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].

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Eliseo

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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).

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:

Substituting the values Po = (3, 2, 2) and u = [1, -3, 2], we get:

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:

Simplifying this equation gives us:

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:

Substituting the coordinates of P = (2,-7,10) and the equation of the plane, we get:

Step 4: Find the symmetric point (P') using the distance and the normal vector.

To find the symmetric point P', we use the formula:

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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