CE Board Exam Randomizer

Question 1

Vector Operations

Vector operations include addition, scalar multiplication, magnitude, dot product, and cross product.

$$\vec{a}\cdot\vec{b}=|\vec{a}||\vec{b}|\cos\theta$$

Answer

$$\vec{a}\cdot\vec{b}=2(1)+(-1)(2)+2(2)=4$$

$$|\vec{a}|=3, \qquad |\vec{b}|=3$$

$$\theta=\cos^{-1}\left(\frac{4}{9}\right)=63.6^\circ$$

Final answer: $63.6^\circ$.

Answer

Magnitude:

$$|\vec{v}| = \sqrt{3^2 + (-4)^2 + 0^2} = \sqrt{9 + 16} = \sqrt{25} = 5$$

Unit vector (divide each component by the magnitude):

$$\hat{v} = \frac{\vec{v}}{|\vec{v}|} = \frac{1}{5}\langle 3,\,-4,\,0\rangle = \left\langle\frac{3}{5},\,-\frac{4}{5},\,0\right\rangle$$

Verify: $|\hat{v}| = \sqrt{(3/5)^2 + (4/5)^2} = \sqrt{9/25 + 16/25} = 1$ ✓

Answer

Expand the determinant:

$$\vec{a}\times\vec{b} = \begin{vmatrix}\vec{i}&\vec{j}&\vec{k}\\2&1&-1\\1&-1&2\end{vmatrix}$$

$$= \vec{i}[(1)(2)-(-1)(-1)] - \vec{j}[(2)(2)-(-1)(1)] + \vec{k}[(2)(-1)-(1)(1)]$$

$$= \vec{i}(2-1) - \vec{j}(4+1) + \vec{k}(-2-1) = \langle 1,\,-5,\,-3\rangle$$

Area of parallelogram = magnitude of cross product:

$$\text{Area} = |\vec{a}\times\vec{b}| = \sqrt{1^2+5^2+3^2} = \sqrt{35} \approx 5.92 \text{ sq. units}$$

Question 2

Angle Between Vectors

Find the angle between $\vec{a}=\langle 2,-1,2\rangle$ and $\vec{b}=\langle 1,2,2\rangle$.

Question 3

Vector Magnitude and Unit Vector — CE Board

Find the magnitude and unit vector of $\vec{v} = \langle 3,\,-4,\,0 \rangle$.

Question 4

Cross Product and Parallelogram Area — CE Board

Given $\vec{a} = \langle 2,\,1,\,-1\rangle$ and $\vec{b} = \langle 1,\,-1,\,2\rangle$, find $\vec{a}\times\vec{b}$ and the area of the parallelogram they form.