Find the distance from the plane 3x +2y - z = 3 and the point (1, 2, 3 ) by thre

Question

Find the distance from the plane 3x +2y - z = 3 and the point (1, 2, 3 ) by three methods: 1. Use vector methods - take a point in the plane, find the vector from this point to (1, 2, 3 ). Find the component of this vector onto the normal vector. Explain why this will lead you to the answer. Draw pictures. 2. Let D = (x-1)^2 + (y-2)^2 + (z-3)^2. This is the square of the distance from a point (x, y, z) to (1, 2, 3). Eliminate one of the variables by using the equation of the plane. Find the minimum value of D. Explain why this will lead you to the answer. 3. Using D as above and use the equation of the plane as the constraint equation. Then use the method of Lagrange multipliers. Explain why this will lead you to the answer.

Explanation / Answer

1)

3x + 2y - z = 3

(1 , 2 , 3)

A point on 3x +2y - z = 3 is : (1 , 0 , 0)

<1 , 2 , 3> DOT <1 , 0 , 0> / |<1 , 2 , 3>|
to find the component

<1,2,3> DOT <1,0,0> is 1*1 + 2*0 + 3*0 ---> 1

|<1,2,3> = sqrt(1 + 4 + 9) = sqrt(14)

So, the distance is :

1 / sqrt(14) units ----> ANSWER using method 1

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2)

D = (x - 1)^2 + (y - 2)^2 + (z -3)^2
3x + 2y - z = 3
z = 3x + 2y - 3
Plug this in :

D = (x - 1)^2 + (y - 2)^2 + (3x + 2y - 6)^2 --> to be minimized

F(x,y) : (x - 1)^2 + (y - 2)^2 + (3x + 2y - 6)^2

G(x,y) : 3x + 2y - 3

Fx = 2(x - 1) + 2(3x + 2y - 6)(3) ---> 20x + 12y - 38

Fy = 2(y - 2) + 2(3x + 2y - 6)(2) ---> 12x + 10y - 28

Gx = 3

Gy = 2

Fx = lamb * Gx
20x + 12y - 38 = lamb * 3

Fy = lamb * Gy
12x + 10y - 28 = lamb * 2

20x + 12y - 38 = 3lamb --> multiply by 10
200x + 120y - 380 = 30lamb

12x + 10y - 28 = 2lamb --> multiply by -12
-144x - 120y + 336= -24lamb

Adding :

200x + 120y - 380 = 30lamb + -144x - 120y + 336= -24lamb

gives us :

56x - 44 = 6lamb
x = (3lamb + 22) / 28

12x + 10y - 28 = lamb * 2

12((3lamb + 22) / 28) + 10y - 28 = 2lamb

12*3lamb/28 + 12*22/28 + 10y - 28 = 2lamb

9lamb/7 + 66/7 + 10y - 28 = 2lamb

9lamb + 66 + 70y - 196 = 14lamb
70y = 5lamb + 130

y = (lamb + 26) / 14

z = 3x + 2y - 3

z = 3((3lamb + 22) / 28) + 2( (lamb + 26) / 14) - 3

z = (9lamb + 66)/28 + (2lamb + 52)/14 - 3

z = (13lamb + 170) / 28 - 3

z = (13lamb + 170 - 84) / 28

z = (13lamb + 86) / 28

So, now, we have the x,y,z in terms of lambda, we can find the distance.

Plugging in, we get 1/sqrt(14) units ---> ANSWER using method 2

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3)

F(x,y,z) : (x - 1)^2 + (y - 2)^2 + (z -3)^2
G(x,y,z) : 3x + 2y - z = 3

Fx = 2(x - 1)
Fy = 2(y - 2)
Fz = 2( z- 3)

Gx = 3
Gy = 2
Gz = -1

Fx = lamb * Gx
2(x - 1) = lamb * 3
x = (3lamb)/2 + 1

Fy = lamb * Gy
2(y - 2) = lamb * 2
y = lamb + 2

Fz = lamb * Gz
2(z - 3) = lamb * -1
z = -lamb/2 + 3

USing 3x + 2y - z = 3, lets find lamb :

3((3lamb)/2 + 1) + 2(lamb + 2) - (-lamb/2 + 3) = 3

9lamb/2 + 3 + 2lamb + 4 + lamb/2 - 3 = 3

7lamb + 4 = 3

lamb = -1/7

So, x = 3lamb/2 + 1
x = 3(-1/7)/2 + 1
x = -3/14 + 1
x = 11/14

y = lamb + 2
y = -1/7 + 2
y = 13/7

z = -lamb/2 + 3
z = 1/14 + 3
z = 43/14

(x - 1)^2 + (y - 2)^2 + (z -3)^2 now becomes :

D = (11/14 - 1)^2 + (13/7 - 2)^2 + (43/14 - 3)^2

D = (-3/14)^2 + (-1/7)^2 + (1/14)^2

D = 9/196 + 1/49 + 1/196

D = 9/196 + 4/196 + 1/196

D = 14/196

D = 1/14

Now, to find the distance, d, we have to sqrt D :

D = sqrt(1/14)

D = 1/sqrt(14) ---> ANSWER using method 3

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