Free math worksheets

- Read the horizontal and vertical components from the diagram. - The magnitude is the distance from the origin to the vector's endpoint. - Use the Pythagorean relationship between the two components and the magnitude.

From the diagram, \(\mathbf{v}=\begin{pmatrix}3\\4\end{pmatrix}\). Therefore, \(\lVert\mathbf{v}\rVert=\sqrt{3^2+4^2}=5\).

\(5\)

- Use the endpoint to determine which axis the vector lies on. - The vector points left from the origin. - Recall the standard-position angle of the negative \(x\)-axis.

From the diagram, \(\mathbf{v}\) points directly along the negative \(x\)-axis. Its standard direction angle is \(180^\circ\).

\(180^\circ\)

- 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: \(\vec{v}=\begin{pmatrix}2\cdot1-3\cdot1+3\\2\cdot0-3\cdot(-1)+0\\2\cdot4-3\cdot2+4\end{pmatrix}=\begin{pmatrix}2\\3\\6\end{pmatrix}\). 2. Find the magnitude: \(\|\vec{v}\|=\sqrt{2^2+3^2+6^2}=\sqrt{49}=7\).

\(\vec{v}=\begin{pmatrix}2\\3\\6\end{pmatrix}\); \(\|\vec{v}\|=7\)

- 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: \(3\begin{pmatrix}-1\\2\\1\end{pmatrix}=\begin{pmatrix}-3\\6\\3\end{pmatrix}\), \(2\begin{pmatrix}3\\0\\4\end{pmatrix}=\begin{pmatrix}6\\0\\8\end{pmatrix}\), and \(4\begin{pmatrix}0.5\\0.5\\0.75\end{pmatrix}=\begin{pmatrix}2\\2\\3\end{pmatrix}\). 2. Combine the results: \(\vec{w}=\begin{pmatrix}-3+6-2\\6+0-2\\3+8-3\end{pmatrix}=\begin{pmatrix}1\\4\\8\end{pmatrix}\). 3. Find the magnitude: \(\|\vec{w}\|=\sqrt{1^2+4^2+8^2}=\sqrt{81}=9\).

\(\vec{w}=\begin{pmatrix}1\\4\\8\end{pmatrix}\); \(\|\vec{w}\|=9\)

- 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: \(\begin{pmatrix}1\\-5\\12\end{pmatrix}+\begin{pmatrix}5\\13\\12\end{pmatrix}=\begin{pmatrix}6\\8\\24\end{pmatrix}\). 2. Multiply by \(0.5\): \(\vec{a}=\begin{pmatrix}3\\4\\12\end{pmatrix}\). 3. Find the magnitude: \(\|\vec{a}\|=\sqrt{3^2+4^2+12^2}=\sqrt{169}=13\).

\(\vec{a}=\begin{pmatrix}3\\4\\12\end{pmatrix}\); \(\|\vec{a}\|=13\)

- 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: \(\vec{b}=(1.4+0.6)\begin{pmatrix}2\\5\\-14\end{pmatrix}=2\begin{pmatrix}2\\5\\-14\end{pmatrix}\). 2. Thus, \(\vec{b}=\begin{pmatrix}4\\10\\-28\end{pmatrix}\). 3. Its magnitude is \(\|\vec{b}\|=\sqrt{4^2+10^2+(-28)^2}=\sqrt{900}=30\).

\(\vec{b}=\begin{pmatrix}4\\10\\-28\end{pmatrix}\); \(\|\vec{b}\|=30\)

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

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

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

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

1. The squared magnitudes of \(\vec{a}\), \(\vec{b}\), \(\vec{c}\), and \(\vec{d}\) are all \(9\), so each has magnitude \(3\). 2. The squared magnitudes of \(\vec{e}\), \(\vec{f}\), \(\vec{g}\), and \(\vec{h}\) are all \(25\), so each has magnitude \(5\).

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

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

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

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

- 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. \(\|\vec{a}\|=\sqrt{2^2+(-3)^2+6^2}=\sqrt{49}=7\). 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. \(\overrightarrow{PQ_k}=\begin{pmatrix}k-3\\4\\4\end{pmatrix}\). 2. When \(k=3\), \(\overrightarrow{PQ_3}=\begin{pmatrix}0\\4\\4\end{pmatrix}\), so \(\|\overrightarrow{PQ_3}\|=\sqrt{32}=4\sqrt{2}\). 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) \(\begin{pmatrix}k-3\\4\\4\end{pmatrix}\) b) \(4\sqrt{2}\) 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.

- 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 \(\|\vec{v}\|^2=(2k)^2+k^2+10^2=5k^2+100\). 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 \(k=-5\)

- Read the normal vector from the equation. - Compute its magnitude before choosing a scale factor. - Include both possible directions for a normal vector of the required magnitude. - Check what happens when \(x=z=0\).

1. A normal vector is \(\vec{n}=\langle 3,0,-4\rangle\), so the dot-product equation is \(\langle 3,0,-4\rangle\cdot\vec{x}=0\). 2. Its magnitude is \(\lVert\vec{n}\rVert=\sqrt{3^2+0^2+(-4)^2}=5\). A parallel normal vector of magnitude \(10\) must be \(2\vec{n}\) or \(-2\vec{n}\). Thus the vectors are \(\langle 6,0,-8\rangle\) 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) \(\langle 3,0,-4\rangle\cdot\vec{x}=0\) b) \(\langle 6,0,-8\rangle\) and \(\langle -6,0,8\rangle\) c) The plane contains the \(y\)-axis.

- 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 \(S_1=P+\langle 2,2,-1\rangle=(5, 4, 3)\). The corresponding sunlight direction is \(\overrightarrow{QS_1}=\langle 2,2,-3\rangle\). 5. The second possible shadow point is \(S_2=P-\langle 2,2,-1\rangle=(1, 0, 5)\). The corresponding sunlight direction is \(\overrightarrow{QS_2}=\langle -2,-2,-1\rangle\). 6. Both vectors point from the rod’s top toward a shadow point and are consistent with the stated shadow length.

\(\vec{s}=\langle 2,2,-3\rangle\) or \(\vec{s}=\langle -2,-2,-1\rangle\)

- 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