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.
Answer:
The correct answer AC = 11.52
Step-by-step explanation:
From the figure attached with this answer shows the triangle ABC.
BC=4 centimeters, m angle B=m angle c, and m angle a = 20 degrees
To find AC
From the figure we can see that D is the mid point and AD⊥BC
Then BD = CD = 4/2 = 2 cm
<BAD = 10° and <CAD = 10°
By using trigonometric ratio,
sin 10 = CD/AC = 2/AC
AC = 2/sin 10 = 2/0.1736 = 11.52
Therefore the value of AC = 11.52 cm
Answer: just follow these simple steps!
Step-by-step explanation:
1. Estimate the quotient.
2. Perform the division. Remember to place a zero in the quotient when the divisor is larger than the dividend. ...
3. Compare your estimate with your quotient to verify that the answer makes sense.
