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Unit vectors and component form

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55184812
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

\(\mathbf{a}=\langle0,1,0\rangle\)
55184912
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

\(\langle3,4\rangle\)
55609712
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

\(3\mathbf{i}-4\mathbf{j}+5\mathbf{k}\)
55609812
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.
52547212
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) \(P=(0, 0, 4)\), \(\mathbf{u}=\langle1,1,0\rangle\)
52548112
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) \(\left\langle\frac{1}{\sqrt{14}},\frac{2}{\sqrt{14}},-\frac{3}{\sqrt{14}}\right\rangle\)
53038012
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; \(\mathbf{q}=2\mathbf{u}\).
53042512
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) \(\left\langle\frac23,\frac23,\frac13\right\rangle\)
53043312
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

\(\mathbf{v}=9\left\langle-\frac49,\frac79,\frac49\right\rangle\)
55185012
The diagram shows the segment from the origin to point \(P\). Interpret \(\mathbf{v}\) as the nonzero vector directed from the origin to \(P\). 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 diagram, \(\mathbf{v}=\langle-4,3\rangle\) and \(\|\mathbf{v}\|=\sqrt{(-4)^2+3^2}=5\). 2. Therefore, the unit vector is \(\frac15\mathbf{v}=\left\langle-\frac45,\frac35\right\rangle\).

Answer

\(\left\langle-\frac45,\frac35\right\rangle\)
55185112
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

\(\langle4,-6,12\rangle\)
55185212
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

\(x=-\frac5{13}\)
55609912
The diagram shows segment \(OP\). Interpret it as the vector directed from \(O\) to \(P\). a) Write the vector 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 \(\langle 5,-2\rangle\). 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: \(-2\)
55610012
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

\(\langle 7,7\sqrt3\rangle\)
52777012
Let \(\mathbf{v}\in\mathbb{R}^3\) be any nonzero vector. 1. Using \(\lVert r\mathbf{a}\rVert=|r|\lVert\mathbf{a}\rVert\), show that \(\mathbf{u}=\frac{1}{\lVert\mathbf{v}\rVert}\mathbf{v}\) 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. 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 \(\mathbf{v}\).

Answer

1. \(\lVert\mathbf{u}\rVert=1\) 2. \(\mathbf{u}\) is the unit vector in the direction of \(\mathbf{v}\).
52777312
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

\((17.2, 29.6, 0.5)\)
52777412
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 \(52\,\text{m}\).
55610112
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) \(\frac{5}{13}\mathbf{i}-\frac{12}{13}\mathbf{j}+0\mathbf{k}\)

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