Math, asked by mannikalanah5646, 1 year ago

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

Answered by knjroopa
1

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)

Answered by ndahir7262
1

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

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