To find the equation of the plane
Ï€ that passes through the points
(2,−3,0) and
(3,0,4), and is parallel to the vector
⟨2,3,−4⟩, we start with the general equation of a plane:
a(x−2)+b(y+3)+c(z−0)=0.Substituting the point
(3,0,4) into the plane equation gives:
a(1)+b(3)+c(4)=0Since the plane is parallel to the vector
⟨2,3,−4⟩, the plane's normal vector must be orthogonal to
⟨2,3,−4⟩. Thus,
2a+3b−4c=0To find
a,b,c such that both conditions are satisfied, compare ratios:
‌=‌=‌.Simplifying gives:
‌=‌=‌.From this, solve for proportional constants:
‌=‌=‌.Choosing
c=1, we find
a=8 and
b=−4.
The equation of the plane becomes:
8(x−2)−4(y+3)+z=0,or simplified:
8x−4y+z−28=0Next, find the line equation that passes through points
(1,2,0) and
(0,1,−2). The direction vector of the line is
⟨−1,−1,−2⟩.
The parametric equation of the line is:
‌=‌=‌=λ.Therefore, the line is:
P(x,y,z)=(−λ+1,−λ+2,−2λ).Substitute these into the plane equation:
8(−λ+1)−4(−λ+2)+(−2λ)−28=0.Simplifying:
‌−8λ+8+4λ−8−2λ−28=0‌−6λ=28‌‌⇒‌‌λ=−‌Substituting
λ=−‌ into the line equations gives the intersection point
P(a,b,c)=(‌,‌,‌)Finally,
a+b+2c is:
a+b+2c=‌+‌+‌=‌=31.