Question #64850

2 Answers
Aug 25, 2016

#P = {3, 4, 2}#

For #Q# there are two feasible solutions

#Q_1 = {2,2,0}#
#Q_2 = {4,6,4}#

Explanation:

Let #L->p_0+lambda vec v# where

#p_0 = {1,0,-2}# and #vec v = {1,2,2}#

If the segment #bar(AP)# is orthogonal to #L# then #norm(A-P)# represents the minimum distance between #A# and #L#. The point #P# of minimum distance is obtained by minimizing

#d = min_{lambda}norm(A-L(lambda)) = min_{lambda}norm(A-p_0-lambda vec v)#

but

#d^2(lambda) = norm(A-p_0)^2-2 lambda << A-p_0, vec v >> + lambda^2 norm(vec v)^2#

Its minimum is located at a #lambda# such that

#d/(d lambda) d^2(lambda) = -2 << A-p_0, vec v >> + 2 lambda norm(vec v)^2 = 0# for

#lambda_P = << A-p_0, vec v >>/norm(vec v)^2 = 2# so #P# is located at

#P = p_0 + lambda_P vec v = {3, 4, 2}#.

Once obtained #P# the next point, #Q# obeys the relationship

#norm(A-P)=norm(P-Q)# but #Q in L# so

#norm(A-P) = norm(P-p_0-lambda vec v)#

Squaring both sides

#norm(A-P)^2 = norm(P-p_0)^2-2lambda << P-p_0, vec v >> + lambda^2 norm(vec v)^2#.

Solving for #lambda# we have two outcomes:

#lambda_Q = {1, 3}#

So

#Q_1 = p_0 + lambda_{Q_1} vec v = {2,2,0}#
#Q_2 = p_0 + lambda_{Q_2} vec v ={4,6,4}#

Aug 25, 2016

The Reqd. Pts. are #P(3,4,2), Q_1(4,6,4), and, Q_2(2,2,0)#.

Explanation:

Reqd. pt. #P in L={(1,0,-2)+lambda(1,2,2) | lambda in RR}#

#rArr EE t in RR, s.t., P=P(1+t,2t,2t-2)#

Now, #A(1,3,4), P(1+t,2t,2t-2) rArr vec(AP)=(t,2t-3,2t-6)#.

Since, #vec(AP) bot L rArr vec(AP) bot "the direction vector of" L#

#rArr vec(AP) bot (1,2,2) rArr vec(AP).(1,2,2)=0#

#:. (t,2t-3,2t-6).(1,2,2)=0. :. t+4t-6+4t-12=0#

#:. t=2 rArr P(1+t,2t,2t-2)=P(3,4,2)#.

As regards #Q in L#, like #P#, #Q# has to be #Q(1+s,2s,2s-2)# for

some #s in RR#

Because in right-isosceles #DeltaAPQ, m/_APQ=90^@, AP=PQ#.

With #A(1,3,4), P(3,4,2), Q(1+s,2s,2s-2), and, AP^2=PQ^2#, we have,

#(1-3)^2+(3-4)^2+(4-2)^2=(1+s-3)^2+(2s-4)^2+(2s-2-2)^2#

#:.4+1+4=(s-2)^2+4(s-2)^2+4(s-2)^2#.

#:. 9(s-2)^2=9 rArr s-2=+-1 rArr s=3, or, s=1#.

Accordingly, we have #2# pts. #Q_1, Q_2 in L# satisfying the given cond., namely,

for #s=3, Q_1(1+s,2s,2s-2)=Q_1(4,6,4), &, "for" s=1, Q_2(2,2,0)#.

Thus, the Reqd. Pts. are #P(3,4,2), Q_1(4,6,4), and, Q_2(2,2,0)#, as

obtained by Respected Cesareo R.

Enjoy Maths.!.