Free math worksheets

- Recall the defining magnitude of a unit vector. - Check the magnitude of each candidate. - Axis-aligned vectors can often be judged without much computation.

A unit vector has magnitude \(1\). Here, \(\lVert\mathbf{a}\rVert=1\), while \(\lVert\mathbf{b}\rVert=\sqrt{2}\) and \(\lVert\mathbf{c}\rVert=2\). Therefore, \(\mathbf{a}\) is the unit vector.

\(\mathbf{a}=\begin{pmatrix}0\\1\\0\end{pmatrix}\)

- A vector from one point to another records the coordinate changes. - Subtract the starting point from the ending point. - Keep the horizontal and vertical changes in the same order as the coordinates.

Subtract the coordinates of \(A\) from the coordinates of \(B\): \(\overrightarrow{AB}=\begin{pmatrix}2-(-1)\\6-2\end{pmatrix}=\begin{pmatrix}3\\4\end{pmatrix}\).

\(\begin{pmatrix}3\\4\end{pmatrix}\)

- Subtract the coordinates of \(A\) from those of \(B\). - Find the vector's magnitude. - Divide the vector by its magnitude.

1. \(\overrightarrow{AB}=\begin{pmatrix}4\\4\\2\end{pmatrix}\). 2. Its magnitude is \(\sqrt{4^2+4^2+2^2}=\sqrt{36}=6\). 3. Divide by the magnitude: \(\frac{\overrightarrow{AB}}{\|\overrightarrow{AB}\|} =\frac{1}{6}\begin{pmatrix}4\\4\\2\end{pmatrix} =\begin{pmatrix}\frac{2}{3}\\\frac{2}{3}\\\frac{1}{3}\end{pmatrix}\).

a) \(\begin{pmatrix}4\\4\\2\end{pmatrix}\) b) \(6\) c) \(\begin{pmatrix}\frac{2}{3}\\\frac{2}{3}\\\frac{1}{3}\end{pmatrix}\)

- Find the magnitude of \(\vec{v}\). - Divide by that magnitude to obtain a unit vector. - The original vector is its magnitude times the unit vector.

1. \(\|\vec{v}\|=\sqrt{(-4)^2+7^2+4^2}=\sqrt{81}=9\). 2. The unit vector in the direction of \(\vec{v}\) is \(\vec{u}=\frac{1}{9}\vec{v} =\begin{pmatrix}-\frac{4}{9}\\\frac{7}{9}\\\frac{4}{9}\end{pmatrix}\). 3. Therefore, \(\vec{v}=9\vec{u}\).

\(\vec{v}=9\begin{pmatrix}-\frac{4}{9}\\\frac{7}{9}\\\frac{4}{9}\end{pmatrix}\)

- Read the vector's components from the endpoint in the diagram. - A unit vector keeps the direction but has magnitude \(1\). - Divide the vector by its magnitude.

From the diagram, \(\mathbf{v}=\begin{pmatrix}-4\\3\end{pmatrix}\) and \(\lVert\mathbf{v}\rVert=\sqrt{(-4)^2+3^2}=5\). Therefore, the unit vector is \(\frac{1}{5}\mathbf{v}=\begin{pmatrix}-\frac{4}{5}\\\frac{3}{5}\end{pmatrix}\).

\(\begin{pmatrix}-\frac{4}{5}\\\frac{3}{5}\end{pmatrix}\)

- A vector can be written as its magnitude times a unit vector in its direction. - The given direction vector already has magnitude \(1\). - Scale every component by the required magnitude.

Multiply the unit direction vector by the required magnitude: \(\mathbf{v}=14\mathbf{u}=\begin{pmatrix}4\\-6\\12\end{pmatrix}\).

\(\begin{pmatrix}4\\-6\\12\end{pmatrix}\)

- The magnitude of a nonzero vector is positive. - Apply the scalar-magnitude rule directly. - A vector of magnitude \(1\) is called a unit vector. - A positive scalar does not reverse direction.

1. Since \(\mathbf{v}\neq\mathbf{0}\), \(\lVert\mathbf{v}\rVert>0\). Therefore, \(\lVert\mathbf{u}\rVert =\left|\frac{1}{\lVert\mathbf{v}\rVert}\right|\lVert\mathbf{v}\rVert =\frac{1}{\lVert\mathbf{v}\rVert}\lVert\mathbf{v}\rVert =1\). 2. The scalar \(\frac{1}{\lVert\mathbf{v}\rVert}\) is positive, so \(\mathbf{u}\) points in the same direction as \(\mathbf{v}\). It is the unit vector in the direction of \(\mathbf{v}\).

1. \(\lVert\mathbf{u}\rVert=1\) 2. \(\mathbf{u}\) is the unit vector in the direction of \(\mathbf{v}\).

- Find the magnitude of the direction vector. - Scale the direction vector to the stated speed. - Convert minutes to hours. - Add displacement to the starting position.

1. \(\lVert\mathbf{v}\rVert=\sqrt{3^2+4^2}=5\). 2. The velocity vector is \(\frac{120}{5}\mathbf{v}=\begin{pmatrix}72\\96\\0\end{pmatrix}\) miles per hour. 3. \(6\) minutes is \(0.1\) hour, so the displacement is \(0.1\begin{pmatrix}72\\96\\0\end{pmatrix} =\begin{pmatrix}7.2\\9.6\\0\end{pmatrix}\). 4. Add the displacement to the starting position: \((10, 20, 0.5)+(7.2, 9.6, 0)=(17.2, 29.6, 0.5)\).

\((17.2, 29.6, 0.5)\)

- Normalize the direction vector before applying the speed. - Multiply the velocity vector by the elapsed time. - Add the displacement to the starting point. - The \(z\)-coordinate gives the height.

1. \(\lVert\mathbf{r}\rVert=\sqrt{2^2+1^2+2^2}=3\). 2. The velocity vector is \(\frac{5}{3}\mathbf{r}=\begin{pmatrix}\frac{10}{3}\\\frac{5}{3}\\\frac{10}{3}\end{pmatrix}\) meters per second. 3. In \(15\) seconds, the displacement is \(15\begin{pmatrix}\frac{10}{3}\\\frac{5}{3}\\\frac{10}{3}\end{pmatrix} =\begin{pmatrix}50\\25\\50\end{pmatrix}\). 4. The new position is \((12, 8, 2)+(50, 25, 50)=(62, 33, 52)\). 5. The height is the \(z\)-coordinate, so the drone is \(52\,\text{m}\) high.

The drone is at \((62, 33, 52)\), and its height is \(52\,\text{m}\).

- Use the condition that the vector's magnitude is \(1\). - The unit-length equation will produce two possible signs for the missing component. - Use the stated quadrant to choose the correct sign.

Since \(\mathbf{u}\) is a unit vector, \(x^2+\left(\frac{12}{13}\right)^2=1\). Thus, \(x^2=1-\frac{144}{169}=\frac{25}{169}\), so \(x=\pm\frac{5}{13}\). A Quadrant II vector has a negative \(x\)-component, so \(x=-\frac{5}{13}\).

\(x=-\frac{5}{13}\)