• vu6ng g6c vdi mp (R) ntn giao : tuye'n ( A ) ciia chung phai
vu6ng goc vdi (R). Do do, . vecto chi phuong u cua ( A ) la
^ tich CO hu6ng ciia hai vecto phap tuy^n ciia (P) va (Q).
= (10; 15; 5) = 5(2; 3; 1)
No cung la vecto phap tuyen ciia mp (R) nen PT mp (R) la: 2x + 3y + z = G
Hinh 105
c - 1 1 1 1 1 - 1
2 -12 5 -12 3 3 2
133. 1 ) ^ 5 = ( 2 ; 0 ; 2 ) = 2(1;0;1)
Dudng thang AB c6 vecto chi phuong M= (1; 0; 1) di qua diem A(0; 0; - 3 ) , nfen AB c6 phuong trinh:
[ x = 0 + t y = 0 + Ot z = - 3 + t '
I r r
The'vaopt ciia (P)tac6:
3 t - 8 ( 0 + 0.t) + 7 ( - 3 + t ) - l = 0 lO.t = 22 o t = —
5
1 = 11 4^
2) Gia sir tim ducfc C(x, y, z) G (P) vfiy:
3 x - 8 y + 7 z - l = 0
AB' = 8, BC' = (x-2)' + y' + (z+1)' AC' = x' + y' + (z +3)'
Vi A ABC diu nen AB' = AC' = BC' ta c6 h6 pt sau:
3 x - 8 y + 7 z - l = 0 (1) x' + y ' + z ' +6z = - l (2) + y2 + z' - 4x + 2z = 3 (3) (2)-(3) 3x - 8y + 7x - 1 = 0 (1) x + z = - l (2) x' + y' + z' + 6z - 1 = 0 (3) The z = -1-x vao (1)', (3)' ta c6 he Pt. -4x - 8y - 8 = 0 (1)" ' 2 x ' + y ' - 4 x - 4 = 0 (2)"
Rut X tiir (1)" ta CO X = -2y - 2. The vao (2)" ta duac:
2 (-2y -2f + f - 4(-2y - 2) - 4 = 0
9y'+ 24y +12 = 0 « 3y' + 8y +4 = 0 y = - 2
2 y = - -
The' vao (1)" ta tim duac x. V6i x tim duac the vao (2)' ta tim duac z.
Cu6'i cung ta dugc nghiern cua he pt la:
z = — 3
Vay CO hai diem C tren (P) de A ABC deu la: C ( 2 ; - 2 ; - 3 ) ; C ( - | ; - | ; - l )
134. Trong hinh hop chii nhat cac du5ng cheo deu bang nhau, nen
ta CO the lay dudng cheo BD'.
Theo tinh chat hinh h6p thi D'D ± day nen BD la hinh chieu cua