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3D Geometry question

2003 · Shift 0 · Q92
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3D Geometry question

2003 · Shift 0 · Q92

JEE MainMathematics3D GeometryMCQ+4 / −1
The two lines x=ay+b,z=cy+dx=ay+b,z=cy+dx=ay+b,z=cy+d and x=a′y+b′,z=c′y+d′x = a'y + b',z = c'y + d'x=a′y+b′,z=c′y+d′ will be perpendicular, if and only if :
  1. A
    aa′+cc′+1=0aa' + cc' + 1 = 0aa′+cc′+1=0
  2. B
    aa′+bb′cc′+1=0aa' + bb'cc' + 1 = 0aa′+bb′cc′+1=0
  3. C
    aa′+bb′cc′=0aa' + bb'cc' = 0aa′+bb′cc′=0
  4. D
    (a+a′)(b+b′)+(c+c′)=0\left( {a + a'} \right)\left( {b + b'} \right) + \left( {c + c'} \right) = 0(a+a′)(b+b′)+(c+c′)=0
View written solutionFree

Correct answer: A

  1. Write each line in parametric/vector form

Given x=ay+b,z=cy+dx=ay+b,\quad z=cy+dx=ay+b,z=cy+d and yyy is the free parameter, let y=ty=ty=t. Then the first line becomes x=at+b,y=t,z=ct+d.x=at+b,\quad y=t,\quad z=ct+d.x=at+b,y=t,z=ct+d. So its direction ratios are d⃗1=(a,1,c).\vec{d}_1=(a,1,c).d1​=(a,1,c).

Similarly, for the second line x=a′y+b′,z=c′y+d′,x=a'y+b',\quad z=c'y+d',x=a′y+b′,z=c′y+d′, put y=sy=sy=s. Then x=a′s+b′,y=s,z=c′s+d′.x=a's+b',\quad y=s,\quad z=c's+d'.x=a′s+b′,y=s,z=c′s+d′. So its direction ratios are d⃗2=(a′,1,c′).\vec{d}_2=(a',1,c').d2​=(a′,1,c′).

  1. Condition for perpendicular lines

Two lines are perpendicular if their direction vectors are perpendicular, i.e. their dot product is zero: d⃗1⋅d⃗2=0.\vec{d}_1\cdot \vec{d}_2=0.d1​⋅d2​=0.

Compute: (a,1,c)⋅(a′,1,c′)=aa′+1⋅1+cc′.(a,1,c)\cdot (a',1,c')=aa'+1\cdot 1+cc'.(a,1,c)⋅(a′,1,c′)=aa′+1⋅1+cc′. Thus, aa′+1+cc′=0.aa'+1+cc'=0.aa′+1+cc′=0. So the required condition is aa′+cc′+1=0.aa'+cc'+1=0.aa′+cc′+1=0.

  1. Match with the options

This is exactly Option A.

  1. Comparison with stored answer

Stored correct answer: A

Our derived answer: A

They agree.

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