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.
162 / 2 = 81
the answer is D
<span>B. ∠1 and ∠9</span>
........................
Answer:
Multiply both sides by 
Step-by-step explanation:
Given
Required
Get an equivalent of 
To do this, we simply multiply through by
Since BA is half of YX, the scale factor is 2.
Find the values of the other line segments.
BA = 11
BC = 44/2 = 22
AC = 32 / 2 = 16
Find the perimeter.
p = 11 + 22 + 16
p = 33 + 16
p = 49
Therefore, the answer is 49.
Best of Luck!
