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3D Geometry question

2013 · Shift 2 · Q23
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3D Geometry question

2013 · Shift 2 · Q23

JEE AdvancedMathematics3D GeometryMCQ+4 / −1
Consider the lines L1:x−12=y−1=z+31,L2:x−41=y+31=z+32{L_1}:{{x - 1} \over 2} = {y \over { - 1}} = {{z + 3} \over 1},{L_2} : {{x - 4} \over 1} = {{y + 3} \over 1} = {{z + 3} \over 2}L1​:2x−1​=−1y​=1z+3​,L2​:1x−4​=1y+3​=2z+3​ and the planes P1:7x+y+2z=3,P2=3x+5y−6z=4.{P_1}:7x + y + 2z = 3,{P_2} = 3x + 5y - 6z = 4.P1​:7x+y+2z=3,P2​=3x+5y−6z=4. Let ax+by+cz=dax+by+cz=dax+by+cz=d be the equation of the plane passing through the point of intersection of lines L1{L_1}L1​ and L2,{L_2},L2​, and perpendicular to planes P1{P_1}P1​ and P2.{P_2}.P2​. Match List III with List IIIIII and select the correct answer using the code given below the lists: List III(P.) a=a=a=(Q.) b=b=b=(R.) c=c=c=(S.) d=d=d= List IIIIII(1.) 131313(2.) −3-3−3(3.) 111(4.) −2-2−2
  1. A
    P=3,Q=2,R=4,S=1P = 3,Q = 2,R = 4,S = 1P=3,Q=2,R=4,S=1
  2. B
    P=1,Q=3,R=4,S=2P = 1,Q = 3,R = 4,S = 2P=1,Q=3,R=4,S=2
  3. C
    P=3,Q=2,R=1,S=4P = 3,Q = 2,R = 1,S = 4P=3,Q=2,R=1,S=4
  4. D
    P=2,Q=4,R=1,S=3P = 2,Q = 4,R = 1,S = 3P=2,Q=4,R=1,S=3
View written solutionFree

Correct answer: A

  1. Write the lines in parametric form

For L1L_1L1​: x−12=y−1=z+31=t\frac{x-1}{2}=\frac{y}{-1}=\frac{z+3}{1}=t2x−1​=−1y​=1z+3​=t So, x=1+2t,y=−t,z=t−3x=1+2t,\quad y=-t,\quad z=t-3x=1+2t,y=−t,z=t−3

For L2L_2L2​: x−41=y+31=z+32=s\frac{x-4}{1}=\frac{y+3}{1}=\frac{z+3}{2}=s1x−4​=1y+3​=2z+3​=s So, x=4+s,y=s−3,z=2s−3x=4+s,\quad y=s-3,\quad z=2s-3x=4+s,y=s−3,z=2s−3

  1. Find the point of intersection of L1L_1L1​ and L2L_2L2​

At intersection, 1+2t=4+s...(1)1+2t=4+s \quad ...(1)1+2t=4+s...(1) −t=s−3...(2)-t=s-3 \quad ...(2)−t=s−3...(2) t−3=2s−3⇒t=2s...(3)t-3=2s-3 \Rightarrow t=2s \quad ...(3)t−3=2s−3⇒t=2s...(3)

From (2): s=3−ts=3-ts=3−t Using in (3): t=2(3−t)t=2(3-t)t=2(3−t) t=6−2tt=6-2tt=6−2t 3t=63t=63t=6 t=2t=2t=2 Then, s=3−2=1s=3-2=1s=3−2=1

Hence intersection point is (1+2(2),−2,2−3)=(5,−2,−1)\left(1+2(2),-2,2-3\right)=(5,-2,-1)(1+2(2),−2,2−3)=(5,−2,−1)

  1. Find the normal vector of the required plane

The required plane is perpendicular to both planes: P1:7x+y+2z=3P_1:7x+y+2z=3P1​:7x+y+2z=3 P2:3x+5y−6z=4P_2:3x+5y-6z=4P2​:3x+5y−6z=4

So its normal vector must be perpendicular to both normals: n⃗1=(7,1,2),n⃗2=(3,5,−6)\vec n_1=(7,1,2),\quad \vec n_2=(3,5,-6)n1​=(7,1,2),n2​=(3,5,−6)

Thus required normal is parallel to n⃗1×n⃗2\vec n_1\times \vec n_2n1​×n2​

Compute:

\begin{vmatrix} \hat i & \hat j & \hat k\\ 7 & 1 & 2\\ 3 & 5 & -6 \end{vmatrix}$$ $$=\hat i(1\cdot(-6)-2\cdot5)-\hat j(7\cdot(-6)-2\cdot3)+\hat k(7\cdot5-1\cdot3)$$ $$=\hat i(-6-10)-\hat j(-42-6)+\hat k(35-3)$$ $$=(-16,48,32)$$ Divide by $16$: $$(-1,3,2)$$ So we can take $$a=-1,\quad b=3,\quad c=2$$ --- 4. **Find the plane through $(5,-2,-1)$** Equation: $$-1(x-5)+3(y+2)+2(z+1)=0$$ $$-x+5+3y+6+2z+2=0$$ $$-x+3y+2z+13=0$$ So, $$x-3y-2z=13$$ Hence, $$a=1,\quad b=-3,\quad c=-2,\quad d=13$$ --- 5. **Match with List II** List II: 1. $13$ 2. $-3$ 3. $1$ 4. $-2$ Therefore, - $a=1 \Rightarrow P=3$ - $b=-3 \Rightarrow Q=2$ - $c=-2 \Rightarrow R=4$ - $d=13 \Rightarrow S=1$ So the correct matching is: $$P=3,\ Q=2,\ R=4,\ S=1$$ This corresponds to **Option A**. --- 6. **Comparison with stored answer** Derived answer: **Option A** Stored correct answer: **A** They agree.
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