Răspuns:
2 vectori sunt perpendiculari daca produsul lor scalar e 0.
pui conditia ca v1*v2-0
v1*v2=(3i-4j)[(1-m)i+(2m-3)j]=0
3(1-m)i²-4(2m-3)j²= i²=j²=1 ij=ji=0
3(1-m)i²-8m+12=
3-3m-8m+12=
-11m+15=0
-11m=-15
m=-15/(-11)=15/11
b)m= -1
v2=(1-(-1))i+(2*(-1)-3)j=
(1+1)i+(-2-3)j=
2i-5j
v1*v2=(3i-4j)(2i-5j)=6i²-4*(-5)j²=6+20=26 (A
dar v1*v2=lv1l*lv2l*cos<(v1*v2)
lv1l=√[3²+(-4)²]=√(9+16)=√25=5
lv2l=√(2²+5²)=√29
26=5*√29*cos(<v1*v2)
cos(v1;v2)=26/5√29
Explicație pas cu pas:
