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.
3 dozen = 36
1/3 ÷ x = 5 ÷ 36
1/3 × 36 = x × 5
12 = 5x
12/5 = x
12/5 cups of flour is needed
Answer:
No solution
Step-by-step explanation:
So we have the equation:
First, subtract 20 from both sides:
Now, we can stop. Recall that absolute value will always <em>always</em> give a positive answer.
Since this equals -16, this means that there is no solution.
And we're done!
Answer:
D 6
Step-by-step explanation:
Normally when I multiply mixed numbers, I change them to improper fractions first so that it is easier.
1
× 5 = 
× 5
Multiply 5 by the numerator of the fraction and leave the denominator alone.
× 5 = 
Change back to simplest form.
= 6
