Find an equation of the plane. The plane that passes through the point (2, 3, 4)

Question

Find an equation of the plane.

The plane that passes through the point (2, 3, 4) and contains the line

x = 4t, y = 2 + t, z = 3 ? t.

Find an equation of the plane. The plane that passes through the line of intersection of the planes x ? z = 3 and y + 3z = 2 and is perpendicular to the plane

x + y ? 3z = 5.

Find the point at which the line intersects the given plane.

x = 4 + 4t,  y = 3t,  z = 5 ? 4t;     x + 3y ? z + 1 = 0

(x, y, z) =

Find the distance from the point to the given plane.

(?5, 8, 6),     x ? 2y ? 4z = 8

Explanation / Answer

Find an equation of the plane.
The plane that passes through the point (2, 3, 4) and contains the line
x = 4t, y = 2 + t, z = 3 - t.

When t = 0 , point = (0 , 2 , 3)

So, points (2,3,4) and (0,2,3) are on the plane.

Setting t = 1, we get (4 , 3 , 2) which is also on the plane

Let a be the vector from (4,3,2) to (0,2,3) --> a = (-4 , -1 , 1)

Let b be the vector from (4,3,2) to (2,3,4) --> b = (-2 , 0 , 2)

The normal of these 2 vectors will be the normal to the plane and can be found by cross-product of a x b

(-4 , -1 , 1) CROSS (-2 , 0 , 2) is <-2 , 6 , -2>

Now, the plane is n . <x - x0 , y - y0 , z - z0> = 0

<-2 , 6 , -2> DOT <x - 2 , y - 3 , z - 4> = 0

-2(x - 2) + 6(y - 3) - 2(z - 4) = 0

-2x + 4 + 6y - 18 - 2z + 8 = 0

2x - 6y + 2z = -6

x - 3y + z = -3 ----> ANSWER

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Find an equation of the plane. The plane that passes through the line of intersection of the planes x - z = 3 and y + 3z = 2 and is perpendicular to the plane x + y - 3z = 5.

Eq of a plane E, passing thru the line of intersection of the planes u : x - z - 3 = 0 and v : y + 3z - 2 = 0
is u + kv = 0 for some real k.

x - z - 3 + k(y + 3z - 2) = 0

1x + ky + (3k - 1)z - (3 + 2k) = 0

the normal to this plane is along N = <1 , k , 3k - 1>

This normal is perpendicular to the normal to the given plane x + y - 3z - 5 = 0, which is <1 , 1 , -3>

So, their dot product = 0

<1 , 1 , -3> DOT <1 , k , 3k - 1> = 0

1 + k - 9k + 3 = 0

k = 1/2

So, 1x + ky + (3k - 1)z - (3 + 2k) = 0 becomes :

1x + (1/2)y + (3/2 - 1)z - (3 + 2(1/2)) = 0

1x + (1/2)y + (1/2)z - 4 = 0

Multiply all over by 2 :

2x + y + z - 8 = 0

2x + y + z = 8 ----> ANSWER

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Find the point at which the line intersects the given plane.
x = 4 + 4t, y = 3t, z = 5 - 4t;     x + 3y - z + 1 = 0
(x, y, z) =

Plug in x,y,z :

4 + 4t + 3(3t) - (5 - 4t) + 1 = 0
4 + 4t + 9t - 5 + 4t + 1 = 0
t = 0

So, the point is x = 4 + 4t --> 4 + 0 --> 4
y = 3t = 3(0) = 0
z = 5 - 4t = 5 - 4(0) = 5

So, (4 , 0 , 5) ---> ANSWER

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Find the distance from the point to the given plane.
(-5, 8, 6),     x - 2y - 4z = 8

ax + by + cz + d = 0
1x - 2y - 4z - 8 = 0

So, a = 1 , b = -2 , c = -4 and d = -8

x0 , y0 , z0 = -5 , 8 , 6

The distance is given by :

(ax0 + by0 + cz0 + d) / sqrt(a^2 + b^2 + c^2)

(-5 - 16 - 24 - 8) / sqrt(1^2 + (-2)^2 + (-4)^2)

(-53) / sqrt(21) --> since distance is positive, take absolute value of this to get the answer

So, the distance is 53 / sqrt(21) units ----> ANSWER

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