Find an equation of the plane that contains the points p(5,−1,1),q(9,1,5),and r(8,−6,0)p(5,−1,1),q(9,1,5),and r(8,−6,0).
topjm [15]
Given plane passes through:
p(5,-1,1), q(9,1,5), r(8,-6,0)
We need to find a plane that is parallel to the plane through all three points, we form the vectors of any two sides of the triangle pqr:
pq=p-q=<5-9,-1-1,1-5>=<-4,-2,-4>
pr=p-r=<5-8,-1-6,1-0>=<-3,5,1>
The vector product pq x pr gives a vector perpendicular to both pq and pr. This vector is the normal vector of a plane passing through all three points
pq x pr
=
i j k
-4 -2 -4
-3 5 1
=<-2+20,12+4,-20-6>
=<18,16,-26>
Since the length of the normal vector does not change the direction, we simplify the normal vector as
N = <9,8,-13>
The required plane must pass through all three points.
We know that the normal vector is perpendicular to the plane through the three points, so we just need to make sure the plane passes through one of the three points, say q(9,1,5).
The equation of the required plane is therefore
Π : 9(x-9)+8(y-1)-13(z-5)=0
expand and simplify, we get the equation
Π : 9x+8y-13z=24
Check to see that the plane passes through all three points:
at p: 9(5)+8(-1)-13(1)=45-8-13=24
at q: 9(9)+8(1)-13(5)=81+9-65=24
at r: 9(8)+8(-6)-13(0)=72-48-0=24
So plane passes through all three points, as required.
<span>22.6666666667, in other words D. 22 2/3</span>
Answer:
Step-by-step explanation:
The general direct variation equation is
y = kx
Plug in x = 4 and y = 6 and solve for k.
6 = k * 4
4k = 6
k = 6/4 = 3/2
9514 1404 393
Answer:
±√58 ≈ ±7.616
Step-by-step explanation:
The linear combination of sine and cosine functions will have an amplitude that is the root of the sum of the squares of the individual amplitudes.
|x| = √(3² +7²) = √58
The motion is bounded between positions ±√58.
_____
Here's a way to get to the relation used above.
The sine of the sum of angles is given by ...
sin(θ+c) = sin(θ)cos(c) +cos(θ)sin(c)
If this is multiplied by some amplitude A, then we have ...
A·sin(θ+c) = A·sin(θ)cos(c) +A·cos(θ)sin(c)
Comparing this to the given expression, we find ...
A·cos(c) = 3 and A·sin(c) = -7
We know that sin²+cos² = 1, so the sum of the squares of these values is ...
(A·cos(c))² +(A·sin(c))² = A²(cos(c)² +sin(c)²) = A²(1) = A²
That is, A² = (3)² +(-7)² = 9+49 = 58. This tells us the position function can be written as ...
x = A·sin(θ +c) . . . . for some angle c
x = (√58)sin(θ +c)
This has the bounds ±√58.
Answer:
930
Step-by-step explanation:
600% of x = 5580
6x = 5580
x = 5580/6
x = 930