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Math  /  Geometry

QuestionExample 3 Determine the point on the plane 4x−2y+z=14 x-2 y+z=1 that is closest to the point (−2,−1,5)(-2,-1,5).

Studdy Solution

STEP 1

What is this asking? Find the point on the plane 4x−2y+z=14x - 2y + z = 1 that is the shortest distance from the point (−2,−1,5)(-2, -1, 5). Watch out! Don't forget that the shortest distance between a point and a plane is along a line perpendicular to the plane!

STEP 2

1. Find the normal vector.
2. Define the line equation.
3. Find the intersection point.

STEP 3

The **normal vector** of a plane is super handy because it points perpendicularly to the plane.
Since we're looking for the shortest distance, we want a line perpendicular to the plane, and the normal vector gives us just that direction!
The coefficients of xx, yy, and zz in the plane equation give us the components of the normal vector.

STEP 4

So, for our plane 4x−2y+z=14x - 2y + z = 1, the **normal vector** is n=(4,−2,1)\mathbf{n} = (4, -2, 1).
Awesome!

STEP 5

Now, we need a line that passes through our point (−2,−1,5)(-2, -1, 5) and is parallel to our **normal vector** n=(4,−2,1)\mathbf{n} = (4, -2, 1).
The equation of a line in 3D uses a **point** and a **direction vector**.
Lucky us, we have both!

STEP 6

The **parametric equations** for our line are: x=−2+4t x = -2 + 4t y=−1−2t y = -1 - 2t z=5+t z = 5 + t Here, tt is our parameter, and as it changes, we move along the line.

STEP 7

This line we've created intersects the plane at the closest point.
To find that point, we **substitute** the line equations into the plane equation.
This will let us solve for tt, and then we can find the coordinates of the intersection point.

STEP 8

**Substituting**, we get: 4(−2+4t)−2(−1−2t)+(5+t)=1 4(-2 + 4t) - 2(-1 - 2t) + (5 + t) = 1 −8+16t+2+4t+5+t=1 -8 + 16t + 2 + 4t + 5 + t = 1 21t−1=1 21t - 1 = 1 21t=2 21t = 2 t=221 t = \frac{2}{21}

STEP 9

Now, we plug this **value of** tt back into our line equations to find the coordinates of the **closest point**: x=−2+4⋅221=−2+821=−42+821=−3421 x = -2 + 4 \cdot \frac{2}{21} = -2 + \frac{8}{21} = \frac{-42 + 8}{21} = -\frac{34}{21} y=−1−2⋅221=−1−421=−21−421=−2521 y = -1 - 2 \cdot \frac{2}{21} = -1 - \frac{4}{21} = \frac{-21 - 4}{21} = -\frac{25}{21} z=5+221=105+221=10721 z = 5 + \frac{2}{21} = \frac{105 + 2}{21} = \frac{107}{21}

STEP 10

The point on the plane 4x−2y+z=14x - 2y + z = 1 closest to the point (−2,−1,5)(-2, -1, 5) is (−3421,−2521,10721)(-\frac{34}{21}, -\frac{25}{21}, \frac{107}{21}).

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