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3D Geometry question

2003 · Shift 0 · Q93
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3D Geometry question

2003 · Shift 0 · Q93

JEE MainMathematics3D GeometryMCQ+4 / −1
The lines x−21=y−31=z−4−k{{x - 2} \over 1} = {{y - 3} \over 1} = {{z - 4} \over { - k}}1x−2​=1y−3​=−kz−4​ and x−1k=y−42=z−51{{x - 1} \over k} = {{y - 4} \over 2} = {{z - 5} \over 1}kx−1​=2y−4​=1z−5​ are coplanar if :
  1. A
    k=3k=3k=3 or −2-2−2
  2. B
    k=0k=0k=0 or −1-1−1
  3. C
    k=1k=1k=1 or −1-1−1
  4. D
    k=0k=0k=0 or −3-3−3
View written solutionFree

Correct answer: D

  1. Write the lines in parametric form

For the first line, x−21=y−31=z−4−k=t\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}=t1x−2​=1y−3​=−kz−4​=t so x=2+t,y=3+t,z=4−kt.x=2+t,\quad y=3+t,\quad z=4-kt.x=2+t,y=3+t,z=4−kt.

Hence, a point on line L1L_1L1​ is A(2,3,4)A(2,3,4)A(2,3,4) and its direction vector is d⃗1=(1,1,−k).\vec d_1=(1,1,-k).d1​=(1,1,−k).

For the second line, x−1k=y−42=z−51=s\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}=skx−1​=2y−4​=1z−5​=s so x=1+ks,y=4+2s,z=5+s.x=1+ks,\quad y=4+2s,\quad z=5+s.x=1+ks,y=4+2s,z=5+s.

Hence, a point on line L2L_2L2​ is B(1,4,5)B(1,4,5)B(1,4,5) and its direction vector is d⃗2=(k,2,1).\vec d_2=(k,2,1).d2​=(k,2,1).

  1. Condition for two lines in space to be coplanar

Two lines are coplanar iff the scalar triple product of:

  • one direction vector of the first line,
  • one direction vector of the second line,
  • the vector joining a point on one line to a point on the other line,

is zero.

So we need [d⃗1,d⃗2,AB→]=0.[\vec d_1,\vec d_2,\overrightarrow{AB}]=0.[d1​,d2​,AB]=0.

Now, AB→=B−A=(1−2, 4−3, 5−4)=(−1,1,1).\overrightarrow{AB}=B-A=(1-2,\,4-3,\,5-4)=(-1,1,1).AB=B−A=(1−2,4−3,5−4)=(−1,1,1).

Thus,

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

\begin{vmatrix} 1 & 1 & -k\ k & 2 & 1\ -1 & 1 & 1 \end{vmatrix} =1\begin{vmatrix}2 & 1\1 & 1\end{vmatrix} -1\begin{vmatrix}k & 1\-1 & 1\end{vmatrix} +(-k)\begin{vmatrix}k & 2\-1 & 1\end{vmatrix}.

Now, $$\begin{vmatrix}2 & 1\\1 & 1\end{vmatrix}=2\cdot 1-1\cdot 1=1,$$ $$\begin{vmatrix}k & 1\\-1 & 1\end{vmatrix}=k\cdot 1-1(-1)=k+1,$$ $$\begin{vmatrix}k & 2\\-1 & 1\end{vmatrix}=k\cdot 1-2(-1)=k+2.$$ So, $$1-(k+1)-k(k+2)=0.$$ Simplify: $$1-k-1-k^2-2k=0$$ $$-k^2-3k=0$$ $$k^2+3k=0$$ $$k(k+3)=0.$$ Therefore, $$k=0 \quad \text{or} \quad k=-3.$$ --- 4. **Match with the options** This corresponds to: **Option D: $k=0$ or $-3$** --- 5. **Comparison with stored answer** Stored correct answer: **D** Our derived answer: **D** So the derived answer agrees with the stored answer.
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