can you answer as many questions as possible please with the right answers and s

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can you answer as many questions as possible please with the right answers and steps
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Find the symmetric formn for the line through P = (-3,1,-1) and Q = (7,1 1,8). Find a parametric equation of a line of intersection of the 12. 13. 14. 15. 16. two planes, 2x + 5y + 16z 13-x-2y-6z =-5. Find the distance from P = (1,1,2); to the plane 3x-7y + z-5=0. Show that the two planes are parallel and find the distance between the planes: 3x + 12y-6z =-2 and 5x + 20y-10z = 7. Find the equation of the plane that contains the point P = (4,-3,0) and the line x = t + 5, y = 2t-1, z =-t + 7. Find the distance from the point P = (2,1,2) to the line x-3-2t,y=-1+3t, z = 1 + 2t.

Explanation / Answer

Q.11

Solution:- first the vector be calculate <7-(-3),11-1,(-8)-(-1)>= <10,10,-7>

Either of the points can be used for symmetric equation

So symmetric equation is:-

(x-3)/10 =(y-1)/10=(z+1)/-7

Q 12

Solution:- the planes have normal vector a=(2,5,16) and b=(1,-2,-6) respectively .let Love denote the line of intersection then v=a×b=|I j k|

|2 5 16|

|1 -2 -6|

=(-30+32 , 12+16 , -4-5)

=(2 , 18 , -9) is parallel to L.we only need to find a point P on line L.

To find P solve the system of equation of planes:

2x + 5y + 16z= 13

x - 2y - 6z = -5

We consider P to be the point on L on the plane z=0, in the system above to get

2x + 5y = 13

x -2y = -5

Hence we get (-1,5,0) and so the equation of the line are

{x = -1+2t

{y = 5+18t

{z =-9t

Q.13

Solution: from the equation for the plane we have

A = 3

B = -7

C = 1

and x = 1 , y = 1 , z = 2

We have the formula:

d = | Ax + By + Cz | /(A^2 +B^2 + C^2)

d =-7/59

D= -0.9111

Q.14

Solution:

The normal vectors of the planes are

n1 = < 3, 12 ,-6>

N2 = <5, 20, -10>

Two vectors are proportional to each other, thus the planes are parallel.

Two planes are parallel so we can find the distance between two:

Multiply the equation if first plane by 5

and multiply the equation of second plane by 3

We get 15x + 60y - 30z + 10= 0

And 15x + 60y - 30z - 21 =0

Now d = |D2 - D1|(A^2 + B^2 + C ^2)

d =|-21 - 10|(225+3600+900)

d = 31/113.24

d = 0.2737

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