    Question

# Match the following : coloumn1coloumn2(A)The shortest distance between the lines(P)13x−33=y−8−1=z−31 and x+3−3=y+72=z−64(B)The distance of the point of intersection of the line(Q)√2109110x−23=y+14=z−212 and the plane x−y+z=5 from the point (−1,−5,−10) is(C) The length of the perpendicular drawn from the point (5,4,−1)(R)2 on the line x−12=y9=z5is(D) The distance between the points P and Q is d and the length of(S)3√30 the projection of PQ on the co−ordinate planes are d1, d2, d3 then d12+d22+kd2 when k is

A

A P, B R, C S, D Q

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B

A Q, B S, C P, D R

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C

A S, B P, C Q, D R

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D

A R, B Q, C R, D P

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Solution

## The correct option is C A S, B P, C Q, D R A→S, B→P, C→Q, D→R (A)¯r=¯a+t¯bshortest distance between these lines=∣∣[¯c−¯a ¯b ¯d]∣∣–––––––––––––¯r=¯c+s¯d∣∣¯b×¯d∣∣ (B) (2,-1,2) satisfies the plane find distance between (2,-1,2) and (-1,-5,-10) (C) Find d.r's of A(5,4,-1) and B (1+2k, 9k, 5k) ((2k-4)¯i + (9k -4) + (5k + 1) ¯k).(2¯i+9¯j+5¯k) =0 find k. (d) projection of PQ on xy- plane= √(x1−x2)2+(y1−y2)2 projection of PQ on xy- plane = √(y1−y2)2+(z1−z2)2 projection of PQ on xy- plane = √(x1−x2)2+(z1−z2)2  Suggest Corrections  1      Similar questions
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