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.
The area of the shape shown in the image which is in the shape of kite figure is 168 squared centimeters.
<h3>What is the area of the kite figure?</h3>
The area of the kite figure is half of the product of its diagonals. Area of kite figure can be find out using the following formula.
Here, p and q are the diagonals of the kite.
The length of the first diagonal of the kite figure is,
p=6+15
p=21 cm
The length of the second diagonal of the kite figure is,
q=8+8
q=16 cm
Thus, the area of kite figure is:
Thus, the area of the shape shown in the image which is in the shape of kite figure is 168 squared centimeters.
Learn more about the area of kite here;
brainly.com/question/16424656
#SPJ1
I'm not sure which numbers you are referring to, but 3 has the same absolute value as -3, and 7 has the same absolute value as -7, just for an example.
Answer:
Step-by-step explanation:
6v + 12b = 504 eq1
2v + 5b = 204 eq2
6v + 15b = 612 eq2 times 3
0 -3b = 504 - 612 subtraction of the two bolded eqs
-3b = -108 solve for b
b = 36
Another way to solve the problem
6v + 12b = 504 I would eliminate the v term by multiplying the bottom
2v + 5b = 204 equation by -3 on BOTH sides and then add the two eqs
2v + 5b = 204
-3(2v + 5b) = -3(204)
-6v - 15b = -612
6v + 12b = 504
-6v - 15b = -612 add the like terms
(6v + (-6v)) + (12b + (-15b) = 504 + (-612)
(6v -6v)) + (12b - 15b) = 504 - 612)
0 + -3b = - 108 solve for b divide both sides by -3
b = -108/-3
b = 36
use eq 2v + 5b = 204 to solve for v and knowing b = 36
2v + 5b = 204 b=36
2v + 5(36) = 204 substract 5 times 36 from both sides
2v = 204 - 180
v = 24 / 2
v = 12
NOW CHECK the values for b and v using the OTHER eq
6v + 12b = 504
6(12) + 12(36) = 504
72 + 432 = 504
504 = 504 IT CHECKS
