Question #64850

2 Answers
Aug 25, 2016

P = {3, 4, 2}P={3,4,2}

For QQ there are two feasible solutions

Q_1 = {2,2,0}Q1={2,2,0}
Q_2 = {4,6,4}Q2={4,6,4}

Explanation:

Let L->p_0+lambda vec vLp0+λv where

p_0 = {1,0,-2}p0={1,0,2} and vec v = {1,2,2}v={1,2,2}

If the segment bar(AP)¯¯¯¯¯¯AP is orthogonal to LL then norm(A-P)AP represents the minimum distance between AA and LL. The point PP 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.!.