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Distance Between a Point and a Plane

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Distance Between a Point and a Plane

The distance between a specific point and a plane is important to a number of different activities.

For instance, a bungee jumping tower would not be very safe if the distance to the ground were not measured at the point directly under the tower, since any angle away from straight down would make the distance measure further and lead to a cord too long!

A computer game programmer needs to know how to calculate the distance between the location of a character on the screen and the walls around it to tell the game how to identify when a projectile hits a target, or when the character hits a wall.

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Guidance

Identifying the Point Closest to the Origin No matter what the orientation of a plane, there will always be one point located closer to the origin than any other point on the plane. This means that the position vector for this point is shorter than any other point in the plane. The diagram below shows a two-dimensional projection of a plane, in grey, near a point not on the plane, in black. The position vectors to a variety of points are shown in the diagram. The position vector marked in blue is shorter than the position vectors for the other points. This shortest vector is perpendicular to the plane. You can also see that the blue line is the vector projection of any orange vector onto the perpendicular direction.

This orthagonality (i.e. being perpendicular) is useful for us because it means that the position vector for this special point is parallel to the normal vector. Therefore, if we know the equation for a normal vector and the position vector for any point on the plane, we can determine the location of the point on the plane closest to the origin by finding the projection of the given point’s position vector onto the normal direction.

The Dihedral Angle

The angle between two planes is called the dihedral angle . The angle between two planes is the same as the angle between their normal vectors. If we want to determine the dihedral angle between two planes, we identify normal vectors to the two planes, then we can use the dot-product of the two normal vectors to determine the angle between the two normals which is also the two planes. Recall $\overrightarrow{A} \times \overrightarrow{B} =|\overrightarrow{A}||\overrightarrow{B}| \ \mbox{cos} \ \theta$

Example A

The three points P = (3, 7, 2), Q = (1, 4, 3), and R = (2, 3, 4) define a plane. Determine the point on the plane which is closest to the origin.

Solution

First find the vectors between two pairs of the points.

$\overrightarrow{PQ} = \left \langle (Q_x - P_x), (Q_y - P_y), (Q_z - P_z) \right \rangle = \left \langle (1 - 3), (4 - 7), (3 - 2) \right \rangle =$ $\left \langle -2, -3, 1 \right \rangle$

$\overrightarrow{PR} = \left \langle (R_x - P_x), (R_y - P_y), (R_z - P_z) \right \rangle = \left \langle (2 - 3), (3 - 7), (4 - 2) \right \rangle =$ $\left \langle -1, -4, 2 \right \rangle$

The cross product of these two vectors is normal to the plane.

$\overrightarrow{PQ} \times \overrightarrow{PR} = \left \langle (PQ_yPR_z - PQ_zPR_y), (PQ_zPR_x - PQ_xPR_z), (PQ_xPR_y - PQ_yPR_x)\right \rangle$

$\overrightarrow{PQ} \times \overrightarrow{PR} = \left \langle [(-3 \cdot 2) - (1 \cdot -4)], [(1 \cdot -1) - (-2 \cdot 2)], [(-2 \cdot -4) - (-1 \cdot -3)] \right \rangle$

$\overrightarrow{n} = \overrightarrow{PQ} \times \overrightarrow{PR} = \left \langle [(-6) - (-4)], [(-1) - (-4)], [(8) - (3)] \right \rangle = \left \langle -2, 3, 5 \right \rangle$

The point on the plane which is nearest to the origin can be found by determining the projection of the position vector of either of these three points onto the normal vector. Remember that the vector projection of one vector onto the direction of another, is given by the dot-product of the first vector onto the unit vector defining the direction of the second vector: $\left (\overrightarrow{A} \times \overrightarrow{B} \right ) \overrightarrow{B}$ .

Since we know three points on the plane, we can use one of them to solve the problem. Let’s start with point P. The vector projection of $\overrightarrow{P}$ onto $\hat{n}$ is given by $\left (\overrightarrow{P} \times \hat{n} \right ) \hat{n}$ , so first we need to determine the unit vector $\hat{n}$ which is given by $\hat{n} = \frac{\hat{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle -2, 3, 5 \right \rangle}{\sqrt{(-2)^2 + 3^2 + (5)^2}} = \frac{\left \langle -2, 3, 5 \right \rangle}{\sqrt{38}} = \left \langle -0.32, 0.49, 0.81 \right \rangle$

$\overrightarrow{P} \times \hat{n} = P_x\hat{n_x} + P_y\hat{n_y} + P_z\hat{n_z} = (3)(-0.32) + (7)(0.49) + (2)(0.81) =$ $-0.96 + 3.43 + 1.62 = 4.09$

$\left (\overrightarrow{P} \times \hat{n} \right )\hat{n} = (4.09) \left \langle -0.32, 0.49, 0.81 \right \rangle = \left \langle -1.3088, 2.0041, 3.3129 \right \rangle$

Therefore, the point on the plane closest to the origin is (-1.3088, 2.0041, 3.3129).

Example B

The three points P = (3, 7, 2), Q = (1, 4, 3), and R = (2, 3, 4) define a plane. Determine the dihedral angle between this plane and the x-y plane.

Solution As we saw in the example above, these three points define a plane which has a normal vector

$\overrightarrow{n} = \left \langle -2, 3, 5 \right \rangle$

The normal to the x-y plane is the unit vector $\hat{z} = \left \langle 0, 0, 1 \right \rangle$ . To find the angle between these two vectors we use the fact that $\overrightarrow{A} \times \overrightarrow{B} =|\overrightarrow{A}||\overrightarrow{B}| \ \mbox{cos} \ \theta$ and that $\overrightarrow{A} \times \overrightarrow{B} = A_xB_x + A_yB_y + A_zB_z$

First find a numerical value for the dot product:

$\overrightarrow{n} \times \hat{z} = n_xz_x + n_yz_y + n_zz_z = (-2 \cdot 0) + (3 \cdot 0) + (5 \cdot 1) = 5$

$|\overrightarrow{n}| = \sqrt{n_x^2 + n_y^2 + n_z^2} = \sqrt{(-2)^2 + (3)^2 + (5)^2} = \sqrt{4 + 9 + 25} = \sqrt{38}$

$|\hat{z}| = \sqrt{z_x^2 + z_y^2 + z_z^2} = \sqrt{0^2 + 0^2 + 1^2} = 1$

Then find the cosine version of the dot product:

$\overrightarrow{n} \times \hat{z} = \sqrt{38} \ \mbox{cos} \ \theta$

Now equate the two and solve for the angle, θ

$\overrightarrow{n} \times \hat{z} = 5 = \sqrt{38} \ \mbox{cos} \ \theta$

$\theta = \mbox{cos}^{-1} \left (\frac{5}{\sqrt{38}} \right ) = 62.5^\circ$

Example C

Determine the dihedral angle between the two planes 12 x + 23 y + 14 z - 5 = 0 and 7 x + 3 y + z + 12 = 0.

Solution

The dihedral angle is defined as the angle between two planes. This angle is also equal to the angle between the normals to the two planes. In two of the previous problems we determined the unit vectors which are perpendicular to these two planes $\overrightarrow{n_1} = \left \langle \frac{12}{29.5}, \frac{23}{29.5}, \frac{14}{29.5} \right \rangle$ and $\overrightarrow{n_2} = \left \langle \frac{7}{\sqrt{59}}, \frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}}, \right \rangle$ . We can then use the dot-product of these two normal vectors to determine the angle between the two. The dot-product is defined as $\overrightarrow{A} \times \overrightarrow{B} = A_xB_x + A_yB_y + A_zB_z + ...$ and as $\overrightarrow{A} \times \overrightarrow{B} = |A||B| \ \mbox{cos} \ \theta$ . First, we need to find the component version of the dot product and the magnitudes of the two normal vectors.

$\overrightarrow{n_1} \times \overrightarrow{n_2} = \overrightarrow{n_{1x}}\overrightarrow{n_{2x}} + \overrightarrow{n_{1y}}\overrightarrow{n_{2y}} + \overrightarrow{n_{1z}}\overrightarrow{n_{2z}} = \frac{12\cdot7}{29.5\sqrt{59}} + \frac{23\cdot3}{29\cdot5\sqrt{59}} + \frac{14\cdot1}{29.5\sqrt{59}}$

$\overrightarrow{n_1} \times \overrightarrow{n_2} = \frac{12\cdot7}{29.5\sqrt{59}} + \frac{23\cdot3}{29.5\sqrt{59}} + \frac{14\cdot1}{29.5\sqrt{59}} = \frac{119}{226.6} + \frac{69}{226.6} + \frac{14}{226.6} = \frac{202}{226.6} = 0.891$

Since these two vectors are unit-vectors, their magnitudes are both equal to 1.

$\mbox{cos} \ \theta = \frac{\overrightarrow{n_1} \times \overrightarrow{n_2}}{|\overrightarrow{n_1}||\overrightarrow{n_2}|} = \frac{0.891}{(1)(1)} = 0.891$

$\theta = cos^{-1} 0.891 = 27.0^\circ$

Vocabulary

The Dihedral Angle is the angle between two planes in a 3D space.

The origin is the point (0, 0, 0) in a 3D coordinate system.

A perpendicular line is oriented at 90 o to a given line or plane.

Guided Practice

Questions

1) Determine the angle between the plane 2 x - 5 y + 8 - 10 = 0 and the y - z plane.

2) The three points $\overrightarrow{P} = \left \langle -2, 3, 4 \right \rangle, \ \overrightarrow{Q} = \left \langle 5, -6, 7 \right \rangle,$ and $\overrightarrow{R} = \left \langle 8, 9, -1 \right \rangle$ identify a plane. Determine the point on the plane which is closest to the origin.

3) Determine the point on the plane 7 x + 3 y + z + 12 = 0 which is closest to the origin.

Solutions

1) The dihedral angle is defined as the angle between the two planes and is also equal to the angle between the two normal unit vectors. In this case, we already know the normal unit vector for the y-z plane, $\hat{x} = \left \langle 1, 0, 0 \right \rangle$ . We still need to determine, however, the unit vector for the plane 2 x - 5 y + 8 z - 10 = 0.

Comparing this equation to $n_xx + n_yy + n_zz + d = 0$ , we can see that $\overrightarrow{n} = \left \langle 2, -5, 8 \right \rangle$ .
Now we can use the definition of the unit vector
$\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle 2, -5, 8 \right \rangle}{\sqrt{2^2 + (-5)^2 + 8^2}} = \frac{\left \langle 2, -5, 8 \right \rangle}{\sqrt{4 + 25 + 64}} = \frac{\left \langle 2, -5, 8 \right \rangle}{9.64} = \left \langle \frac{2}{9.64}, \frac{-5}{9.64}, \frac{8}{9.64} \right \rangle$
The angle between the two planes is equal to the angle between the two normal vectors.
We can then use the dot-product of these two normal vectors to determine the angle between the two. The dot-product is defined as $\overrightarrow{A} \times \overrightarrow{B} = A_xB_x + A_yB_y + A_zB_z + ...$ and as $\overrightarrow{A} \times \overrightarrow{B} = |A||B| \ \mbox{cos} \ \theta$ . First, we need to find the component version of the dot product and the magnitudes of the two normal vectors.
$\overrightarrow{n_1} \times \overrightarrow{n_2} = \overrightarrow{n_{1x}}\overrightarrow{n_{2x}} + \overrightarrow{n_{1y}}\overrightarrow{n_{2y}} + \overrightarrow{n_{1z}}\overrightarrow{n_{2z}} = \frac{2 \cdot 1}{9.64} + \frac{-5\cdot0}{9.64} + \frac{8\cdot0}{9.64} = \frac{2}{9.64} = 0.2074$
Since these two vectors are unit-vectors, their magnitudes are both equal to 1.
$\mbox{cos} \ \theta = \frac{\overrightarrow{n_1} \times \overrightarrow{n_2}}{|\overrightarrow{n_1}||\overrightarrow{n_2}|} = \frac{0.2074}{(1)(1)} = 0.2074$
$\theta = cos^{-1} 0.2074 = 16.18^\circ$

2) The point on the plane nearest to the origin can be found by determining the projection of the position vector of one of these three points onto the normal vector. Remember that the vector projection of one vector onto the direction of another is given by the dot-product of the first vector onto the unit vector defining the direction of the second vector: $\left ( \overrightarrow{P} \times \hat{n} \right )\hat{n}$ .

We can use the position vectors for the three points to determine two vectors within the plane. Once we have those two vectors, their cross-product will define the direction normal to the plane. First find the two equations in the plane:
$\overrightarrow{A} = \overrightarrow{Q} - \overrightarrow{P} = \left \langle 5, -6, 7 \right \rangle - \left \langle -2, 3, 4 \right \rangle = \left \langle 7, -9, 3 \right \rangle$
$\overrightarrow{B} = \overrightarrow{R} - \overrightarrow{P} = \left \langle 8, 9, -1 \right \rangle - \left \langle -2, 3, 4 \right \rangle = \left \langle 10, 6, -5 \right \rangle$
Now determine the cross product of the two vectors
$\overrightarrow{n} = \overrightarrow{A} \times \overrightarrow{B} = \left \langle (A_yB_z - A_zB_y), (A_zB_x - A_xB_z), (A_xB_y - A_yB_x) \right \rangle$
$\overrightarrow{n} = \overrightarrow{A} \times \overrightarrow{B} = \left \langle (45 - 18), (30 - (-35)), (42 + 90) \right \rangle$
$\overrightarrow{n} = \overrightarrow{A} \times \overrightarrow{B} = \left \langle 27, 65, 132 \right \rangle$
Now we need to determine the unit vector associated with this normal vector
$\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle 27, 65, 132 \right \rangle}{\sqrt{27^2 + (65)^2 + (132)^2}} = \frac{\left \langle 27, 65, 132 \right \rangle}{\sqrt{22378}}$
$\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \left \langle 0.181, 0.435, 0.882 \right \rangle$
Now we determine the vector progression of one of the three initial position vectors onto the direction of this normal unit-vector: $\left (\overrightarrow{P} \times \hat{n} \right )\hat{n}$ . Remember that the dot product is given by $\overrightarrow{A} \times \overrightarrow{B} = A_xB_x + A_yB_y + A_zB_z + ...$ .
$\left (\overrightarrow{P} \times \hat{n} \right )\hat{n} = (-2(0.181) + 3(0.435) + 4(0.882))\left \langle 0.181, 0.435, 0.882 \right \rangle$
$\left (\overrightarrow{P} \times \hat{n} \right )\hat{n} = (4.471)\left \langle 0.181, 0.435, 0.882 \right \rangle$
$\left (\overrightarrow{P} \times \hat{n} \right )\hat{n} = (4.471)\left \langle 0.181, 0.435, 0.882 \right \rangle = \left \langle 0.809, 1.945, 3.943 \right \rangle$

3) The point on the plane nearest to the origin can be found by determining the projection of the position vector of any point on the plane onto the normal vector. The vector projection of one vector onto the direction of another is given by the dot-product of the first vector onto the unit vector defining the direction of the second vector: $\left (\overrightarrow{P} \times \hat{n} \right )\hat{n}$ .

In this case, we can determine a normal vector using the equation of the plane. Comparing 7 x + 3 y + z + 12 = 0 to the generic equation $n_xx + n_yy + n_zz + d = 0$ , we can see that $\overrightarrow{n} = \left \langle 7, 3, 1 \right \rangle$ and
$\hat{n} = \frac{\overrightarrow{n}}{|\overrightarrow{n}|} = \frac{\left \langle n_x, n_y, n_z \right \rangle}{\sqrt{n_x^2 + n_y^2 + n_z^2}} = \frac{\left \langle 7, 3, 1 \right \rangle}{\sqrt{(7)^2 + (3)^2 + (1)^2}} = \frac{\left \langle 7, 3, 1 \right \rangle}{\sqrt{49 + 9 + 1}} = \frac{\left \langle 7, 3, 1 \right \rangle}{\sqrt{59}} = \left \langle \frac{7}{\sqrt{59}}, \frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}} \right \rangle$
We also need to know the location of a point on the plane. If we write the equation of the plane in intercept form, we can determine the position vector for the x-, y-, and z-intercepts of the plane.
The equation $1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}$ must be true for all points on a plane. Therefore, we should first rearrange 7 x +3 y + z + 12 = 0 into the form $1 = \frac{x}{a} + \frac{y}{b} + \frac{z}{c}$ .
7 x + 3 y + z = -12 becomes $\frac{7}{-12}x + \frac{3}{-12}y + \frac{1}{-12}z = 1$
Therefore, $a = \frac{-12}{7}, \ b = \frac{-12}{3} = -4$ , and $c = \frac{-12}{1} = -12$ and the position vectors of the three intercepts are $\overrightarrow{A} = \left \langle -1.714, 0, 0 \right \rangle, \ \overrightarrow{B} = \left \langle 0, -4, 0 \right \rangle$ , and $\overrightarrow{C} = \left \langle 0, 0, -12 \right \rangle$ .
To complete the problem, compute the dot product.
$\left (\overrightarrow{B} \times \hat{n} \right )\hat{n} = \left (B_x \hat{n}_x + B_y \hat{n}_y + B_z \hat{n}_z \right )\hat{n} = \left ( 0 \left (\frac{7}{\sqrt{59}} \right ) - 4 \left (\frac{3}{\sqrt{59}} \right ) + 0 \left (\frac{1}{\sqrt{59}} \right ) \right ) \left \langle \frac{7}{\sqrt{59}}, \frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}} \right \rangle .$
$\left (\overrightarrow{B} \times \hat{n} \right )\hat{n} = \frac{-12}{\sqrt{59}}\left \langle \frac{7}{\sqrt{59}}, \frac{3}{\sqrt{59}}, \frac{1}{\sqrt{59}} \right \rangle = \left \langle \frac{-84}{59}, \frac{-36}{59}, \frac{-12}{59} \right \rangle = \left \langle -1.424, -0.610, -0.203 \right \rangle$

Practice

The three points define a plane. Determine the point on the plane which is closest to the origin

1. $P = (3, 6, 9), Q = (9, 6, 3)$ and $R = (6, -9, 9)$
2. $P = (1, -7, 2), Q = (4, 2, 9)$ and $R = (3, -5, 1)$
3. $P = (3, 8, 10), Q = (-2, 5, 8)$ and $R = (7, 4, 8)$
4. $P = (9, -1, 4), Q = (6, 2, -8)$ and $R = (12 , 9, 10)$
5. $P = (5, 8,-9), Q = ( -5, 3, 9)$ and $R = (10, 4, -6)$

Determine the dihedral angle between each of the planes in questions 1-5 and the x-y plane, use the $|\overrightarrow{n}|$ you calculated for each plane and recall that the normal to the x-y plane is the unit vector $\hat{z} = \left \langle 0, 0, 1 \right \rangle$

1. $P = (3, 6, 9), Q = (9, 6, 3)$ and $R = (6, -9, 9)$
2. $P = (1, -7, 2), Q = (4, 2, 9)$ and $R = (3, -5, 1)$
3. $P = (3, 8, 10), Q = (-2, 5, 8)$ and $R = (7, 4, 8)$
4. $P = (9, -1, 4), Q = (6, 2, -8)$ and $R = (12 , 9, 10)$
5. $P = (5, 8,-9), Q = ( -5, 3, 9)$ and $R = (10, 4, -6)$

Determine the dihedral angle between the two planes

1. $9x + 17y - 4z - 7 = 0$ and $-17x + 24y + 14z + 2 = 0$
2. $2x - 4y + 10z - 11 = 0$ and $2x - 9y + 4z + 12 = 0$
3. $-7x + 20y + 6z + 4 = 0$ and $-19x - 3y + z + 5 = 0$
4. $5x - 8y + 20z - 5 = 0$ and $6x + y + 19z - 7 = 0$
5. $14x + 11y - 5z - 16 = 0$ and $11x - 13y + 8z + 4 = 0$
6. $-10x + 9y + 2z + 8 = 0$ and $21x + 7y - 4z + 15 = 0$