<span>(2x + 1) • (x - 3) • (x + 4)</span>
Given plane Π : f(x,y,z) = 4x+3y-z = -1
Need to find point P on Π that is closest to the origin O=(0,0,0).
Solution:
First step: check if O is on the plane Π : f(0,0,0)=0 ≠ -1 => O is not on Π
Next:
We know that the required point must lie on the normal vector <4,3,-1> passing through the origin, i.e.
P=(0,0,0)+k<4,3,-1> = (4k,3k,-k)
For P to lie on plane Π , it must satisfy
4(4k)+3(3k)-(-k)=-1
Solving for k
k=-1/26
=>
Point P is (4k,3k,-k) = (-4/26, -3/26, 1/26) = (-2/13, -3/26, 1/26)
because P is on the normal vector originating from the origin, and it satisfies the equation of plane Π
Answer: P(-2/13, -3/26, 1/26) is the point on Π closest to the origin.
6... I think.
I hope that this is right
BC is the length of AD because they are parallel. Have a nice day
The phrase that would complete the proof is this:
Angle 2 is equal to Angle 4.
Vertical angles are angles formed when two lines intersect each other. The angles that are opposite each other are equal by a series of supplementary, complementary, and transitive property proofs.
