CE Board Exam Randomizer

⬅ Back to Vectors and Solid Analytic Geometry Topics

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.

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