Aimathic
Login | English | Deutsch

Precalculus math worksheets

Build your own math worksheets from 35,000+ problems for grades 3 to 12, from fractions to AP Calculus. Every problem comes with step-by-step solutions.

Unit vectors and component form

Learning goals

  • Writes vectors in component form.
  • Computes and uses standard unit vectors.
  • Normalizes a nonzero vector to obtain a unit vector.
  • Reconstructs a vector from magnitude and direction when appropriate.

Click problems to add them to your worksheet.

Unit vectors and component form 1
Which of these vectors is a unit vector? \(\mathbf{a}=\langle0,1,0\rangle\), \(\mathbf{b}=\langle1,1,0\rangle\), \(\mathbf{c}=\langle2,0,0\rangle\)

Hints

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

Solution

1. A unit vector has magnitude \(1\). 2. Here, \(\|\mathbf{a}\|=1\), while \(\|\mathbf{b}\|=\sqrt2\) and \(\|\mathbf{c}\|=2\). 3. Therefore, \(\mathbf{a}\) is the unit vector.

Answer

\(\mathbf{a}=\langle0,1,0\rangle\)
Unit vectors and component form 2
Points \(A=(-1, 2)\) and \(B=(2, 6)\) are given. Write \(\overrightarrow{AB}\) in component form.

Hints

- 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.

Solution

1. Subtract the coordinates of \(A\) from the coordinates of \(B\): \(\overrightarrow{AB}=\langle2-(-1),6-2\rangle=\langle3,4\rangle\).

Answer

\(\langle3,4\rangle\)
Unit vectors and component form 3
Write \(\mathbf{v}=\langle 3,-4,5\rangle\) in \(\mathbf{i},\mathbf{j},\mathbf{k}\) notation.

Hints

- Match the first component with \(\mathbf{i}\), the second with \(\mathbf{j}\), and the third with \(\mathbf{k}\). - Keep the sign of each component. - The coefficients in \(\mathbf{i},\mathbf{j},\mathbf{k}\) notation are the same components as in angle-bracket form.

Solution

1. The x-, y-, and z-components multiply \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\), respectively. 2. Therefore, \(\mathbf{v}=3\mathbf{i}-4\mathbf{j}+5\mathbf{k}\).

Answer

\(3\mathbf{i}-4\mathbf{j}+5\mathbf{k}\)
Unit vectors and component form 4
Can the zero vector \(\mathbf{0}=\langle 0,0,0\rangle\) be normalized to produce a unit vector? Explain.

Hints

- Recall the formula used to normalize a vector. - Find the magnitude of \(\mathbf{0}\). - Check whether the required division is defined.

Solution

1. Normalizing a nonzero vector \(\mathbf{v}\) requires dividing by its magnitude: \(\frac{\mathbf{v}}{\|\mathbf{v}\|}\). 2. The zero vector has magnitude \(\|\mathbf{0}\|=0\). 3. Dividing by \(0\) is undefined, so the zero vector cannot be normalized and has no associated unit direction vector.

Answer

No. The zero vector has magnitude \(0\), so normalization would require division by zero.
Unit vectors and component form 5
The diagram shows vector \(\mathbf{v}\). a) Write \(\mathbf{v}\) in \(\mathbf{i},\mathbf{j}\) notation. b) State its horizontal and vertical components.
Figure for problem 556099

Hints

- Read the horizontal and vertical changes from the initial point to the endpoint. - The horizontal component multiplies \(\mathbf{i}\), and the vertical component multiplies \(\mathbf{j}\). - Keep the sign that indicates whether the vertical change is upward or downward.

Solution

1. Reading the endpoint from the coordinate grid gives the component form \(\langle5,-2\rangle\). 2. Therefore, the vector is \(5\mathbf{i}-2\mathbf{j}\). 3. Its horizontal component is \(5\), and its vertical component is \(-2\).

Answer

a) \(5\mathbf{i}-2\mathbf{j}\) b) Horizontal component: \(5\); vertical component: \(-2\)
Unit vectors and component form 6
Which standard unit vector points one unit in the positive \(z\)-direction: \(\mathbf{i}\), \(\mathbf{j}\), or \(\mathbf{k}\)? Give its components.

Hints

- Each standard unit vector corresponds to exactly one coordinate axis. - Which component changes when you move only along the \(z\)-axis?

Solution

1. The standard basis directions correspond to the \(x\)-, \(y\)-, and \(z\)-axes, respectively. 2. The positive \(z\)-direction is \(\mathbf{k}=\langle0,0,1\rangle\).

Answer

\(\mathbf{k}=\langle0,0,1\rangle\)
Unit vectors and component form 7
Write \(\mathbf{v}=4\mathbf{i}-2\mathbf{j}+3\mathbf{k}\) in component form.

Hints

- The coefficients of the three standard unit vectors give the three components. - Keep the negative sign on the middle coefficient.

Solution

1. Read the coefficients of \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) as the \(x\)-, \(y\)-, and \(z\)-components. 2. The component form is \(\mathbf{v}=\langle4,-2,3\rangle\).

Answer

\(\langle4,-2,3\rangle\)
Unit vectors and component form 8
Find a unit vector in the same direction as \(\mathbf{v}=\langle0,0,-7\rangle\).

Hints

- A unit vector has length \(1\) but keeps the direction of the original nonzero vector. - How can you change the length without changing the sign of its nonzero component?

Solution

1. The magnitude of \(\mathbf{v}\) is \(7\). 2. Divide its components by the magnitude: \(\frac{\mathbf{v}}{\|\mathbf{v}\|}=\langle0,0,-1\rangle\).

Answer

\(\langle0,0,-1\rangle\)
Unit vectors and component form 9
For each condition, give one point \(P\) and one nonzero direction vector \(\mathbf{u}\) in component form that satisfy it. Do not write a line equation. a) The direction is along the x-axis. b) The direction is parallel to the z-axis, and the point is \((2, 5, 0)\). c) Both the point and direction lie in the xz-plane, the point is not the origin, and the direction is not parallel to either coordinate axis. d) The point lies in the horizontal plane \(z=4\), and the direction is parallel to that plane but not parallel to either the x-axis or the y-axis.

Hints

- Read each coordinate condition as a restriction on vector components. - A direction parallel to a coordinate axis has only the corresponding component nonzero. - A direction parallel to a horizontal plane has zero vertical component.

Solution

1. Along the x-axis, a valid choice is \(P=(0, 0, 0)\) and \(\mathbf{u}=\langle1,0,0\rangle\). 2. Parallel to the z-axis, use the required point \(P=(2, 5, 0)\) and a vector with only a z-component, such as \(\mathbf{u}=\langle0,0,1\rangle\). 3. To stay in the xz-plane, both y-components must be zero. One valid choice is \(P=(1, 0, 0)\) and \(\mathbf{u}=\langle1,0,1\rangle\). 4. For a direction parallel to \(z=4\), the z-component of the direction must be zero. One valid choice is \(P=(0, 0, 4)\) and \(\mathbf{u}=\langle1,1,0\rangle\).

Answer

Answers vary. One set of valid choices is: a) \(P=(0, 0, 0)\), \(\mathbf{u}=\langle1,0,0\rangle\) b) \(P=(2, 5, 0)\), \(\mathbf{u}=\langle0,0,1\rangle\) c) \(P=(1, 0, 0)\), \(\mathbf{u}=\langle1,0,1\rangle\) d) \(P=(0, 0, 4)\), \(\mathbf{u}=\langle1,1,0\rangle\)
Unit vectors and component form 10
Let \(\mathbf{v}=\langle 2,4,-6\rangle\). a) Find \(\|\mathbf{v}\|\). b) Find a vector in the same direction as \(\mathbf{v}\) with magnitude \(\sqrt{14}\). c) Find a unit vector in the same direction as \(\mathbf{v}\).

Hints

- Magnitude comes from the square root of the sum of the squared components. - Scaling a vector by a positive scalar preserves its direction and scales its magnitude by the same factor. - A unit vector has magnitude \(1\).

Solution

1. \(\|\mathbf{v}\|=\sqrt{2^2+4^2+(-6)^2}=\sqrt{56}=2\sqrt{14}\). 2. A vector in the same direction with half the magnitude is \(\frac12\mathbf{v}=\langle 1,2,-3\rangle\), whose magnitude is \(\sqrt{14}\). 3. Divide \(\mathbf{v}\) by its magnitude: \(\frac{\mathbf{v}}{\|\mathbf{v}\|}=\left\langle\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},-\frac{3}{\sqrt{14}}\right\rangle\).

Answer

a) \(2\sqrt{14}\) b) \(\langle 1,2,-3\rangle\) c) \(\left\langle\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},-\frac{3}{\sqrt{14}}\right\rangle\)
Unit vectors and component form 11
A nonzero vector \(\mathbf{u}=\langle u_x,u_y,u_z\rangle\) lies in the xz-plane and satisfies \(5u_x-2u_z=0\). a) Give a direction vector with the smallest positive integer x- and z-components that satisfies both conditions. b) Determine whether \(\mathbf{q}=\langle4,0,10\rangle\) is a scalar multiple of your vector from part a. If it is, state the scalar.

Hints

- A vector in the xz-plane must have a zero y-component. - Rewrite the linear relation as a ratio between the x- and z-components. - In part b, the same scalar must work for all three components.

Solution

1. Because \(\mathbf{u}\) lies in the xz-plane, \(u_y=0\). The relation \(5u_x=2u_z\) is satisfied by the smallest positive integer pair \(u_x=2\), \(u_z=5\). Thus, one direction vector is \(\langle2,0,5\rangle\). 2. \(\langle4,0,10\rangle=2\langle2,0,5\rangle\), so \(\mathbf{q}\) is a scalar multiple with scalar \(2\).

Answer

a) \(\langle2,0,5\rangle\) b) Yes; \(\mathbf{q}=2\mathbf{u}\).
Unit vectors and component form 12
The points are \(A=(3, -2, 1)\) and \(B=(7, 2, 3)\). a) Find \(\overrightarrow{AB}\). b) Find the distance \(AB\). c) Give a unit vector in the direction of \(\overrightarrow{AB}\).

Hints

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

Solution

1. \(\overrightarrow{AB}=\langle4,4,2\rangle\). 2. Its magnitude is \(\sqrt{4^2+4^2+2^2}=\sqrt{36}=6\). 3. Divide by the magnitude: \(\frac{\overrightarrow{AB}}{\|\overrightarrow{AB}\|}=\frac16\langle4,4,2\rangle=\left\langle\frac23,\frac23,\frac13\right\rangle\).

Answer

a) \(\langle4,4,2\rangle\) b) \(6\) c) \(\left\langle\frac23,\frac23,\frac13\right\rangle\)
Unit vectors and component form 13
Write \(\mathbf{v}=\langle-4,7,4\rangle\) in the form \(k\mathbf{u}\), where \(k>0\) and \(\mathbf{u}\) is a unit vector in the same direction as \(\mathbf{v}\).

Hints

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

Solution

1. \(\|\mathbf{v}\|=\sqrt{(-4)^2+7^2+4^2}=\sqrt{81}=9\). 2. The unit vector in the direction of \(\mathbf{v}\) is \(\mathbf{u}=\frac19\mathbf{v}=\left\langle-\frac49,\frac79,\frac49\right\rangle\). 3. Therefore, \(\mathbf{v}=9\mathbf{u}\).

Answer

\(\mathbf{v}=9\left\langle-\frac49,\frac79,\frac49\right\rangle\)
Unit vectors and component form 14
Find the unit vector in the same direction as \(\mathbf{v}\).
Figure for problem 551850

Hints

- Read the endpoint coordinates of the origin-based vector from the diagram. - A unit vector keeps the direction but has magnitude \(1\). - Divide the vector by its magnitude.

Solution

1. From the graph, \(\mathbf{v}=\langle-4,3\rangle\) and \(\|\mathbf{v}\|=\sqrt{(-4)^2+3^2}=5\). 2. The unit vector is \(\frac{1}{5}\mathbf{v}=\left\langle-\frac{4}{5},\frac{3}{5}\right\rangle\).

Answer

\(\left\langle-\frac{4}{5},\frac{3}{5}\right\rangle\)
Unit vectors and component form 15
A vector \(\mathbf{v}\) has magnitude \(14\) and points in the direction of the unit vector \(\mathbf{u}=\left\langle\frac27,-\frac37,\frac67\right\rangle\). Find \(\mathbf{v}\) in component form.

Hints

- 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.

Solution

1. Multiply the unit direction vector by the required magnitude: \(\mathbf{v}=14\mathbf{u}=\langle4,-6,12\rangle\).

Answer

\(\langle4,-6,12\rangle\)
Unit vectors and component form 16
The vector \(\mathbf{u}=\left\langle x,\frac{12}{13}\right\rangle\) is a unit vector that points into Quadrant II. Find \(x\).

Hints

- 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.

Solution

1. Since \(\mathbf{u}\) is a unit vector, \(x^2+\left(\frac{12}{13}\right)^2=1\). 2. Thus, \(x^2=1-\frac{144}{169}=\frac{25}{169}\), so \(x=\pm\frac5{13}\). 3. A Quadrant II vector has a negative x-component, so \(x=-\frac5{13}\).

Answer

\(x=-\frac5{13}\)
Unit vectors and component form 17
A vector has magnitude \(14\) and direction angle \(60^\circ\). Write the vector in exact component form.

Hints

- Treat the magnitude as the hypotenuse of a component triangle. - The x-component uses cosine of the direction angle. - The y-component uses sine of the direction angle.

Solution

1. For magnitude \(r\) and direction angle \(\theta\), the components are \(\langle r\cos\theta,r\sin\theta\rangle\). 2. Substitute \(r=14\) and \(\theta=60^\circ\): \(\langle 14\cos60^\circ,14\sin60^\circ\rangle\). 3. Therefore, the vector is \(\langle 7,7\sqrt3\rangle\).

Answer

\(\langle 7,7\sqrt3\rangle\)
Unit vectors and component form 18
Find a unit vector in the direction of \(\mathbf{v}=\langle3,4,0\rangle\). Show that your answer has length \(1\).

Hints

- Find the length of the original vector first. - Divide every component by that length. - Use the magnitude formula on your result to check it.

Solution

1. The magnitude is \(\|\mathbf{v}\|=\sqrt{3^2+4^2}=5\). 2. Divide by \(5\): \(\mathbf{u}=\langle\frac{3}{5},\frac{4}{5},0\rangle\). 3. Check: \(\|\mathbf{u}\|=\sqrt{\frac{9}{25}+\frac{16}{25}}=1\).

Answer

\(\mathbf{u}=\langle\frac{3}{5},\frac{4}{5},0\rangle\); its length is \(1\) because \(\frac{9}{25}+\frac{16}{25}=1\).
Unit vectors and component form 19
A vector has magnitude \(6\) and points in the unit direction \(\mathbf{u}=\langle\frac{1}{3},\frac{2}{3},\frac{2}{3}\rangle\). Find the vector in component form.

Hints

- Unit direction gives the orientation but not the full length. - What scalar stretches a unit vector to the stated magnitude? - Check by computing the norm of your result.

Solution

1. A vector with magnitude \(6\) and unit direction \(\mathbf{u}\) equals \(6\mathbf{u}\). 2. Multiply the components to obtain \(\langle2,4,4\rangle\). 3. The length \(\sqrt{2^2+4^2+4^2}=6\) confirms the given magnitude.

Answer

\(\langle2,4,4\rangle\)
Unit vectors and component form 20
Let \(\mathbf{v}\in\mathbb{R}^3\) be any nonzero vector. 1. Using \(\|r\mathbf{a}\|=|r|\|\mathbf{a}\|\), show that \(\mathbf{u}=\frac{1}{\|\mathbf{v}\|}\mathbf{v}\) has magnitude \(1\). 2. What is the geometric meaning of \(\mathbf{u}\) relative to \(\mathbf{v}\)?

Hints

- 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.

Solution

1. Because \(\mathbf{v}\neq\mathbf{0}\), its magnitude is positive and the scaling factor is defined. 2. The magnitude rule gives \(\|\mathbf{u}\|=\left|\frac{1}{\|\mathbf{v}\|}\right|\|\mathbf{v}\|=\frac{1}{\|\mathbf{v}\|}\|\mathbf{v}\|=1\). 3. The scale factor is positive, so \(\mathbf{u}\) points in the same direction as \(\mathbf{v}\) and has unit length.

Answer

1. Since \(\mathbf{v}\neq\mathbf{0}\), \(\|\mathbf{v}\|>0\) and \(\|\mathbf{u}\|=\left|\frac{1}{\|\mathbf{v}\|}\right|\|\mathbf{v}\|=1\). 2. \(\mathbf{u}\) is the unit vector pointing in the same direction as \(\mathbf{v}\).
Unit vectors and component form 21
A helicopter starts at \(P=(10, 20, 0.5)\), with coordinates measured in miles. It flies at a constant speed of \(120\) miles per hour in the direction \(\mathbf{v}=\langle3,4,0\rangle\). Find its position after \(6\) minutes.

Hints

- 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.

Solution

1. \(\|\mathbf{v}\|=\sqrt{3^2+4^2}=5\). 2. The velocity vector is \(\frac{120}{5}\mathbf{v}=\langle72,96,0\rangle\) miles per hour. 3. \(6\) minutes is \(0.1\) hour, so the displacement is \(0.1\langle72,96,0\rangle=\langle7.2,9.6,0\rangle\). 4. Add the displacement to the starting position: \((10, 20, 0.5)+\langle7.2,9.6,0\rangle=(17.2, 29.6, 0.5)\).

Answer

\((17.2, 29.6, 0.5)\)
Unit vectors and component form 22
A research drone starts at \(A=(12, 8, 2)\), with coordinates measured in meters. It moves in a straight line at a constant speed of \(5\,\text{m/s}\) in the direction \(\mathbf{r}=\langle2,1,2\rangle\). Find the drone's coordinates after \(15\) seconds and its height if the z-axis represents height.

Hints

- 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.

Solution

1. \(\|\mathbf{r}\|=\sqrt{2^2+1^2+2^2}=3\). 2. The velocity vector is \(\frac53\mathbf{r}=\left\langle\frac{10}{3},\frac53,\frac{10}{3}\right\rangle\) meters per second. 3. In \(15\) seconds, the displacement is \(15\left\langle\frac{10}{3},\frac53,\frac{10}{3}\right\rangle=\langle50,25,50\rangle\). 4. The new position is \((12, 8, 2)+\langle50,25,50\rangle=(62, 33, 52)\). 5. The height is the z-coordinate, so the drone is \(52\,\text{m}\) high.

Answer

The drone is at \((62, 33, 52)\), and its height is \(52\,\text{m}\).
Unit vectors and component form 23
The vector \(\mathbf{v}=a\mathbf{i}-12\mathbf{j}+0\mathbf{k}\) has magnitude \(13\), where \(a>0\). a) Find \(a\). b) Find the unit vector in the direction of \(\mathbf{v}\), in \(\mathbf{i},\mathbf{j},\mathbf{k}\) notation.

Hints

- Use the magnitude equation to determine the missing coefficient before normalizing. - The sign condition on \(a\) selects one of the two square-root possibilities. - A unit vector is obtained by dividing each component by the vector's magnitude.

Solution

1. The magnitude condition gives \(a^2+(-12)^2+0^2=13^2\). 2. Thus \(a^2=25\). Since \(a>0\), \(a=5\). 3. Therefore, \(\mathbf{v}=5\mathbf{i}-12\mathbf{j}\). 4. Divide by the magnitude \(13\): \(\frac{\mathbf{v}}{\|\mathbf{v}\|}=\frac{5}{13}\mathbf{i}-\frac{12}{13}\mathbf{j}+0\mathbf{k}\).

Answer

a) \(a=5\) b) \(\frac{5}{13}\mathbf{i}-\frac{12}{13}\mathbf{j}+0\mathbf{k}\)

All problems may be used, copied and printed free of charge for school and tutoring, including paid tutoring. Commercial adaptations as well as publication or redistribution on the internet are not permitted.

Building free math worksheets