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MHT CET202019 Oct 2020Morning ShiftMathematicsThree Dimensional GeometryActual

The equation of the line passing through the point (1,2,3) and perpendicular to the lines x-1 1 = y-2 2 = z-3 3 and r = (-3 +2 +5 k ) is

Options

  1. Ar =( +2 +3 k )+ (2 +7 -4 k )
  2. Br =( +2 +3 k )+ (2 +7 +4 k )
  3. Cr =( +2 +3 k )+ (2 -7 -4 k )
  4. Dr =( +2 +3 k )+ (2 -7 +4 k )

Correct answer

D. r =( +2 +3 k )+ (2 -7 +4 k )

Step-by-step solution

(D) The vector perpendicular to both the given lines is given by | array lll i & j & k 1 & 2 & 4 2 & 2 & 1 array |= i (-6)- j (-7)+ k (-2)=-6 i +7 j -2 k Hence d.r.s. of required line are 6,-7,2 . Thus eq. of required line is x-1 6 = y-2 -7 = 2-3 2 i.e. x-1 6 = 2-y 7 = z-3 2

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