At exactly what point perform some curves r1(t) = t, 5 − t, 63 + t2 and r2(s)

= 9 − s, s − 4, s2 intersect? At exactly what point perform some curves r1(t) (t, 5-t, 63 t2) and r20s) (9 s, S 4, s2) in (x, y, z) Get a hold of their particular perspective of intersection, 0, proper to your closest level.

Both curves r1 and r2 come in the 3

dimentional (x,y,z) jet

All Curves cannot satisfy. To confirm whether or not they satisfy at an original

point, we have to equate the x, y and z aspects of a spot

Equating the x aspects of things for the curves

t=9-s s+t

=9………….(1)

Equating the y aspects of the things both for curves

5-t = s-4 s+t

=9 …………..(2) (the exact same appearance formerly gotten)

Equating the z elements

63+ t2 =

s2………………………….(3)

From equation (2) we are able to

compose s = 9-t

placing the worthiness of s in

equation (3), we have 63+ t2 =81 -18t +

t2

Cancelling the

t2 on both edges associated with preceding equation, we have 63 =

81-18t

Thus the worthiness of t is available is 1

From equation (1), the worthiness of s = 8 whenever t = 1

Through the use of these values of t and s, we have the point of

intersection is (1,4,64)

To obtain the perspective involving the curves, we must discover dot

services and products of pitch associated with tangent. Thus we are able to get a hold of cosine of perspective

among them.

The differenciation of r1 wrt t and r2 wrt

s provides <1,-1,2t> and <-1,1,2s> as element of

mountains associated with tangent

<1,-1,2> and <-1,1,16> correspondingly

Dot item of 2 amounts u and v is u.v =|u||v cos =

1*(-1) + -1*1+ 2*16 = 30

Thus cos = u.v/|u||v| =

0.7624

Cos-1 (0.7624) = 40.3

Thus the perspective of intersection proper to closest level is

40