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.
Answer:
9
Step-by-step explanation:
The factors of 18 are : 1, 2, 3, 6, 9, 18
The factors of 27 are : 1, 3, 9, 27
The common factors are : 1, 3, 9
The gcf is 9
n, n + 2 - two consecutive odd integers
196 - the sum
The equation:
n + (n + 2) = 196
n + n + 2 = 196 |subtract 2 from both sides
2n = 194 |divide both sides by 2
n = 97
n + 2 = 97 + 2 = 99
Answer: 97, 99.
4 because doubled it equals eight, add four it equals twelve, subtract four it equals eight, subtract three and it equals five.
