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# Find the symmetric point to a point P = $2,-7,10$ with respect to a plane containing a point Po = $3, 2, 2$ and perpendicular to a vector u = [1, -3, 2].

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## Answer to a math question Find the symmetric point to a point P = $2,-7,10$ with respect to a plane containing a point Po = $3, 2, 2$ and perpendicular to a vector u = [1, -3, 2].

Eliseo
4.6
To find the symmetric point to point P = $2,-7,10$ with respect to a plane containing point Po = $3, 2, 2$ and perpendicular to vector u = [1, -3, 2], we can use the formula for reflecting a point P across a plane.

Let's break it down into steps:

Step 1: Find the equation of the plane.
To find the equation of the plane, we need a point on the plane $Po$ and the normal vector $u$.
The equation of the plane is given by:
Ax + By + Cz + D = 0
Substituting the values Po = $3, 2, 2$ and u = [1, -3, 2], we get:
x - 3y + 2z + D = 0 \quad \text{$1$}

Step 2: Finding the value of D.
Since point Po = $3, 2, 2$ lies on the plane, we substitute its coordinates into equation $1$ to find the value of D:
3 - 3$2$ + 2$2$ + D = 0
Simplifying this equation gives us:
D = -3Answer: D = -3.

Step 3: Find the distance between point P and the plane.
We can find the distance between a point and a plane using the formula:
d = \left| \frac{{Ax + By + Cz + D}}{{\sqrt{{A^2 + B^2 + C^2}}}} \right|

Substituting the coordinates of P = $2,-7,10$ and the equation of the plane, we get:
d = \left| \frac{{2 - 3$-7$ + 2$10$ - 3}}{{\sqrt{{1^2 + $-3$^2 + 2^2}}}} \right|= \left| \frac{{2 + 21 + 20 - 3}}{{\sqrt{{14}}}} \right|= \left| \frac{{40}}{{\sqrt{{14}}}} \right|Answer: The distance between the point P and the plane is \left| \frac{{40}}{{\sqrt{{14}}}} \right|.

Step 4: Find the symmetric point $P'$ using the distance and the normal vector.
To find the symmetric point P', we use the formula:
P' = P - 2 \left$\frac{{d}}{{\sqrt{{A^2 + B^2 + C^2}}}} \right$ u
where P is the given point $2, -7, 10$, d is the distance between the point and the plane, and u is the normal vector.

Substituting the values into the equation, we get:
\[
P' = $2, -7, 10$ - 2 \left$\frac{{40}}{{\sqrt{{14}}}} \right$ [1, -3, 2]
= $2, -7, 10$ - \left$\frac{{80}}{{\sqrt{{14}}}} \right$ [1, -3, 2]
= $2, -7, 10$ - \left$\frac{{80}}{{\sqrt{{14}}}} \right$ [1, -3, 2]
= \left$2 - \frac{{80}}{{\sqrt{{14}}}} , -7 + \frac{{240}}{{\sqrt{{14}}}}, 10 - \frac{{160}}{{\sqrt{{14}}}} \right$

Answer: The symmetric point P' is \left$2 - \frac{{80}}{{\sqrt{{14}}}} , -7 + \frac{{240}}{{\sqrt{{14}}}}, 10 - \frac{{160}}{{\sqrt{{14}}}} \right$.

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