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:
25% decreased by 60% is 10
Step-by-step explanation:
New value =
25 - Percentage decrease =
25 - (60% × 25) =
25 - 60% × 25 =
(1 - 60%) × 25 =
(100% - 60%) × 25 =
40% × 25 =
40 ÷ 100 × 25 =
40 × 25 ÷ 100 =
1,000 ÷ 100 = 10
hope this helps :)
Answer:
8.31
Step-by-step explanation:
Using the Pythagorean Theorem:
The side length for 
, rounded to the nearest hundredths, would be 8.31.
-(-3) = +3
-6 + 3 is just removing 3 from -6 which leaves us with -3.
