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Let d be the distance between the foot of perpendiculars of the points P(1, 2, -1) and Q(2, -1, 3) on the plane -x + y + z = 1. Then d2 is equal to…………..

0 2 Views | Posted 2 months ago
Asked by Shiksha User

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    Answered by

    alok kumar singh | Contributor-Level 10

    2 months ago

    Let π≡−x+y+z−1=0

    1 2 > 0 as both pt.lies on same side

    now P1=|−1+2−1−13|=13

    P2 = |−2+ (−1)+3−13|=13

    as P1 = P2 so distance between foot of perpendicular will be same as distance between the points

    d =  (1−2)2+ (2+1)2+ (−1−3)2

    = 1+9+16=

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