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3D Geometry question

2006 · Shift 0 · Q66
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3D Geometry question

2006 · Shift 0 · Q66

JEE MainMathematics3D GeometryMCQ+4 / −1
The two lines x=ay+b,z=cy+d;x=ay+b, z=cy+d;x=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′ are perpendicular to each other if :
  1. A
    aa′+cc′=−1aa'+cc'=-1aa′+cc′=−1
  2. B
    aa′+cc′=1aa'+cc'=1aa′+cc′=1
  3. C
    aa′+cc′=−1{a \over {a'}} + {c \over {c'}} = - 1a′a​+c′c​=−1
  4. D
    aa′+cc′=1{a \over {a'}} + {c \over {c'}} = 1a′a​+c′c​=1
View written solutionFree

Correct answer: A

  1. Write each line in symmetric/vector form

The lines are given as x=ay+b,z=cy+dx=ay+b,\quad z=cy+dx=ay+b,z=cy+d and 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′.

Here, yyy is the parameter for both lines.

So the first line can be written as x=b+ay,y=y,z=d+cy.x=b+ay,\quad y=y,\quad z=d+cy.x=b+ay,y=y,z=d+cy. Hence its direction ratios are d⃗1=(a,1,c).\vec{d}_1=(a,1,c).d1​=(a,1,c).

Similarly, the second line is x=b′+a′y,y=y,z=d′+c′y,x=b'+a'y,\quad y=y,\quad z=d'+c'y,x=b′+a′y,y=y,z=d′+c′y, 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. d⃗1⋅d⃗2=0.\vec{d}_1\cdot \vec{d}_2=0.d1​⋅d2​=0.

So, (a,1,c)⋅(a′,1,c′)=0.(a,1,c)\cdot(a',1,c')=0.(a,1,c)⋅(a′,1,c′)=0.

  1. Compute the dot product

aa′+1⋅1+cc′=0aa' + 1\cdot 1 + cc' = 0aa′+1⋅1+cc′=0 aa′+cc′+1=0aa'+cc'+1=0aa′+cc′+1=0 aa′+cc′=−1.aa'+cc'=-1.aa′+cc′=−1.

  1. Match with the options

This is exactly Option A.

Therefore, the two lines are perpendicular if aa′+cc′=−1.aa'+cc'=-1.aa′+cc′=−1.

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