Free math worksheets

- Read the horizontal and vertical components from point \(P\). - The magnitude is the distance from the origin to the endpoint. - Use the Pythagorean relationship between the two components and the magnitude.

1. From the endpoint, \(\mathbf{v}=\langle 3,4\rangle\). 2. Therefore, \(\|\mathbf{v}\|=\sqrt{3^2+4^2}=5\).

\(5\)

- Use point \(P\) to determine which axis contains the vector. - The specified direction is from the origin toward \(P\). - Recall the standard-position angle of the negative \(x\)-axis.

1. Point \(P\) lies directly to the left of the origin on the negative \(x\)-axis. 2. Therefore, the vector directed from the origin to \(P\) has standard direction angle \(180^\circ\).

\(180^\circ\)

- Read the horizontal and vertical changes from the coordinate grid. - Use the component changes to find the magnitude. - Determine the quadrant before converting a reference angle into a direction angle.

1. Reading the endpoint from the grid gives \(\mathbf{v}=\langle -3,4\rangle\). 2. Its magnitude is \(\|\mathbf{v}\|=\sqrt{(-3)^2+4^2}=5\). 3. The vector lies in quadrant II. Its reference angle is \(\tan^{-1}\left(\frac{4}{3}\right)\approx53.1^\circ\). 4. Therefore, the direction angle is \(180^\circ-53.1^\circ=126.9^\circ\).

a) \(5\) b) \(126.9^\circ\)

- A standard direction angle starts from east. - A quadrant bearing starts from north or south. - Use the complementary angle in the northeast quadrant.

1. A direction angle of \(30^\circ\) means \(30^\circ\) north of east. 2. Bearings are measured from north or south, so the angle from north is \(90^\circ-30^\circ=60^\circ\). 3. The direction is therefore \(N60^\circ E\).

\(N60^\circ E\).

- Subtract corresponding vector components. - Use the magnitude formula for \(a+bi\). - Interpret the magnitude of a difference vector.

1. \(w=(7-3)+(5-2)i=4+3i\). 2. \(|w|=\sqrt{4^2+3^2}=5\). 3. Because \(w=z_1-z_2\), \(|w|=|z_1-z_2|\) is the distance between the points represented by \(z_1\) and \(z_2\). The distance is \(5\) units.

1. \(w=4+3i\) 2. \(|w|=5\) 3. The two points are \(5\) units apart.

- Use \(\sqrt{x^2+y^2+z^2}\) for vector magnitude. - Add the vectors before finding the magnitude of the sum. - Equality in the triangle inequality occurs when the vectors point in the same direction.

1. \(\lVert\mathbf{u}\rVert=\sqrt{3^2+4^2}=5\), and \(\lVert\mathbf{v}\rVert=12\). 2. \(\mathbf{u}+\mathbf{v}=\begin{pmatrix}3\\4\\12\end{pmatrix}\), so \(\lVert\mathbf{u}+\mathbf{v}\rVert=\sqrt{3^2+4^2+12^2}=13\). 3. Since \(13\leq5+12=17\), the triangle inequality holds. 4. Equality requires the vectors to be parallel and point in the same direction. Here, \(\mathbf{u}\) and \(\mathbf{v}\) are perpendicular, so the inequality is strict.

a) \(\lVert\mathbf{u}\rVert=5\), \(\lVert\mathbf{v}\rVert=12\), \(\lVert\mathbf{u}+\mathbf{v}\rVert=13\) b) \(13\leq17\) c) Equality would require the vectors to be parallel and point in the same direction.

- Compute each coordinate separately. - Watch the sign when subtracting a negative component. - Use the three-dimensional magnitude formula after finding the vector.

1. Combine corresponding components: \(\mathbf{v}=\langle2\cdot1-3\cdot1+3,2\cdot0-3(-1)+0,2\cdot4-3\cdot2+4\rangle=\langle2,3,6\rangle\). 2. Find the magnitude: \(\|\mathbf{v}\|=\sqrt{2^2+3^2+6^2}=\sqrt{49}=7\).

\(\mathbf{v}=\langle2,3,6\rangle\); \(\|\mathbf{v}\|=7\)

- Perform the scalar multiplications first. - Apply the subtraction to every component of the last vector. - Use the magnitude formula after simplifying the vector.

1. Scale each vector: \(3\langle-1,2,1\rangle=\langle-3,6,3\rangle\), \(2\langle3,0,4\rangle=\langle6,0,8\rangle\), and \(4\langle0.5,0.5,0.75\rangle=\langle2,2,3\rangle\). 2. Combine the results: \(\mathbf{w}=\langle-3+6-2,6+0-2,3+8-3\rangle=\langle1,4,8\rangle\). 3. Find the magnitude: \(\|\mathbf{w}\|=\sqrt{1^2+4^2+8^2}=\sqrt{81}=9\).

\(\mathbf{w}=\langle1,4,8\rangle\); \(\|\mathbf{w}\|=9\)

- Evaluate the expression inside the brackets first. - Multiply every component by \(0.5\). - Use the magnitude formula for a vector in three dimensions.

1. Add inside the brackets: \(\langle1,-5,12\rangle+\langle5,13,12\rangle=\langle6,8,24\rangle\). 2. Multiply by \(0.5\): \(\mathbf{a}=\langle3,4,12\rangle\). 3. Find the magnitude: \(\|\mathbf{a}\|=\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\).

\(\mathbf{a}=\langle3,4,12\rangle\); \(\|\mathbf{a}\|=13\)

- Notice that both terms contain the same vector. - Combine the scalar coefficients before multiplying. - Squaring removes the sign of a negative component in the magnitude formula.

1. Combine the scalar coefficients: \(\mathbf{b}=(1.4+0.6)\langle2,5,-14\rangle=2\langle2,5,-14\rangle\). 2. Thus, \(\mathbf{b}=\langle4,10,-28\rangle\). 3. Its magnitude is \(\|\mathbf{b}\|=\sqrt{4^2+10^2+(-28)^2}=\sqrt{900}=30\).

\(\mathbf{b}=\langle4,10,-28\rangle\); \(\|\mathbf{b}\|=30\)

- A vector in the \(xy\)-plane has no vertical component. - A vector normal to the \(xy\)-plane points along the \(z\)-axis. - Add vectors componentwise, then use the magnitude formula.

1. A vector in the \(xy\)-plane has third component \(0\). Adding a zero component does not change the magnitude, so its three-dimensional magnitude remains \(8\). 2. A vector perpendicular to the \(xy\)-plane is parallel to the \(z\)-axis. One possible vector is \(\mathbf{v}=\langle 0,0,6\rangle\). 3. \(\mathbf{s}=\langle 3,4,12\rangle\), so \(\|\mathbf{s}\|=\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\).

1. Third component \(0\); magnitude \(8\) 2. \(\langle 0,0,6\rangle\) 3. \(13\)

- Square each component and add. - Negative signs disappear when squared. - Squaring a square root returns its radicand.

1. \(\|\mathbf{a}\|=\sqrt{(-8)^2+12^2+9^2}=\sqrt{289}=17\). 2. \(\|\mathbf{b}\|=\sqrt{13+15+21}=\sqrt{49}=7\). 3. \(\|\mathbf{c}\|=\sqrt{1.5^2+(-2)^2+6^2}=\sqrt{42.25}=6.5\).

a) \(17\) b) \(7\) c) \(6.5\)

- Use the squared magnitude to avoid carrying a square root. - Set the result equal to \(15^2\). - Include both square-root solutions.

1. The squared magnitude is \(\|\mathbf{v}\|^2=(2k)^2+k^2+10^2=5k^2+100\). 2. Set it equal to \(15^2\): \(5k^2+100=225\). 3. Thus, \(k^2=25\), so \(k=5\) or \(k=-5\).

\(k=5\) or \(k=-5\)

- Compute squared magnitudes first. - Squaring a negative component makes it positive. - Squaring a square root gives its radicand.

1. Compute squared magnitudes: \(\|\mathbf a\|^2=9\), \(\|\mathbf b\|^2=25\), \(\|\mathbf c\|^2=9\), \(\|\mathbf d\|^2=25\), \(\|\mathbf e\|^2=9\), \(\|\mathbf f\|^2=25\), \(\|\mathbf g\|^2=9\), and \(\|\mathbf h\|^2=25\). 2. Therefore, \(\mathbf a,\mathbf c,\mathbf e,\mathbf g\) have magnitude \(3\), while \(\mathbf b,\mathbf d,\mathbf f,\mathbf h\) have magnitude \(5\).

\(\mathbf a,\mathbf c,\mathbf e,\mathbf g\) have magnitude \(3\). \(\mathbf b,\mathbf d,\mathbf f,\mathbf h\) have magnitude \(5\).

- Compare squared magnitudes. - Square radical components carefully. - Look for a total of \(169\).

1. \(\|\mathbf{u}\|^2=12^2+4^2+3^2=169\), so \(\|\mathbf{u}\|=13\). 2. The squared magnitudes are \(\|\mathbf{v}_1\|^2=169\), \(\|\mathbf{v}_2\|^2=100+50+19=169\), \(\|\mathbf{v}_3\|^2=1+144+12=157\), \(\|\mathbf{v}_4\|^2=25+144=169\), \(\|\mathbf{v}_5\|^2=9+16+144=169\), and \(\|\mathbf{v}_6\|^2=64+64+16=144\). 3. Therefore, \(\mathbf{v}_1\), \(\mathbf{v}_2\), \(\mathbf{v}_4\), and \(\mathbf{v}_5\) have magnitude \(13\).

\(\mathbf{v}_1\), \(\mathbf{v}_2\), \(\mathbf{v}_4\), and \(\mathbf{v}_5\)

- Use both components in the magnitude formula. - Determine the vector's quadrant from the component signs. - An inverse tangent gives a reference angle unless its quadrant is interpreted correctly.

1. The magnitude is \(\sqrt{(-8)^2+(-6)^2}=10\). 2. Both components are negative, so the vector lies in quadrant III. 3. The reference angle is \(\tan^{-1}\left(\frac{6}{8}\right)\approx36.9^\circ\). 4. The direction angle is \(180^\circ+36.9^\circ=216.9^\circ\).

a) \(10\) b) \(216.9^\circ\)

- A bearing is measured clockwise from north, not counterclockwise from east. - Compare the eastward component with the northward component. - The displacement lies in the northeast quadrant, so the bearing is between \(000^\circ\) and \(090^\circ\).

1. The displacement vector is \(\langle 6,8\rangle\), where the first component is east and the second is north. 2. The vector lies east of north, so the bearing angle \(\beta\) satisfies \(\tan\beta=\frac{6}{8}\). 3. Thus \(\beta\approx36.9^\circ\). 4. In three-digit bearing notation, the bearing is \(036.9^\circ\).

\(036.9^\circ\)

- Think about vector addition using the parallelogram rule. - The magnitude of a complex number is its distance from the origin. - Compare the three vector lengths as sides of a triangle.

1. Add corresponding parts: \(s=(3-1)+(2+4)i=2+6i\). Geometrically, the vector for \(s\) is the diagonal of the parallelogram formed by the vectors for \(z_1\) and \(z_2\). 2. The magnitudes are \(|z_1|=\sqrt{3^2+2^2}=\sqrt{13}\), \(|z_2|=\sqrt{(-1)^2+4^2}=\sqrt{17}\), and \(|s|=\sqrt{2^2+6^2}=\sqrt{40}=2\sqrt{10}\). 3. Numerically, \(2\sqrt{10}\approx6.32\) and \(\sqrt{13}+\sqrt{17}\approx7.73\). Since \(6.32\le7.73\), the triangle inequality holds.

1. \(s=2+6i\); it is the diagonal vector from the parallelogram rule. 2. \(|z_1|=\sqrt{13}\), \(|z_2|=\sqrt{17}\), and \(|s|=2\sqrt{10}\) 3. \(2\sqrt{10}\le\sqrt{13}+\sqrt{17}\), so the triangle inequality is satisfied.

- Use the magnitude formula for each vector. - Equal magnitudes have equal squared magnitudes. - Apply the condition that \(z\) is positive.

1. \(\|\mathbf{a}\|=\sqrt{2^2+(-3)^2+6^2}=\sqrt{49}=7\). 2. Equal magnitudes require \(6^2+3^2+z^2=7^2\). 3. Thus, \(45+z^2=49\), so \(z^2=4\). 4. Since \(z>0\), \(z=2\).

a) \(7\) b) \(z=2\)

- Imagine changing the angle between the two vectors. - What happens when both vectors point along the same line in the same direction? - Apply the triangle inequality to the head-to-tail vector triangle.

1. The largest possible magnitude is \(7+4=11\), reached when the vectors point in the same direction. The smallest is \(|7-4|=3\), reached when they point in opposite directions. 2. The maximum occurs when \(\mathbf{a}\) and \(\mathbf{b}\) are parallel and point in the same direction. 3. In the head-to-tail vector triangle, the side representing \(\mathbf{a}+\mathbf{b}\) cannot be longer than the sum of the other two sides. Thus, \(\lVert\mathbf{a}+\mathbf{b}\rVert\leq\lVert\mathbf{a}\rVert+\lVert\mathbf{b}\rVert=11\).

1. Minimum: \(3\); maximum: \(11\) 2. The vectors are parallel and point in the same direction. 3. The triangle inequality gives \(\lVert\mathbf{a}+\mathbf{b}\rVert\leq7+4=11\).

- Subtract the coordinates of \(P\) from those of \(Q_k\). - Use the three-dimensional magnitude formula. - Solve both cases of the resulting squared equation.

1. \(\overrightarrow{PQ_k}=\langle k-3,4,4\rangle\). 2. When \(k=3\), \(\overrightarrow{PQ_3}=\langle 0,4,4\rangle\), so \(\|\overrightarrow{PQ_3}\|=\sqrt{32}=4\sqrt2\). 3. For magnitude \(6\), \((k-3)^2+4^2+4^2=36\). 4. Thus, \((k-3)^2=4\), so \(k=1\) or \(k=5\).

a) \(\langle k-3,4,4\rangle\) b) \(4\sqrt2\) c) \(k=1\) or \(k=5\)

- Compare vectors for opposite sides. - Under the inclusive definition, every parallelogram is a trapezoid. - In a parallelogram, equal adjacent side lengths establish a rhombus.

1. \(\overrightarrow{AB}=(3, 2, 1)\) and \(\overrightarrow{DC}=C-D=(3, 2, 1)\). 2. Since \(\overrightarrow{AB}=\overrightarrow{DC}\), one pair of opposite sides is equal and parallel. Thus, \(ABCD\) is a parallelogram and, under the inclusive definition, also a trapezoid. 3. \(\overrightarrow{BC}=(-1, 3, 2)\). 4. \(\|\overrightarrow{AB}\|=\sqrt{14}\) and \(\|\overrightarrow{BC}\|=\sqrt{14}\). 5. A parallelogram with two adjacent sides of equal length is a rhombus. Therefore, \(ABCD\) is also a rhombus.

\(ABCD\) is a trapezoid, a parallelogram, and a rhombus.

- Read the normal vector from the coordinate coefficients. - Compare its current magnitude with the required magnitude before choosing scale factors. - For the axis relationship, test what happens when \(x=z=0\).

1. A normal vector is \(\mathbf{n}=\langle3,0,-4\rangle\), so the dot-product equation is \(\langle3,0,-4\rangle\cdot\mathbf{x}=0\). 2. Its magnitude is \(\|\mathbf{n}\|=\sqrt{3^2+0^2+(-4)^2}=5\). A parallel normal vector of magnitude \(10\) must be \(2\mathbf{n}\) or \(-2\mathbf{n}\). Thus, the vectors are \(\langle6,0,-8\rangle\) and \(\langle-6,0,8\rangle\). 3. The \(y\)-component of every normal vector is \(0\), so the \(y\)-axis direction lies in the plane. The equation also contains the origin, so the entire \(y\)-axis lies in the plane.

a) \(\langle3,0,-4\rangle\cdot\mathbf{x}=0\) b) \(\langle6,0,-8\rangle\) and \(\langle-6,0,8\rangle\) c) The plane contains the \(y\)-axis.

- Use the magnitude equation before deciding the sign of the missing component. - The stated quadrant determines which square-root choice is valid. - Convert the reference angle to the standard direction angle using the quadrant.

1. The magnitude condition gives \((-6)^2+y^2=10^2\), so \(y^2=64\). 2. Thus \(y=\pm8\). Because the vector lies in quadrant III, its y-component must be negative, so \(y=-8\). 3. The reference angle is \(\tan^{-1}\left(\frac{8}{6}\right)\approx53.1^\circ\). 4. A quadrant-III direction angle is \(180^\circ+53.1^\circ=233.1^\circ\).

a) \(y=-8\) b) \(233.1^\circ\)

- Find the top of the vertical rod. - Compute the magnitude of the shadow line's direction vector. - Locate points at the stated distance from \(P\) in both directions along the line. - A sunlight direction points from the rod's top to a possible shadow point.

1. The top of the rod is \(Q(3,2,6)\). 2. The direction vector of \(h\) has magnitude \(\sqrt{2^2+2^2+(-1)^2}=3\). 3. Because the shadow length is \(3\), the shadow point of the rod's top can be one direction-vector length from \(P\) in either direction along \(h\). 4. The first possible shadow point is \(S_1=P+\langle 2,2,-1\rangle=(5,4,3)\). One representative sunlight direction is \(\overrightarrow{QS_1}=\langle 2,2,-3\rangle\). 5. The second possible shadow point is \(S_2=P-\langle 2,2,-1\rangle=(1,0,5)\). One representative sunlight direction is \(\overrightarrow{QS_2}=\langle -2,-2,-1\rangle\). 6. Positive scalar multiples of either representative describe the same corresponding direction.

One representative for each supported direction is \(\langle 2,2,-3\rangle\) or \(\langle -2,-2,-1\rangle\).

- Count ordered pairs of distinct vertices. - Equal vectors must have the same magnitude and direction. - Separate base edges, lateral edges, and diagonals of the rectangular faces. - A vector and its opposite count as different vectors.

1. For the initial vertex there are \(6\) choices, and for the terminal vertex there are \(5\) remaining choices. Thus, there are \(6\cdot5=30\) directed segments. 2. Corresponding lateral edges of a prism are equal and parallel with the same orientation, so \(\overrightarrow{AD}=\overrightarrow{BE}=\overrightarrow{CF}\). 3. Count the distinct vectors by type. The two orientations of the lateral-edge vector contribute \(2\) vectors. The three edge directions in the triangular bases, together with their opposites, contribute \(6\) vectors. For segments joining noncorresponding vertices of opposite bases, there are six vectors with the upward lateral component and six with the downward lateral component. Their in-plane components are the six distinct directed base-edge vectors, and the opposite lateral components keep the upward and downward groups distinct. Thus these segments contribute \(12\) more vectors. Therefore, the total is \(2+6+12=20\).

a) \(30\) directed segments b) \(\overrightarrow{AD}\), \(\overrightarrow{BE}\), and \(\overrightarrow{CF}\) c) \(20\) distinct vectors

- Count ordered pairs of distinct vertices. - Equal vectors must match in both magnitude and direction. - Consider base edges, base diagonals, and lateral edges separately. - Check whether segments outside the same plane can have identical component vectors.

1. The pyramid has \(5\) vertices, so the number of ordered pairs of distinct vertices is \(5\cdot4=20\). 2. In the square base, opposite sides are equal and parallel with matching orientation: \(\overrightarrow{AB}=\overrightarrow{DC}\), \(\overrightarrow{BA}=\overrightarrow{CD}\), \(\overrightarrow{AD}=\overrightarrow{BC}\), and \(\overrightarrow{DA}=\overrightarrow{CB}\). No other directed segments represent equal vectors. 3. The \(8\) directed base edges represent \(4\) distinct vectors. The \(4\) directed base diagonals are distinct, and the \(8\) directed lateral edges are distinct. Therefore, the total is \(4+4+8=16\).

a) \(20\) directed segments b) \(\{\overrightarrow{AB},\overrightarrow{DC}\}\), \(\{\overrightarrow{BA},\overrightarrow{CD}\}\), \(\{\overrightarrow{AD},\overrightarrow{BC}\}\), and \(\{\overrightarrow{DA},\overrightarrow{CB}\}\) c) \(16\) distinct vectors

- Determine the quadrant from the signs of the two components before interpreting an inverse tangent. - A standard direction angle is measured counterclockwise from the positive x-axis and lies between \(0^\circ\) and \(360^\circ\). - Use the acute reference angle together with the quadrant.

1. The component signs \((- , +)\) place the vector in quadrant II. 2. The inverse-tangent calculation by itself returns an angle associated with the ratio, but it does not automatically supply the required quadrant-II standard direction angle. 3. A positive reference angle is \(\tan^{-1}\left(\frac{7}{4}\right)\approx60.3^\circ\). 4. Therefore, the correct direction angle is \(180^\circ-60.3^\circ=119.7^\circ\).

Jordan ignored the vector's quadrant. The correct direction angle is \(119.7^\circ\).