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3D Geometry question

2024 · Shift 1 · Q33
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  5. /2024 · Shift 1 · Q33

3D Geometry question

2024 · Shift 1 · Q33

JEE AdvancedMathematics3D GeometryMCQ+3 / −1

Let γ∈R\gamma \in \mathbb{R}γ∈R be such that the lines L1:x+111=y+212=z+293L_1: \frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}L1​:1x+11​=2y+21​=3z+29​ and L2:x+163=y+112=z+4γL_2: \frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}L2​:3x+16​=2y+11​=γz+4​ intersect. Let R1R_1R1​ be the point of intersection of L1L_1L1​ and L2L_2L2​. Let O=(0,0,0)O=(0,0,0)O=(0,0,0), and n^\hat{n}n^ denote a unit normal vector to the plane containing both the lines L1L_1L1​ and L2L_2L2​.

Match each entry in List-I to the correct entry in List-II.

List-I List-II
(P) γ\gammaγ equals (1) −i^−j^+k^-\hat{i} - \hat{j} + \hat{k}−i^−j^​+k^
(Q) A possible choice for n^\hat{n}n^ is (2) 32\sqrt{\frac{3}{2}}23​​
(R) OR1→\overrightarrow{OR_1}OR1​​ equals (3) 111
(S) A possible value of OR1→⋅n^\overrightarrow{OR_1} \cdot \hat{n}OR1​​⋅n^ is (4) 16i^−26j^+16k^\frac{1}{\sqrt{6}} \hat{i} - \frac{2}{\sqrt{6}} \hat{j} + \frac{1}{\sqrt{6}} \hat{k}6​1​i^−6​2​j^​+6​1​k^
(5) 23\sqrt{\frac{2}{3}}32​​

The correct option is :
  1. A
    (P)→(3)(Q)→(4)(R)→(1)(S)→(2)(\mathrm{P}) \rightarrow(3) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(2)(P)→(3)(Q)→(4)(R)→(1)(S)→(2)
  2. B
    (P)→(5)(Q)→(4)(R)→(1)(S)→(2)(\mathrm{P}) \rightarrow(5) \quad(\mathrm{Q}) \rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad(\mathrm{S}) \rightarrow(2)(P)→(5)(Q)→(4)(R)→(1)(S)→(2)
  3. C
    (P)→(3)(\mathrm{P}) \rightarrow(3) \quad(P)→(3)(Q) →(4)(R)→(1)\rightarrow(4) \quad(\mathrm{R}) \rightarrow(1) \quad→(4)(R)→(1)(S) →(5)\rightarrow(5)→(5)
  4. D
    (P)→(3)(Q)→(1)(R)→(4)(\mathrm{P}) \rightarrow(3) \quad(\mathrm{Q}) \rightarrow(1) \quad(\mathrm{R}) \rightarrow(4) \quad(P)→(3)(Q)→(1)(R)→(4)(S) →(5)\rightarrow(5)→(5)
View written solutionFree

Correct answer: C

  1. Write the lines in parametric form

For L1L_1L1​: x+111=y+212=z+293=t\frac{x+11}{1}=\frac{y+21}{2}=\frac{z+29}{3}=t1x+11​=2y+21​=3z+29​=t So, x=t−11,y=2t−21,z=3t−29x=t-11,\qquad y=2t-21,\qquad z=3t-29x=t−11,y=2t−21,z=3t−29 Hence a point on L1L_1L1​ is A=(−11,−21,−29)A=(-11,-21,-29)A=(−11,−21,−29) and direction vector is d⃗1=(1,2,3).\vec d_1=(1,2,3).d1​=(1,2,3).

For L2L_2L2​: x+163=y+112=z+4γ=s\frac{x+16}{3}=\frac{y+11}{2}=\frac{z+4}{\gamma}=s3x+16​=2y+11​=γz+4​=s So, x=3s−16,y=2s−11,z=γs−4x=3s-16,\qquad y=2s-11,\qquad z=\gamma s-4x=3s−16,y=2s−11,z=γs−4 Hence a point on L2L_2L2​ is B=(−16,−11,−4)B=(-16,-11,-4)B=(−16,−11,−4) and direction vector is d⃗2=(3,2,γ).\vec d_2=(3,2,\gamma).d2​=(3,2,γ).

  1. Use intersection condition to find γ\gammaγ

At intersection, coordinates must be equal for some t,st,st,s: t−11=3s−16t-11=3s-16t−11=3s−16 2t−21=2s−112t-21=2s-112t−21=2s−11 3t−29=γs−43t-29=\gamma s-43t−29=γs−4

From the first equation: t−3s=−5...(1)t-3s=-5 \quad ...(1)t−3s=−5...(1) From the second equation: 2t−2s=10⇒t−s=5...(2)2t-2s=10 \Rightarrow t-s=5 \quad ...(2)2t−2s=10⇒t−s=5...(2)

From (2)(2)(2), t=s+5t=s+5t=s+5. Substitute into (1)(1)(1): (s+5)−3s=−5(s+5)-3s=-5(s+5)−3s=−5 −2s=−10⇒s=5-2s=-10 \Rightarrow s=5−2s=−10⇒s=5 So, t=10.t=10.t=10.

Now use the third equation: 3(10)−29=γ(5)−43(10)-29=\gamma(5)-43(10)−29=γ(5)−4 30−29=5γ−430-29=5\gamma-430−29=5γ−4 1=5γ−41=5\gamma-41=5γ−4 5γ=55\gamma=55γ=5 γ=1.\gamma=1.γ=1.

Therefore, (P)→(3)\boxed{(P)\to (3)}(P)→(3)​

  1. Find the point of intersection R1R_1R1​

Using t=10t=10t=10 in L1L_1L1​: x=10−11=−1,y=2(10)−21=−1,z=3(10)−29=1x=10-11=-1,\qquad y=2(10)-21=-1,\qquad z=3(10)-29=1x=10−11=−1,y=2(10)−21=−1,z=3(10)−29=1 So, R1=(−1,−1,1).R_1=(-1,-1,1).R1​=(−1,−1,1). Thus, OR1→=−i^−j^+k^.\overrightarrow{OR_1}=-\hat i-\hat j+\hat k.OR1​​=−i^−j^​+k^.

Therefore, (R)→(1)\boxed{(R)\to (1)}(R)→(1)​

  1. Find a unit normal to the plane containing both lines

Since the lines intersect, the plane containing them has normal vector n⃗=d⃗1×d⃗2.\vec n=\vec d_1\times \vec d_2.n=d1​×d2​. Now γ=1\gamma=1γ=1, so d⃗2=(3,2,1).\vec d_2=(3,2,1).d2​=(3,2,1).

Compute:

\begin{vmatrix} \hat i & \hat j & \hat k\\ 1 & 2 & 3\\ 3 & 2 & 1 \end{vmatrix}$$ $$=\hat i(2\cdot1-3\cdot2)-\hat j(1\cdot1-3\cdot3)+\hat k(1\cdot2-2\cdot3)$$ $$=\hat i(2-6)-\hat j(1-9)+\hat k(2-6)$$ $$=-4\hat i+8\hat j-4\hat k$$ $$=-4(\hat i-2\hat j+\hat k).$$ So a normal vector is proportional to $$(-1,2,-1)$$ (or equivalently $(1,-2,1)$). Magnitude of $(1,-2,1)$ is $$\sqrt{1+4+1}=\sqrt6.$$ Hence a unit normal can be $$\hat n=\frac{1}{\sqrt6}\hat i-\frac{2}{\sqrt6}\hat j+\frac{1}{\sqrt6}\hat k.$$ Therefore, $$\boxed{(Q)\to (4)}$$ --- 5. **Find a possible value of $\overrightarrow{OR_1}\cdot \hat n$** We have $$\overrightarrow{OR_1}=(-1,-1,1), \qquad \hat n=\left(\frac1{\sqrt6},-\frac2{\sqrt6},\frac1{\sqrt6}\right).$$ Then $$\overrightarrow{OR_1}\cdot \hat n =(-1)\frac1{\sqrt6}+(-1)\left(-\frac2{\sqrt6}\right)+(1)\frac1{\sqrt6}$$ $$=\frac{-1+2+1}{\sqrt6}=\frac{2}{\sqrt6}=\sqrt{\frac23}.$$ Also, since the opposite unit normal is also possible, the value could be negative of this. Among the options, the listed possible value is $$\sqrt{\frac23}.$$ Therefore, $$\boxed{(S)\to (5)}$$ --- 6. **Final matching** $$ (P)\to(3),\quad (Q)\to(4),\quad (R)\to(1),\quad (S)\to(5). $$ This corresponds to **Option C**.
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