CE Board Exam Randomizer

Question 1

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.

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

Answer

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

$$d=\frac{7}{3}=2.33$$

Final answer: $2.33$ units.

Answer

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

$$\vec{n_1} = \langle 1,\,2,\,-2\rangle,\qquad \vec{n_2} = \langle 3,\,0,\,-4\rangle$$

$$\vec{n_1}\cdot\vec{n_2} = (1)(3)+(2)(0)+(-2)(-4) = 3+0+8 = 11$$

$$|\vec{n_1}| = \sqrt{1+4+4} = 3,\qquad |\vec{n_2}| = \sqrt{9+0+16} = 5$$

$$\cos\theta = \frac{11}{3\cdot 5} = \frac{11}{15}$$

$$\theta = \cos^{-1}\!\left(\frac{11}{15}\right) \approx \boxed{42.8^\circ}$$

Answer

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:

$$\vec{AP}\times\vec{d} = \begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\1&1&1\\1&2&2\end{vmatrix} = \langle (2-2),\,-(2-1),\,(2-1)\rangle = \langle 0,\,-1,\,1\rangle$$

$$|\vec{AP}\times\vec{d}| = \sqrt{0+1+1} = \sqrt{2}$$

$$|\vec{d}| = \sqrt{1+4+4} = 3$$

$$d = \frac{\sqrt{2}}{3} \approx \boxed{0.471 \text{ units}}$$

Question 2

Distance From Point to Plane

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

Question 3

Angle Between Two Planes — CE Board

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

Question 4

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 $\vec{d} = \langle 1,\,2,\,2\rangle$.