If #P_1=xpbb(i)+ypbb(j)+zpbb(k)# is the position vector of a point in the plane, and #x,y,z# is a point in the plane, with position vector:
#P=xbb(i)+ybb(j)+zbb(k)#
Then #P-P_1# is a vector lying in the plane.
If the normal to the plane is #bb(n)= abb(i)+b bb(j)+cbb(k)#
We know the dot product of a vector in the plane and the normal to the plane is #0#
#P-P_1=(x-xp)bb(i)+(y-yp)bb(j)+(z-zp)bb(k)#
#bb(n)*(P-P_1)=0=ax-axp+by-byp+cz-czp=0#
#ax+by+cz=axp+byp+czp#
Given the equation of a plane in the form:
#ax+by+cz=d#
We can see:
#ax+by+cz=axp+byp+czp#
#ax+by+cz=d#
The purpose of this is so we can find the angle between the normals to the plane. This is the same as the angle between the planes:
Our equations are:
#10x+5y+8z=-7#
#9x+10y+z=-6#
The normals are just:
#bb(n_1)=10bb(i)+5bb(j)+8bb(k)#
#bb(n_2)=9bb(i)+10bb(j)+bb(k)#
Using the dot product:
#a*b=||a||*||b||cos(theta)#
#bb(n_1)*bb(n_2)=(10*9+5*10+8*1)=148#
#||bb(n_1)||=sqrt((10)^2+(5)^2+(8)^2)=sqrt(189)=3sqrt(21)#
#||bb(n_2)||=sqrt((9)^2+(10)^2+(1)^2)=sqrt(182)#
#cos(theta)=148/(3sqrt(21)*sqrt(182))#
#theta=arccos(cos(theta))=arccos(148/(3sqrt(21)*sqrt(182)))=0.64# radians
This is the acute angle.
obtuse angle is:
#pi-0.647# radians
