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.
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