The shortest distance from a point to a plane is actually the length of the perpendicular dropped from the point to touch the plane. This lesson conceptually breaks down the above meaning and helps you learn how to calculate the distance in Vector form as well as Cartesian form, aided with a solved example at the end.

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**Shortest Distance from a Point to a Plane**

The focus of this lesson is to calculate the shortest distance between a point and a plane. This distance is actually the length of the perpendicular from the point to the plane. We can clearly understand that the point of intersection between the point and the line that passes through this point which is also normal to a plane is closest to our original point.

Thus, the line joining these two points i.e. the perpendicular should give us the said shortest distance. So, if we take the normal vector \vec{n} and consider a line parallel to it which meets our original point, then this parallel line is actually the required shortest distance.

*(Source: Song Ho Ahn)*

In the following section, we shall look at how the shortest distance can be calculated in both, Vector form and Cartesian form.

**Browse more Topics Under Three Dimensional Geometry**

- Angle Between a Line and a Plane
- Angle Between Two Lines
- Coplanarity of Two Lines
- Angle Between Two Planes
- Direction Cosines and Direction Ratios of a Line
- Distance Between Parallel Lines
- The Distance Between Two Skew Lines
- Distance of a Point from a Plane
- Equation of a Plane in Normal Form
- Equation of a Plane Perpendicular to a Given Vector and Passing Through a Given Point
- The Equation of Line for Space
- Equation of Plane Passing Through Three Non Collinear Points
- Intercept Form of the Equation of a Plane
- Plane Passing Through the Intersection of Two Given Planes

## Equation in Vector Form

We shall consider a point A whose position vector is given by \(\vec{a} \) and a plane whose equation is given by

\( \vec{r}. \vec{N} = \vec{D} \), where \( \vec{N}\) is the normal to the plane.

If we consider another plane to be parallel to our first plane and let this new plane pass through the point A, then the equation of the second plane, where \(\vec{Nâ€™}\) is normal to the plane Â can be written as

\( ( \vec{r} – \vec{a} ) . \vec{N} = 0 \)

In other words, \( \vec{r}. \vec{N} = \vec{a}. \vec{N} \)

If we consider O to be the origin, then the distance of the first plane from the origin is given by ON. Similarly, the distance of the second plane from the origin is given by ONâ€™. So, we can get the distance between the two planes as ON â€“ ONâ€™.

That is, the distance equals \(|\vec{D} â€“ \vec{a}.\vec{N} | \)

So, for a plane whose equation is given byÂ \( \vec{r}. \vec{N} = \vec{D} \), the distance of a point A whose position vector is given by \vec{a} to the plane is given by â€“

\( d = | \vec{a}. \vec{N} â€“\vec{D} | / | \vec{N} | \)

You can clearly understand that if the distance were to be calculated from the origin, we only need to substitute the position vector \(\vec{a}\) by 0 and the formula for the shortest distance, in that case, will become â€“

\( d = | \vec{D} | / | \vec{N} | \)

## Equation in Cartesian Form

Now let us consider a plane whose Cartesian equation is given by â€“

Ax + By + Cz = D

Then the position vector \(\vec{a}\) of a point whose Cartesian coordinates are given by \((x_1, y_1, z_1)\) can be formulated as â€“

\( \vec{a} = x_1Â \hat{i} + y_1Â \hat{j} + z_1Â \hat{k} \)

Now, the equation of the normal to the plane is â€“

\( \vec{N} = A \hat{i} + B \hat{j} + C \hat{k} \)

Now, we shall simply use the formula in vector form to arrive at the Cartesian formula â€“

\( d = | \vec{a}. \vec{N} â€“ D | / | \vec{N} | \)

\( = | (x)_1Â \hat{i} + y_1Â \hat{j} + z_1Â \hat{k} ) . (A \hat{i} + B \hat{j} + C \hat{k} ) â€“ D | / (A^2Â + B^2Â + C^2)^{1/2}Â \)

\( = | ( Ax_1Â + By_1Â + Cz_1Â â€“ D ) | / (A^2Â + B^2Â + C^2)^{1/2}Â \)

To understand further, how the calculations are actually performed, follow the simple example given below:

**Solved Example for You**

**Question: **Find the distance of the plane whose equation is given by

3x – 4y + 12z = 3 , from the origin.

**Answer: **We can see that the point here is actually the origin (0, 0, 0) while A = 3, B = – 4, C = 12 and D = 3

So, using the formula for the shortest distance in Cartesian form, we have â€“

d = | (3 x 0) + (- 4 x 0) + (12 x 0) â€“ 3 | / (3^{2 }+ (-4)^{2} + (12)^{2})^{1/2}

= 3 / (169)^{1/2}

= 3 / 13 units is the required distance.

Ques. Can a plane be curved?

Ans. A plane curve is a curve inside a plane that might be a Euclidean plane, an affine plane or it can be a projective plane. These plane curves also comprise the Jordan curves, that enclose an area of the plane but need not be smooth, and the graphs of continuous functions.

Ques. What is a plane in science?

Ans. In the vast study of Science a surface produced by a straight line constantly moving at velocity with respect to a point that is fixed. An area of a 2D surface containing determinate extension and spatial direction or position.

Ques. Do two intersecting lines determine a plane?

Ans. If two lines are intersecting, then exactly one plane comprises the lines. In case, 2 lines are intersecting, then they will be intersecting at exactly one point. Lastly, 3 non-collinear points define a plane.

Ques. Why do three points define a plane?

Ans. Because 3 non-collinear points are required for determining a unique plane in the Euclidean geometry. Given 2 points, there is correctly a line that can comprise them, but infinitely numerous planes can comprise that line. As long as 3 points don’t lie all on the same line, they determine a unique plane.

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