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Shortest Distance from origin to plane

https://i.gyazo.com/f76e44ed8ed0dbaab4bdc1790d10ee6a.png

Hi, I don't know how to approach this question. Any help would be appreciated. Thanks.
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B_9710
18
The vector [23−5] \displaystyle \begin{bmatrix} 2\\ 3 \\ -5 \end{bmatrix} is perpendicular to the plane and the shortest distance from a plane to a point is along a line which is perpendicular to the plane. So the line which contains the line segment of shortest distance from the origin to the plane is the line that is perpendicular to the plane (and hence parallel to the vector above) and passes through the origin.
So this line may be written as
r=λ[23−5] \displaystyle \mathbf{r} = \lambda \begin{bmatrix} 2\\ 3 \\ -5 \end{bmatrix} .
So x=2λ,y=3λ,z=−5λ x=2\lambda , y=3\lambda , z=-5\lambda so plug it into the equation for the plane and find the value of λ \lambda and then from there you should be able to find the distance quite simply.
Original post by B_9710
The vector [23−5] \displaystyle \begin{bmatrix} 2\\ 3 \\ -5 \end{bmatrix} is perpendicular to the plane and the shortest distance from a plane to a point is along a line which is perpendicular to the plane. So the line which contains the line segment of shortest distance from the origin to the plane is the line that is perpendicular to the plane (and hence parallel to the vector above) and passes through the origin.
So this line may be written as
r=λ[23−5] \displaystyle \mathbf{r} = \lambda \begin{bmatrix} 2\\ 3 \\ -5 \end{bmatrix} .
So x=2λ,y=3λ,z=−5λ x=2\lambda , y=3\lambda , z=-5\lambda so plug it into the equation for the plane and find the value of λ \lambda and then from there you should be able to find the distance quite simply.


Thank you! I got the distance as (5root38)/19. Would you say this is correct?
Original post by rickonstark999
Thank you! I got the distance as (5root38)/19. Would you say this is correct?
Yes. Note that this is the same as the "10" in 2x+3y-5z = 10, divided by sqrt(2^2 + 3^2 + 5^2).

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