Question 61902: Hi you guys, i need help with this question for sat or sunday, thank you very much.
Find the distance from the point Q(-3,-3,-4) to the line which passes through the point P(3,3,-4) and is parallel to the line whose equation is:
(x,y,z) = (-7,1,-2) + t(-1,-2,2).
Answer by venugopalramana(3286)   (Show Source):
You can put this solution on YOUR website! SEE THE FOLLOWING EXAMPLE AND DO.IT IS EXACTLY SIMILAR PROBLEM.
IF STILL IN DIFFICULTY PLEASE COME BACK.
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Find the distance from the point Q(-3,-3,-8) to the line which passes through the point P(-7,-7,2) and is parallel to the line(L SAY) whose equation is:
(x,y,z) = (-3,-1,0) + t(2,2,-2).
Thank you!
EQN.OF LINE THROUGH P AND PARALLEL TO L IS
R=(-7,-7,2)+T(2,2,-2)=i(-7+2T)+j(-7+2T)+k(2-2T)
LET,R BE A POINT ON LINE L SUCH THAT QR IS PERPENDICULAR TO L FOR A
PARTICULAR T
VECTOR QR = R-Q = i(-7+2T+3)+j(-7+2T+3)+k(2-2T+8)
QR = i(-4+2T)+j(-4+2T)+k(10-2T)
IT IS PER PENDICULAR TO LINE L WHOSE DIRECTION VECTOR = D = 2i+2j-2k
HENCE THEIR DOT PRODUCT SHOULD EQUAL ZERO
D.QR = 2(-4+2T)+2(-4+2T)-2(10-2T)=0
12T = 36
T=3
HENCE QR = 2i+2j+4k
|QR|=SQRT(2^2+2^2+4^2)=SQRT(24)=2SQRT(6)
HENCE PERPENDICULAR DISTANCE = 2SQRT(6)
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