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3D Geometry question

2013 · Shift 2 · Q24
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3D Geometry question

2013 · Shift 2 · Q24

JEE AdvancedMathematics3D GeometryMultiple correct+4 / −1
Two lines L1:x=5,y3−α=z−2{L_1}:x = 5,{y \over {3 - \alpha }} = {z \over { - 2}}L1​:x=5,3−αy​=−2z​ and L2:x=α,y−1=z2−α{L_2}:x = \alpha ,{y \over { - 1}} = {z \over {2 - \alpha }}L2​:x=α,−1y​=2−αz​ are coplanar. Then α\alphaα can take value(s)
  1. A
    111
  2. B
    222
  3. C
    333
  4. D
    444
View written solutionFree

Correct answer: D, A

  1. Interpret the symmetric forms of the lines

Given L1:x=5,y3−α=z−2L_1: x=5,\quad \frac{y}{3-\alpha}=\frac{z}{-2}L1​:x=5,3−αy​=−2z​ and L2:x=α,y−1=z2−α.L_2: x=\alpha,\quad \frac{y}{-1}=\frac{z}{2-\alpha}.L2​:x=α,−1y​=2−αz​.

Let the common parameter on each line be used separately.

For L1L_1L1​, let y3−α=z−2=t.\frac{y}{3-\alpha}=\frac{z}{-2}=t.3−αy​=−2z​=t. Then x=5,y=(3−α)t,z=−2t.x=5,\quad y=(3-\alpha)t,\quad z=-2t.x=5,y=(3−α)t,z=−2t. So a point on L1L_1L1​ is A=(5,0,0),A=(5,0,0),A=(5,0,0), and a direction vector is d⃗1=(0,3−α,−2).\vec d_1=(0,3-\alpha,-2).d1​=(0,3−α,−2).

For L2L_2L2​, let y−1=z2−α=s.\frac{y}{-1}=\frac{z}{2-\alpha}=s.−1y​=2−αz​=s. Then x=α,y=−s,z=(2−α)s.x=\alpha,\quad y=-s,\quad z=(2-\alpha)s.x=α,y=−s,z=(2−α)s. So a point on L2L_2L2​ is B=(α,0,0),B=(\alpha,0,0),B=(α,0,0), and a direction vector is d⃗2=(0,−1,2−α).\vec d_2=(0,-1,2-\alpha).d2​=(0,−1,2−α).

  1. Condition for two lines to be coplanar

Two lines with points A,BA,BA,B and direction vectors d⃗1,d⃗2\vec d_1,\vec d_2d1​,d2​ are coplanar iff [AB→,d⃗1,d⃗2]=0,[\overrightarrow{AB},\vec d_1,\vec d_2]=0,[AB,d1​,d2​]=0, where the scalar triple product is zero.

Here, AB→=B−A=(α−5,0,0).\overrightarrow{AB}=B-A=(\alpha-5,0,0).AB=B−A=(α−5,0,0). So we need

\alpha-5 & 0 & 0\\ 0 & 3-\alpha & -2\\ 0 & -1 & 2-\alpha \end{pmatrix}=0.$$ Since only the first entry of the first row is nonzero, $$ (\alpha-5)\det\begin{pmatrix}3-\alpha & -2\\ -1 & 2-\alpha\end{pmatrix}=0.$$ Now compute the $2\times 2$ determinant: $$ (3-\alpha)(2-\alpha)-(-2)(-1)=0.$$ So $$ (\alpha-5)\left((3-\alpha)(2-\alpha)-2\right)=0.$$ Expand: $$ (3-\alpha)(2-\alpha)=6-5\alpha+\alpha^2.$$ Hence $$6-5\alpha+\alpha^2-2=\alpha^2-5\alpha+4=(\alpha-1)(\alpha-4).$$ Therefore, $$ (\alpha-5)(\alpha-1)(\alpha-4)=0.$$ So the possible values are $$\alpha=5,\;1,\;4.$$ --- 3. **Match with the given options** Options are: - A: $1$ - B: $2$ - C: $3$ - D: $4$ Among the options, the valid values are $$\alpha=1 \quad \text{and} \quad \alpha=4.$$ So the correct options are **A and D**. --- 4. **Compare with stored correct answer** Stored correct answer: **D, A** This matches our result (**A, D**).
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