Let T be the plane 2x−3y = −2. Find the shortest distance d from the point P0=(−1, −2, 1) to T, and the point Q in T that is closest to P0. Use the square root symbol '√' where needed to give an exact value for your answer.

Question

Let T be the plane 2x−3y = −2. Find the shortest distance d from the point P0=(−1, −2, 1) to T, and the point Q in T that is closest to P0. Use the square root symbol '√' where needed to give an exact value for your answer.

I need help with this one question!!

Answer 1

take each value of Q and subtract the respective value from P0

then square the 3 new values you have, add them together and then take the square root of their sum
Pythagorean theorem basically

Answer 2

Q makes sense but d is calculated a different way

Answer 3

To find the shortest distance from a point to a plane, you can use the formula:

d = |ax0 + by0 + cz0 + d| / √(a^2 + b^2 + c^2)

where (x0, y0, z0) is the coordinates of the point, and (a, b, c, d) are the coefficients of the plane equation.

In this case, the plane equation is 2x - 3y = -2. So, we can rewrite it as 2x - 3y + 0z + 2 = 0, where a=2, b=-3, c=0, and d=2.

The coordinates of the point P0 are (-1, -2, 1).

Now, let's substitute the values into the formula:

d = |2*(-1) + (-3)*(-2) + 0*1 + 2| / √(2^2 + (-3)^2 + 0^2)

Simplifying this, we have:

d = |-2 + 6 + 2| / √(4 + 9 + 0)

d = |6| / √13

d = 6 / √13

So, the shortest distance from the point P0 to the plane T is 6 / √13.

To find the point Q on the plane T that is closest to P0, we need to find the coordinates (x, y, z) that satisfy the equation of the plane and minimize the distance between P0 and Q.

Substituting the equation of the plane into the coordinates of Q, we have:

2x - 3y + 0z + 2 = 0

Rearranging this equation, we get:

2x - 3y = -2

To minimize the distance between P0 and Q, we need to find the values of x and y that satisfy this equation.

To solve the equation, we can isolate x:

2x = 3y - 2

x = (3y - 2) / 2

Now, we can substitute this x value back into the equation of the plane:

2((3y - 2) / 2) - 3y = -2

Simplifying this equation, we have:

3y - 4 - 3y = -2

-4 = -2

This equation is inconsistent, which means there are no points on the plane that satisfy the equation. Therefore, there is no point Q on the plane T that is closest to P0.

Answer 4

Use the distance formula from point to plane:

distance from (−1, −2, 1) to 2x - 3y + 0z + 2 = 0 is
= |2(-1) - 3(-2) + 0(1) + 2 |/√(2^2 + 3^2 + 0^2)
= 6/√13

if you actually need to find Q:
normal to the plane is vector (2,-3,0), so the perpendicular line containing (−1, −2, 1) is
x = -1 + 2t
y = -2 - 3t
z = 1

sub those into the plane equation and solve for t
put the value of t into the parametric equations to get Q

Check my answer above by finding PQ using the formula for the distance between two points.