MATH SOLVE

2 months ago

Q:
# Let L be the line passing through the point P(−1, 1, 4) with direction vector d=[−1, 1, −2]T. Find the shortest distance d from the point P0(5, 3, 2) to L, and the point Q on L that is closest to P0. Use the square root symbol '√' where needed to give an exact value for your answer.

Accepted Solution

A:

Answer:the minimum distance is d=2√6 and the point Q = (7,1,6)Step-by-step explanation:Lets make a vector QP , that will be perpendicular to QPoQPo = (x-5, y-3, z-2)QP = (x+1, y-1, z-4)since QPo represents the minimum distance the scalar product respect with QP should be 0QP*QPo = 0(x-5)*(x+1)+ (y-3)(y-1)+ (z-2)(z-4) = 0also knowing that Q passes through the line is parallel to d=[-1,1,-2] and goes through PQ (x,y,z) = P(−1, 1, 4) + d(-1,1,-2) *t x = -1 -t , y=1+t , z= 4 -2treplacing values(-t)*(-6-t) + (t)(t-2)+ (6+2t)(-2t)=0(-t)*(-6-t) + (t)(t-2)+ (6+2t)(-2t)=0t [(6+t) +(t+2) -2(6+2t)]=0since t≠0 because Q≠P(6+t) +(t+2) -2(6+2t) =06+t+t+2-12-4t=0-2t-4=0t=4/(-2) = -2thereforeQ (x,y,z) = Po(5,3,2) + d(-1,1,-2) *(-2) = (7,1,6)regarding the minimum distance, this is the modulus of the vector QPod= √[(x-5)²,+(y-3)²+(z-2)²] = √[(7-5)²,+(1-3)²+(6-2)²] =√(4+4+16)=√24=2√6