Question

$\text{Find the distance from the point}(2,0,-3)\text{to the plane}$

$3x-4y+5z=1$

Short answer

The equation is$7x-13y-15z+22=0$

Step-by-step solution

Step 1:Given information

Plane: $3x-4y+5z=1$

Point:$\left(2,0,-3\right)$

Step 2:Calculation

$\text{Consider the plane that contains the points}P=(2,-3,5),Q=(3,1,0)\text{and}R=(-1,0,7)\text{.}$

$\text{The vector}\overrightarrow{PQ}\text{is:}$

$\overrightarrow{PQ}=⟨3-2,1-(-3),0-5⟩$

$=⟨1,4,-5⟩$

The vector $\overrightarrow{QR}$ is:

$\overrightarrow{QR}=⟨-1-3,0-1,7-0⟩$

$=⟨-4,-1,7⟩$

$\text{The vector}\overrightarrow{RP}\text{is:}$

$\overrightarrow{RP}=⟨2-(-1),-3-0,5-7⟩$

$=⟨3,-3,-2⟩$

The vectors are non-collinear as neither of them is scalar multiple of the other.

To find the equation of the plane, compute the normal vector by taking cross-product of any two vectors.

$\text{The normal vector}\mathbf{N}=\overrightarrow{PQ}\times \overrightarrow{RP}\text{is given by:}$

$\mathbf{N}=\left|\begin{array}{ccc}i& j& k\\ 1& 4& -5\\ 3& -3& -2\end{array}\right|$

$=7\mathbf{i}-13\mathbf{j}-15\mathbf{k}$

$=⟨7,-13,-15⟩$

$\text{The equation of the plane is:}$

$7(x-2)-13(y+3)-15(z-5)=0$

width="228" height="20" role="math">7x-13y-15z-14-39+75=0

width="162" height="20" role="math">7x-13y-15z+22=0

$\text{The equation of the plane that contains the points}P=(2,-3,5),Q=(3,1,0)\text{and}R=(-1,0,7)$

$\text{is}7x-13y-15z+22=0\text{.}$