Answer:
Step-by-step explanation:
A tangent vector to the curve of intersection is given by N1 × N2
Where N1 is normal to the graph of
G(x, y, z) = z= x² + y² at the point P(-1,1,2) and
N2 is normal to the level surface
F(x, y, z)=4x² + 3y²+ 7z²= 35 at the point P(-1,1,2).
Now,
G(x, y, z)=x²+y²-z
Then, taking the grad of G will give a vector perpendicular to the paraboloid
∇G = ∂G/∂x •i + ∂G/∂y •j + ∂G/∂z •k
∇G= 2x•i +2y•j -k
At the point (-1,1,2)
N1=(-2,2,2)
Also to get N2
F(x, y, z)=4x² + 3y²+ 7z² - 35
∇F = ∂F/∂x •i + ∂F/∂y •j + ∂F/∂z •k
∇F = 8x•i + 6y•j + 14z•k
At point (-1,1,2)
N2=(-8,6,28)
Computing N1×N2
Note
i×i=j×j=k×k=0
i×j=k, j×i=-k
j×k=i, k×j=-i
k×i=j, i×k=-j
Then,
a×b= (a•i + b•j + c•k) × (x•i + y•j + z•k)
a×b = a•i×(x•i + y•j + z•k) + b•j×(x•i + y•j + z•k) + c•k×(x•i + y•j + z•k)
a×b= (a•i × x•i)+ (a•i × y•j) + (a•i × z•k) + (b•j × x•i) + (b•j × y•j) + (b•j × z•k) + (c•k × x•i) + (c•k × y•j) + (c•k × z•k)
a×b= 0 + ay•k - az•j - bx•k + 0 + bz•i + cx•j - cy•i + 0
a×b= ay•k - az•j - bx•k + bz•i + cx•j -cy•i
Then, rearranging
a×b= (bz - cy)•i+ (cx - az)•j + (ay-bx)•k
Now let assume that
N1=(-2,2,2) a=-2, b=2 and c=2
N2=(-8,6,28) x=-8, y=6 and z=28
N1×N2=(56-12)•i+(-16+56)•j+(-12+16)•k
N1×N2=44i + 40j + 4k
We compute N1 × N2 = (44,40,4) which is tangent to the curve of intersection at the point P.
The equation of the line is given as
x=r + λt
Where r is the point (-1,1,2)
And λ is the direction (44,40,4)
Hence the tangent line is given by
x = 1 + 44t, y = −1 + 40t, z = 2 + 4t.
