=\left(\frac12,1,-1\right).d2​=(21​,1,−1). To avoid fractions, use the parallel direction vector d⃗2=(1,2,−2).\vec d_2=(1,2,-2).d2​=(1,2,−2).

  1. Condition for two lines to be coplanar

Two lines are coplanar iff the scalar triple product of:

  • the vector joining a point on one line to a point on the other, and
  • the two direction vectors

is zero.

Now, AB→=B−A=(0−1, 1−(−3), 2−1)=(−1,4,1).\overrightarrow{AB}=B-A=(0-1,\ 1-(-3),\ 2-1)=(-1,4,1).AB=B−A=(0−1, 1−(−3), 2−1)=(−1,4,1).

Thus the coplanarity condition is [AB→,d⃗1,d⃗2]=0.[\overrightarrow{AB},\vec d_1,\vec d_2]=0.[AB,d1​,d2​]=0.

So we compute

-1 & 4 & 1\\ 1 & -\lambda & \lambda\\ 1 & 2 & -2 \end{vmatrix}=0.$$ 3. **Evaluate the determinant** Expanding along the first row:

-1\begin{vmatrix}-\lambda & \lambda\ 2 & -2\end{vmatrix} -4\begin{vmatrix}1 & \lambda\ 1 & -2\end{vmatrix} +1\begin{vmatrix}1 & -\lambda\ 1 & 2\end{vmatrix}=0.

Now compute each minor: $$\begin{vmatrix}-\lambda & \lambda\\ 2 & -2\end{vmatrix}=(-\lambda)(-2)-\lambda(2)=2\lambda-2\lambda=0,$$ $$\begin{vmatrix}1 & \lambda\\ 1 & -2\end{vmatrix}=1(-2)-\lambda(1)=-2-\lambda,$$ $$\begin{vmatrix}1 & -\lambda\\ 1 & 2\end{vmatrix}=1\cdot 2-(-\lambda)(1)=2+\lambda.$$ Therefore, $$0-4(-2-\lambda)+(2+\lambda)=0.$$ $$8+4\lambda+2+\lambda=0$$ $$10+5\lambda=0$$ $$\lambda=-2.$$ 4. **Check options** - A: $0$ ✗ - B: $-1$ ✗ - C: $-\frac12$ ✗ - D: $-2$ ✓ Hence the correct answer is: $$\boxed{-2}$$
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