An angle between the plane, x+y+z = 5 and the line of intersection of the planes, 3x+4y+z1=0 and 5x+8y+2z+14 = 0, is
Answers
Answer:
Step-by-step explanation:
Given An angle between the plane, x+y+z = 5 and the line of intersection of the planes, 3x+4y+z1=0 and 5x+8y+2z+14 = 0, is
We need to find the line or point of intersection of the planes.
So we need to consider the given equations . consider the two equations
3 x + 4 y + z = 0 multiply by 2
5 x + 8 y + 2 z = - 14
So we get
6 x + 8 y + 2 z = 0
5 x + 8 y + 2 z = - 14
-----------------------------
x = 14
So from first and third equation we get
3 x + 4 y + z = 0
x + y + z = 5
--------------------------------- subtracting we get
2 x + 3 y = - 5
2(14) + 3 y = - 5
28 + 3 y = - 5
3 y = -28 - 5
y = -33 / 3
y = - 11
Now consider the equation
x + y + z = 5
14 - 11 + z = 5
3 = 5 - z
3 - 5 = - z
- 2 = - z
z = 2
So points are ( 14, - 11 , 2)
Answer: 80.7279
Step-by-step explanation:
Given An angle between the plane, x+y+z = 5 and the line of intersection of the planes, 3x+4y+z1=0 and 5x+8y+2z+14 = 0, is
We need to find the line or point of intersection of the planes.
So we need to consider the given equations . consider the two equations
3 x + 4 y + z = 0 multiply by 2
5 x + 8 y + 2 z = - 14
So we get
6 x + 8 y + 2 z = 0
5 x + 8 y + 2 z = - 14
-----------------------------
x = 14
So from first and third equation we get
3 x + 4 y + z = 0
x + y + z = 5
--------------------------------- subtracting we get
2 x + 3 y = - 5
2( 14) + 3 y = - 5
-28 + 3 y = - 5
3 y = -28 - 5
y = -33 / 3
y = -11
Now consider the equation
x + y + z = 5
14 + (-11) + z = 5
z = 2
So points are (14, -11 , 2) of the intersection of plane of line
angle between line and plain
cosФ = (n1·n2)÷||n1||·||·n2||
= (1, 1, 1)·(14, -11, 2)÷(√1²+1²+1²)·(√14²+(-11)²+2²)
=5÷√3·√321
cosФ = 0.161122
Ф = 80.7279