MCV4U Dot and Cross Product Practice Test (with Solutions)

Free Cartesian Vectors Practice Test

8 questions (12 parts), exactly as a free account gets them in the first Cartesian Vectors test on Martian Lab. Try each one on paper first, then open the answer.

1. Consider position vector $\vec{u} = [-2, 3, -4]$,

   1. Draw $\vec{u}$ by showing it with a prism.
      
      Show solution

      
   2. Find the exact value of the magnitude of $\vec{u}$.
      Show answer

      $|\vec{u}| = \sqrt{(-2)^2 + (3)^2 + (-4)^2}$ $= \sqrt{4 + 9 + 16}$ $= \sqrt{29}$
   3. Find the unit vector that is in the direction of $\vec{u}$.
      Show answer

      $\dfrac{1}{|\vec{u}|}\vec{u}$ $= \dfrac{1}{\sqrt{29}}[-2, 3, -4]$ $= \left[-\dfrac{2}{\sqrt{29}}, \dfrac{3}{\sqrt{29}}, -\dfrac{4}{\sqrt{29}}\right]$
2. Given that $\vec{u} = [-4, 1, -8]$ and $\vec{v} = [8, -2, 1]$, determine the angle between $\vec{u}$ and $\vec{v}$.

   Show solution

   $\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta$ $\cos\theta = \dfrac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|}$ $= \dfrac{[-4, 1, -8][8, -2, 1]}{9\sqrt{69}}$ $= \dfrac{-42}{9\sqrt{69}}$ $\theta = \cos^{-1}\!\left(\dfrac{-42}{9\sqrt{69}}\right)$ $\doteq 124.18^\circ$ $|\vec{u}| = \sqrt{(-4)^2 + (1)^2 + (-8)^2}$ $= 9$ $|\vec{v}| = \sqrt{(8)^2 + (-2)^2 + (1)^2}$ $= \sqrt{69}$ $\therefore\ \theta = 124.18^\circ$
3. Find $3\vec{u} - 2\vec{v}$ where $\vec{u} = [-4, 1, -8]$ and $\vec{v} = [8, -2, 1]$.

   Show answer

   $3\vec{u} - 2\vec{v}$ $= 3[-4, 1, -8] - 2[8, -2, 1]$ $= [-12, 3, -24] - [16, -4, 2]$ $= [-12 - 16,\; 3 + 4,\; -24 - 2]$ $= [-28, 7, -26]$ $\therefore\ 3\vec{u} - 2\vec{v} = [-28, 7, -26]$
4. Given $\vec{u} = [2, -3, 3]$ and $\vec{v} = [3, 1, 4]$, determine the vector projection of $\vec{v}$ onto $\vec{u}$.

   Show solution

   $\operatorname{proj}_{\vec{u}}\vec{v} = \left(\dfrac{\vec{v} \cdot \vec{u}}{\vec{u} \cdot \vec{u}}\right)\vec{u}$ $= \dfrac{[3, 1, 4][2, -3, 3]}{[2, -3, 3][2, -3, 3]}\,[2, -3, 3]$ $= \dfrac{15}{22}\,[2, -3, 3]$ $= \left[\dfrac{15}{11},\; -\dfrac{45}{22},\; \dfrac{45}{22}\right]$ $\therefore\ \operatorname{proj}_{\vec{u}}\vec{v} = \left[\dfrac{15}{11},\; -\dfrac{45}{22},\; \dfrac{45}{22}\right]$
5. Determine whether ${(\vec{a} \cdot \vec{b})\vec{c}} \times \vec{a}$ is a vector, is a scalar or is meaningless. Explain your reasoning.

   Show solution

   This gives a vector because $(\vec{a} \cdot \vec{b})$ gives a scalar. $(\vec{a} \cdot \vec{b})\vec{c}$ is scalar times vector, which gives a vector, and the cross product with $\vec{a}$ gives a vector.
6. What is the relationship between $\vec{u}$, $\vec{v}$ and $\vec{w}$ if $(\vec{u} \times \vec{v})\vec{w} = 0$? Explain.

   Show solution

   $\vec{u} \times \vec{v} \cdot \vec{w} \rightarrow \text{the volume of parallelepiped}$ $(\vec{u} \times \vec{v})\vec{w} = 0 \rightarrow \text{volume is } 0$ $\vec{u} \times \vec{v} = 0$ It can be that $\vec{u}$ and $\vec{v}$ are collinear/parallel. $\vec{w}$ is orthogonal with $(\vec{u} \times \vec{v})$. Any two vectors are collinear can result in a flat plane.
7. Given $\vec{a} = [6, 2, 3]$ and $\vec{b} = [-3, -1, z]$. For what values of $z$ are the vectors:

   1. collinear?
      Show solution

      $\vec{a} = k\vec{b}$ $[6, 2, 3] = k[-3, -1, z]$ $6 = -3k$ $2 = -k$ $k = -2$ $3 = kz$ $3 = (-2)z$ $z = -\dfrac{3}{2}$ $\therefore\ z$ is $-\dfrac{3}{2}$ so that $\vec{a}$ is collinear with $\vec{b}$
   2. orthogonal?
      Show answer

      $\vec{a} \cdot \vec{b} = 0$ $[6, 2, 3] \cdot [-3, -1, z] = 0$ $-18 + (-2) + (3z) = 0$ $-20 = -3z$ $z = \dfrac{20}{3}$ $\therefore\ z$ is $\dfrac{20}{3}$ so that $\vec{a} \perp \vec{b}$
8. Consider vectors $\vec{a} = [3, -5, -1]$, $\vec{b} = [4, 1, 0]$ and $\vec{c} = [-2, -5, -1]$.

   1. Find the area of the parallelogram defined by $\vec{b}$ and $\vec{c}$.
      Show solution

      $\vec{b} \times \vec{c} = [4, 1, 0] \times [-2, -5, -1]$ $= [-1 - 0,\; 0 - (-4),\; -20 - (-2)]$ $= [-1, 4, -18]$ $|\vec{b} \times \vec{c}| = \sqrt{(-1)^2 + (4)^2 + (-18)^2} = \sqrt{341}\text{ units}^2$ $\therefore$ area of parallelogram defined by $\vec{b} \times \vec{c}$ is $\sqrt{341}$ units$^2$
   2. Find the volume of the parallelepiped defined by $\vec{a}$, $\vec{b}$, and $\vec{c}$.
      Show solution

      $\left|(\vec{a} \times \vec{b}) \cdot \vec{c}\right|$ $= \left([3, -5, -1] \times [4, 1, 0]\right) \cdot [-2, -5, -1]$ $= [1, -4, 23] \cdot [-2, -5, -1]$ $= -2 + (20) + (-23)$ $= \left|-5\right| = 5\text{ units}^3$ $\therefore$ the volume of the parallelepiped is 5 units$^3$.

This Unit on Martian Lab

The lessons go from algebraic vectors through the dot product and the cross product to their applications and linear combinations.

1. Algebraic Vectors
2. Dot Product
3. Cross Product
4. Applications of the Dot & Cross Products
5. Linear Combination

Then 4 full Cartesian Vectors practice tests to check yourself.

Cartesian Vectors FAQ

Cartesian Vectors Topics

One idea at a time: a short explanation, then five worked examples.

• Dot Product and Projections
• Cross Product, Area and Torque

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