Question

Question #64850

Answer 1

`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}`

Answer 2

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.!.