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Distances and Angles in Space

The distance from a point to a plane is based on substituting the point into the plane equation and dividing by the normal magnitude.

$$d=\frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$$

Distance From Point to Plane

Find the distance from $(1,2,-1)$ to the plane $2x-y+2z-5=0$.

$$d=\frac{|2(1)-2+2(-1)-5|}{\sqrt{2^2+(-1)^2+2^2}}$$
$$d=\frac{7}{3}=2.33$$

Final answer: $2.33$ units.

Angle Between Two Planes — CE Board

Find the acute angle between the planes $x + 2y - 2z = 5$ and $3x - 4z = 1$.

The angle between two planes equals the angle between their normal vectors.

$$\vec{n_1} = \langle 1,\,2,\,-2\rangle,\qquad \vec{n_2} = \langle 3,\,0,\,-4\rangle$$
$$\vec{n_1}\cdot\vec{n_2} = (1)(3)+(2)(0)+(-2)(-4) = 3+0+8 = 11$$
$$|\vec{n_1}| = \sqrt{1+4+4} = 3,\qquad |\vec{n_2}| = \sqrt{9+0+16} = 5$$
$$\cos\theta = \frac{11}{3\cdot 5} = \frac{11}{15}$$
$$\theta = \cos^{-1}\!\left(\frac{11}{15}\right) \approx \boxed{42.8^\circ}$$

Distance from a Point to a Line in 3D — CE Board

Find the distance from point $P(1,1,1)$ to the line through the origin with direction $\vec{d} = \langle 1,\,2,\,2\rangle$.

Use the cross-product formula. Let $A(0,0,0)$ be a point on the line:

$$d = \frac{|\vec{AP}\times\vec{d}|}{|\vec{d}|}$$

$\vec{AP} = \langle 1,1,1\rangle$. Compute the cross product:

$$\vec{AP}\times\vec{d} = \begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&1&1\\1&2&2\end{vmatrix} = \langle (2-2),\,-(2-1),\,(2-1)\rangle = \langle 0,\,-1,\,1\rangle$$
$$|\vec{AP}\times\vec{d}| = \sqrt{0+1+1} = \sqrt{2}$$
$$|\vec{d}| = \sqrt{1+4+4} = 3$$
$$d = \frac{\sqrt{2}}{3} \approx \boxed{0.471 \text{ units}}$$

Problem: Point-to-Plane Distance

Find the distance from P(1, 2, -1) to the plane 2x - y + 2z - 5 = 0.

$$d=\frac{|2(1)-2+2(-1)-5|}{\sqrt{2^2+(-1)^2+2^2}}=\frac{7}{3}=2.33$$

Answer: The distance is 2.33 units.

Problem: Angle Between Planes

Find the acute angle between planes with normals n1 = <1, 2, 2> and n2 = <3, 0, 4>.

$$n_1\cdot n_2=11,\quad |n_1|=3,\quad |n_2|=5$$
$$\cos\theta=\frac{11}{15}\Rightarrow \theta=42.8^\circ$$

Answer: The acute angle is about 42.8°.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: t2115

MSTE - Analytic Geometry / Solid Analytic Geometry / Besavilla CE Pre-Board Math & Surveying

Find the equation of the locus of a point which is at a distance 6 from the point A(5, 3, 2).

  1. $x^2 + y^2 + z^2 - 10x - 6y - 4z + 3 = 0$
  2. $x^2 + y^2 + z^2 - 10x - 6y - 4z + 4 = 0$
  3. $x^2 + y^2 + z^2 - 10x - 6y - 4z + 2 = 0$
  4. $x^2 + y^2 + z^2 - 10x - 6y - 4z + 5 = 0$
  5. $x^2 + y^2 + z^2 - 10x - 6y - 4z + 1 = 0$
The locus is a sphere centered at $A(5,3,2)$ with radius 6.
$(x-5)^2+(y-3)^2+(z-2)^2=6^2$
Expand:
$x^2-10x+25+y^2-6y+9+z^2-4z+4=36$
$x^2+y^2+z^2-10x-6y-4z+2=0$
$\boxed{x^2+y^2+z^2-10x-6y-4z+2=0}$