Can u find the point on the plane&nbsp;x+2y+3z=13&nbsp;that is closest to the point (1,1,1). You need to minimize the function

Question

Can u find the point on the plane&amp;nbsp;x+2y+3z=13&amp;nbsp;that is closest to the point (1,1,1). You need to minimize the function

rogreenhoxa8 · Accepted Answer

Apply Lagrange Multipliers:D=((x−1)2+(y−1)2+(z−1)2)12&amp;nbsp;Now, we can use Constraint Surface:x+2⋅y+3⋅z=13to&amp;nbsp;g(x,y,z)=x+2⋅y+3⋅z−13&amp;nbsp;Formulate Condition (for constant scalarmultiplier &quot;λ&quot;):△downD=λ⋅△downg→(12⋅D−12⋅[2⋅(x−1)⋅i+2⋅(y−1)⋅j+2⋅(z−1)⋅k]&amp;nbsp;=λ⋅[(1)⋅i+(2)⋅j+(3)⋅k](x−1)=(1)⋅λ⋅D12(y−1)=(2)⋅λ⋅D12(z−1)=(3)⋅λ⋅D12→(x−1)=y−12=z−13→[y=2⋅x−1]&amp;nbsp;and&amp;nbsp;[z=3⋅x−2]x+2⋅y+3⋅z=13→x+2⋅[2⋅x−1]+3⋅[3⋅x−2]=13→x+4⋅x−2+9⋅x−6=13→14⋅x=21→x=(32)(x=32)y=2⋅(32)−1=(2)z=3⋅(32)−2=52

Jeffrey Jordon · Accepted Answer

Given x+2y+3z=13 Point (1,1,1)The normal vector to the plane is &amp;lt;1,2,3&amp;gt;The point you seek would have to be multiple of this vector added to (1,1,1)∴P=(1,1,1)+c&amp;lt;1,2,3&amp;gt;=(1+c,1+2c,1+3c)But this point has to satisfy the plane`s equation.(1+c)+2(1+2c)+3(1+3c)=131+c+2+4c+3+9c=1314c=13-6c=714=12∴&amp;nbsp;the point P is&amp;nbsp;(1+12,1+2×12,1+3×12)=(32,2,52)

Store	Original Price of Aquarium ($)
Pets Plus	118
Pet Planet	110

Find the point on the plane x+2y+3z=13 that is closest to the point (1,1,1). How would you minimize the function?

Answered question

Answer & Explanation

New Questions in High School