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Lines and Planes in Space

A plane can be described using a point on the plane and a normal vector perpendicular to it.

$$A(x-x_0)+B(y-y_0)+C(z-z_0)=0$$

Plane From Point and Normal

Find the plane through $(1,2,3)$ with normal vector $\langle 2,-1,4\rangle$.

$$2(x-1)-1(y-2)+4(z-3)=0$$
$$2x-y+4z-12=0$$

Final answer: $2x-y+4z-12=0$.

Parametric Equations of a Line in 3D — CE Board

Write the parametric equations of the line passing through $P(2,\,-1,\,3)$ with direction vector $\vec{d} = \langle 1,\,4,\,-2\rangle$.

A line in 3D through point $(x_0,y_0,z_0)$ with direction $\langle a,b,c\rangle$ is given by:

$$x = x_0 + at,\quad y = y_0 + bt,\quad z = z_0 + ct$$

Substituting:

$$x = 2 + t$$
$$y = -1 + 4t$$
$$z = 3 - 2t$$

These describe every point on the line as $t$ varies over all real numbers.

Plane Through Three Points — CE Board

Find the equation of the plane through the points $P_1(1,0,0)$, $P_2(0,2,0)$, and $P_3(0,0,3)$.

Form two vectors lying in the plane:

$$\vec{P_1P_2} = \langle -1,2,0\rangle,\qquad \vec{P_1P_3} = \langle -1,0,3\rangle$$

Find the normal via cross product:

$$\vec{n} = \vec{P_1P_2}\times\vec{P_1P_3} = \begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\-1&2&0\\-1&0&3\end{vmatrix} = \langle 6,\,3,\,2\rangle$$

Plane equation using point $P_1(1,0,0)$:

$$6(x-1)+3(y-0)+2(z-0)=0$$
$$\boxed{6x + 3y + 2z = 6}$$

Tip: Alternatively, use intercept form $\dfrac{x}{1}+\dfrac{y}{2}+\dfrac{z}{3}=1$ and multiply through by 6.

Problem: Parametric Line Through Two Points

Find parametric equations for the line through A(1, 2, 3) and B(4, 0, 5).

$$\vec d=B-A=\langle3,-2,2\rangle$$

Answer: $x=1+3t,\ y=2-2t,\ z=3+2t$.

Problem: Plane from Point and Normal

Find the plane through (2, -1, 4) with normal vector <3, 1, -2>.

$$3(x-2)+(y+1)-2(z-4)=0$$
$$3x+y-2z+3=0$$

Answer: The plane is $3x+y-2z+3=0$.

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Exam Generator Problems

Additional board-style practice items for this topic.

Question Bank: q703

MSTE - Analytic Geometry / Solid Analytic Geometry / Engr. Janclyde Espinosa (Clidez)

Point P has cylindrical coordinates (8, 30°, 5). Find the value of x in the Cartesian coordinates.

  1. 6.93
  2. 5.21
  3. 6.12
  4. 5.94
For cylindrical coordinates $(r,\theta,z)$, Cartesian $x=r\cos\theta$. Thus:
$x=8\cos30^\circ$
$x=8(0.8660)=6.928$
$\boxed{6.93}$

Question Bank: w258

MSTE - Analytic Geometry / Solid Analytic Geometry / MSTE April 2025

Find the angle between the planes $4x - 2y + 6z + 15 = 0$ and $15x + 10y - 5z - 27 = 0$.

  1. 78°67′
  2. 65°14′
  3. 75°34′
  4. 85°54′
The angle between two planes equals the angle between their normal vectors, whose components are the coefficients of $x$, $y$, and $z$.
$A = 4i - 2j + 6k$
$B = 15i + 10j - 5k$

$\cos\theta = \dfrac{A\cdot B}{|A|\times|B|}$
$\cos\theta = \dfrac{4(15) - 2(10) + 6(-5)}{\sqrt{4^{2}+2^{2}+6^{2}}\ \times\ \sqrt{15^{2}+10^{2}+5^{2}}}$
$\cos\theta = \dfrac{60 - 20 - 30}{\sqrt{56}\ \sqrt{350}}$
$\boxed{\theta = 85^\circ 54\prime 14.24\prime\prime}$