3D Geometry 2 Question 1

1. The distance of the point having position vector i^+2j^+6k^ from the straight line passing through the point (2,3,4) and parallel to the vector, 6i^+3j^4k^ is

(a) 213

(b) 43

(c) 6

(d) 7

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Answer:

Correct Answer: 1. (d)

Solution:

  1. Let point P whose position vector is (i^+2j^+6k^) and a straight line passing through Q(2,3,4) parallel to the vector n=6i^+3j^4k^.

Required distance d= Projection of line segment PQ perpendicular to vector n.

=|PQ×n||n|

Now, PQ=3i^+j^10k^, so

PQ×n=|i^j^k^3110634|=26i^48j^+3k^

 So, d=(26)2+(48)2+(3)2(6)2+(3)2+(4)2

=676+2304+936+9+16=298961=49=7 units 



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