Let T be the plane 3x−2z = 14. Find the shortest distance d from the point P0=(5, 5, −3) 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.
d=?
Q=(?,?,?)
Q is somewhere on the vector
<5,5,-3> + c<3,0,-2> = <5+3c,5,-3-2c>
But that point must lie in the plane, so
3(5+3c) - 2(-3-2c) = 14
c = -7/13
Now you know Q = (5,5,-3) - 7/13 (3,0,-2)
The distance is, as usual
|3*5 + 0*5 - 2(-3)-14|/√(3^2+0^2+2^2) = 7/√13
Thanks!!!!
To find the shortest distance from the point P0 to the plane T, we can use the formula for the distance from a point to a plane.
The formula is given as:
d = |ax + by + cz + d0| / √(a^2 + b^2 + c^2)
where (a, b, c) is the normal vector to the plane, and d0 is the distance from the origin to the plane.
In our case, the equation of plane T is 3x − 2z = 14. Comparing this equation with the general form ax + by + cz + d0 = 0, we can identify the normal vector (a, b, c) as (3, 0, -2).
Next, substitute the coordinates of the given point P0 = (5, 5, -3) into the formula:
d = |3*5 + 0*5 + (-2)*(-3) + 14| / √(3^2 + 0^2 + (-2)^2)
= |15 + 0 + 6 + 14| / √(9 + 4)
= |35| / √13
= 35 / √13
Therefore, the shortest distance from the point P0 to the plane T is 35 / √13 (exact value).
To find the point Q in the plane that is closest to P0, we can substitute this value into the equation of the plane T:
3x - 2z = 14
3x - 2(-3) = 14
3x + 6 = 14
3x = 8
x = 8/3
Plugging this value of x into the equation of the plane, we can solve for z:
3(8/3) - 2z = 14
8 - 2z = 14
-2z = 14 - 8
-2z = 6
z = -3
So, the coordinates of the point Q in the plane T that is closest to P0 are (8/3, ?, -3).
Unfortunately, we do not have enough information to determine the value of the y-coordinate for point Q, as it was not provided in the question.
To find the shortest distance (d) from a point to a plane and the point on the plane (Q) closest to the given point, follow these steps:
Step 1: Determine the equation of the perpendicular line.
We can obtain the equation of the perpendicular line by using the normal vector of the plane. The equation of a line passing through a point (P0) and perpendicular to a plane is (x, y, z) = (x₀, y₀, z₀) + t(a, b, c), where (x₀, y₀, z₀) are the coordinates of P0 and (a, b, c) are the components of the normal vector.
Given the equation of the plane 3x - 2z = 14, we can determine its normal vector as (3, 0, -2). Therefore, the equation for the line is (x, y, z) = (5, 5, -3) + t(3, 0, -2).
Step 2: Solve the system of equations for the point of intersection.
Substitute this equation into the equation of the plane, and solve for the parameter t by equating the expressions for x, y, and z.
3x - 2z = 14
3(5 + 3t) - 2(-3 - 2t) = 14
15 + 9t + 6 + 4t = 14
13t = -7
t = -7/13
Step 3: Find the coordinates of point Q.
Plug the value of t back into the equation for the line to obtain the coordinates of the point Q.
Q = (5, 5, -3) + (-7/13)(3, 0, -2)
Q = (5 - (21/13), 5, -3 - (14/13))
Q = (32/13, 5, -55/13)
Step 4: Calculate the distance between P0 and Q.
The distance (d) between two points can be found using the Euclidean distance formula: d = √[(x₂ - x₁)² + (y₂ - y₁)² + (z₂ - z₁)²].
Comparing the coordinates of P0 = (x₁, y₁, z₁) and Q = (x₂, y₂, z₂):
d = √[(32/13 - 5)² + (5 - 5)² + (-55/13 - (-3))²]
d = √[(32/13 - 65/13)² + 0² + (-55/13 + 39/13)²]
d = √[(-33/13)² + (0)² + (-16/13)²]
d = √[(1089/169) + (256/169)]
d = √(1345/169)
d = √(1345)/√(169)
d = √(1345)/13
Therefore, the shortest distance (d) from P0 to T is √(1345)/13, and the coordinates of Q are (32/13, 5, -55/13).