Vectors and Parametric Equations — Free Pre-Calculus Review Games.

Magnitude = \(\sqrt{9+16} = \sqrt{25} = 5\).

Add components: <2+4, 3+(-1)> = <6, 2>.

Vectors have both magnitude (length) and direction.

Scalar multiplication: <3*2, 3*(-1)> = <6, -3>.

The zero vector <0,0> has magnitude 0.

Dot product = 2*4 + 3*(-1) = 8 - 3 = 5.

Perpendicular (orthogonal) vectors have dot product equal to 0.

Divide by magnitude 5: <3/5, 4/5>.

Both x and y are linear in t, so the path is a straight line.

t = x-1. y = 2(x-1)-3 = 2x-5.

Dot product = 0, so the angle is 90 degrees.

\(x^2+y^2 = 9\cos^2(t)+9\sin^2(t) = 9\). A circle of radius \(3\).

\(\langle 10\cos(30), 10\sin(30) \rangle = \langle 10\sqrt{3}/2, 10/2 \rangle = \langle 5\sqrt{3}, 5 \rangle\).

proj_v(u) = (u.v / |v|^2) * v.

\(\cos(t) = x/2\), \(\sin(t) = y/3\). \(\cos^2+\sin^2=1\): \(x^2/4 + y^2/9 = 1\).

The magnitude is found using \(\sqrt{(-6)^2 + 8^2} = \sqrt{36+64} = \sqrt{100} = 10\). The choice \(14\) is wrong because it incorrectly adds the absolute values of the components instead of using the Pythagorean formula. Always compute vector magnitude with the square root of the sum of squared components.

Vector subtraction is done component-wise: \(5-3=2\) and \(-2-4=-6\), giving \(\langle 2, -6 \rangle\). The choice \(\langle 8, 2 \rangle\) is incorrect because it adds the components instead of subtracting them. Remember that vector subtraction requires subtracting corresponding components in order.

A vector is geometrically represented as an arrow, and its initial point is where it starts while its terminal point is where it ends. The distractor 'magnitude and direction' describes the properties of a vector, not the points used to draw it as an arrow. Understanding that a vector's arrow representation begins at an initial point and ends at a terminal point is foundational to vector geometry.

Since \(\vec{v} = -1 \cdot \vec{u}\), multiplying by a negative scalar reverses the direction while keeping the same magnitude, so the vectors are opposite. The choice 'they have different magnitudes' is wrong because both vectors have magnitude \(\sqrt{5}\). Scalar multiplication by a negative number always reverses a vector's direction without changing its length.

In unit vector notation, the x-component multiplies \(\mathbf{i}\) and the y-component multiplies \(\mathbf{j}\), giving \(4\mathbf{i} - 3\mathbf{j}\). The choice \(4\mathbf{i} + 3\mathbf{j}\) is wrong because it drops the negative sign on the y-component. Always preserve the sign of each component when converting to unit vector notation.

A unit vector is defined as any vector whose magnitude equals exactly \(1\). The distractor 'it points along the x-axis' is incorrect because unit vectors can point in any direction, not just along an axis. Recognizing unit vectors is essential for normalizing directions in vector problems.

First scale each vector: \(2\langle 1,3 \rangle = \langle 2,6 \rangle\) and \(3\langle 2,-1 \rangle = \langle 6,-3 \rangle\), then add component-wise to get \(\langle 8, 3 \rangle\). The choice \(\langle 5, 5 \rangle\) is wrong because it fails to correctly scale both vectors before adding. Always scale each vector fully before performing addition in linear combinations.

In parametric equations, \(x\) and \(y\) are each expressed as separate functions of an independent variable \(t\), which often represents time or another quantity. The distractor 'the slope of the curve' is wrong because slope is derived from the relationship between \(x\) and \(y\), not the parameter itself. Recognizing \(t\) as the driving independent variable is key to understanding parametric representations.

The dot product is calculated as \((0)(3) + (5)(0) = 0 + 0 = 0\). The choice \(15\) is incorrect because it results from multiplying the nonzero components together instead of matching corresponding components. A dot product of zero for nonzero vectors always indicates the vectors are perpendicular.

Speed only measures magnitude and is scalar, while velocity includes both magnitude and direction, making it a vector. The choice 'both speed and velocity are scalar' is wrong because velocity explicitly requires a direction component. Distinguishing scalar from vector quantities is fundamental before performing any vector operations.

Since \(x = t\), substituting \(t = x\) into \(y = t^2\) gives \(y = x^2\) directly. The choice \(x = y^2\) reverses the roles of \(x\) and \(y\) incorrectly. When eliminating a parameter, always solve one equation for \(t\) and substitute it into the other equation.

Negating a vector reverses the sign of every component, so \(-\langle 3,-2 \rangle = \langle -3, 2 \rangle\). The choice \(\langle 3, 2 \rangle\) is wrong because it only negates the second component and leaves the first unchanged. Negating a vector always requires flipping the sign of all components simultaneously.

Adding the vectors gives \(\langle 3, 4 \rangle\), and its magnitude is \(\sqrt{3^2 + 4^2} = \sqrt{25} = 5\). The choice \(7\) is incorrect because it comes from simply adding the individual magnitudes \(3+4\) rather than computing the magnitude of the resultant vector. Vector addition must be performed on components first before finding magnitude, since magnitudes do not add linearly except in special collinear cases.

The dot product is computed as \((-3)(2) + (5)(4) = -6 + 20 = 14\). The choice \(-6\) only accounts for the first term of the calculation and stops there. Always add both products of corresponding components fully when computing a dot product.

Perpendicular vectors have a dot product of zero, so \(2k + 3(-6) = 0\) gives \(2k = 18\), so \(k = 9\). The choice \(-9\) results from a sign error when isolating \(k\) in the equation. Setting the dot product equal to zero is the standard method for finding unknowns that make vectors perpendicular.

Solving \(y = t+1\) for \(t\) gives \(t = y - 1\), and substituting into \(x = t^2\) yields \(x = (y-1)^2\). The choice \(y = x^2+1\) incorrectly substitutes \(x\) directly for \(t\) instead of solving for \(t\) first. Always isolate the parameter in the simpler equation before substituting into the other.

Both \(x\) and \(y\) are linear functions of \(t\), so eliminating the parameter produces a linear equation, which graphs as a straight line. The choice 'a parabola' is incorrect because parabolas require a squared term in the parametric equations, which is absent here. Linear parametric equations in \(t\) always correspond to straight-line paths.

Using \(\cos\theta = \frac{\vec{u}\cdot\vec{v}}{|\vec{u}||\vec{v}|} = \frac{6}{2\sqrt{2}\cdot 3} = \frac{1}{\sqrt{2}}\), so \(\theta = 45^\circ\). The choice \(90^\circ\) is wrong because the dot product is nonzero, ruling out perpendicularity. The dot product formula for angle relies on both the magnitudes and the dot product, not just one factor.

The scalar projection of \(\vec{u}\) onto \(\vec{v}\) is \(\frac{\vec{u}\cdot\vec{v}}{|\vec{v}|}\), which measures how much of \(\vec{u}\) lies along the direction of \(\vec{v}\). The choice \(\frac{\vec{u}\cdot\vec{v}}{|\vec{u}|}\) is wrong because it divides by the wrong vector's magnitude, changing the meaning of the projection. Scalar projections always divide the dot product by the magnitude of the vector being projected onto.

Since \(x^2 + y^2 = \cos^2(t) + \sin^2(t) = 1\), the particle traces a circle of radius \(1\) centered at the origin. The choice 'an ellipse with semi-axes 2 and 3' is wrong because both coefficients on \(\cos(t)\) and \(\sin(t)\) here are \(1\), not \(2\) and \(3\). Equal coefficients on sine and cosine in parametric equations always produce a circle rather than an ellipse.

Adding the x-components gives \(2-1+4 = 5\) and adding the y-components gives \(5+3-2 = 6\), resulting in \(\langle 5, 6 \rangle\). The choice \(\langle 5, 10 \rangle\) incorrectly sums the y-components as \(5+3+2\) rather than \(5+3-2\). When adding multiple vectors, each component must be added separately while preserving the sign of every term.

A parametric line through a point \((x_0, y_0)\) with direction \(\langle a, b \rangle\) is written as \(x = x_0 + at\), \(y = y_0 + bt\), giving \(x = 1+3t\), \(y = 2-t\). The choice \(x = 3 + t\), \(y = -1 + 2t\) incorrectly swaps the roles of the point and direction vector. Always substitute the given point as the starting values and the direction vector components as the coefficients of \(t\).

Adding the vectors gives \(\langle 0, 2 \rangle\), and its magnitude is \(\sqrt{0^2+2^2} = 2\). The choice \(0\) is incorrect because it would only occur if the vectors were exact opposites, which they are not since their x-components cancel but y-components add. Always compute the resultant vector fully before taking its magnitude.

Work is defined as the dot product of force and displacement vectors, \(W = \vec{F}\cdot\vec{d}\), because it measures the component of force acting in the direction of motion. The choice \(W = |\vec{F}||\vec{d}|\) is wrong because it ignores the angle between the vectors, which the dot product formula inherently accounts for through \(\cos\theta\). The dot product is the natural tool for physical quantities that depend on directional alignment, such as work.

The unit vector in the direction of \(\vec{u}\) is \(\left\langle \frac{6}{10}, \frac{8}{10} \right\rangle\), and multiplying by \(5\) gives \(\langle 3, 4 \rangle\). The choice \(\langle 5, 5 \rangle\) is wrong because it does not preserve the original direction ratio of \(6:8\). To scale a vector to a desired magnitude, first find its unit vector, then multiply by the target magnitude.

Since \(y = e^{2t} = (e^t)^2\) and \(x = e^t\), substituting gives \(y = x^2\). The choice \(x = y^2\) reverses which variable is squared, misrepresenting the relationship. When exponential parametric equations share a common base, express one in terms of the other using exponent rules before eliminating the parameter.

These equations track the horizontal position \(x\) and vertical position \(y\) of the ball as functions of time \(t\), which is the standard setup for modeling projectile motion. The choice 'the velocity of the ball at each instant' is wrong because velocity would be represented by the derivatives of these position functions, not the functions themselves. Parametric position equations describe where an object is located at each moment in time.

The dot product of a vector with itself equals the square of its magnitude, so \(\vec{u}\cdot\vec{u} = 4^2 + (-3)^2 = 25\). The choice \(5\) is incorrect because it is the magnitude itself, not the dot product, which equals the magnitude squared. Recognizing that \(\vec{u}\cdot\vec{u} = |\vec{u}|^2\) provides a useful shortcut for magnitude calculations.

The resultant vector is \(\langle 5, 12 \rangle\) with magnitude \(\sqrt{25+144} = 13\) N, and the angle is \(\arctan(12/5) \approx 67.4^\circ\) above the x-axis. The choice \(17\) N at \(45^\circ\) is wrong because it incorrectly adds the magnitudes of the individual forces instead of combining perpendicular components properly. Perpendicular force problems require vector addition and the arctangent function to find both magnitude and direction of the resultant.

Setting up the dot product equation gives \(2c + 3(-4) = -2\), so \(2c - 12 = -2\), leading to \(2c = 10\) and \(c = 5\). The choice \(-5\) results from a sign error when solving the linear equation for \(c\). Careful algebraic manipulation is essential when solving dot product equations for unknown components.

The equations represent a circle shifted so that \(x+1 = 3\cos(t)\) and \(y-2 = 3\sin(t)\), giving center \((-1, 2)\) and radius \(3\) since \((x+1)^2+(y-2)^2 = 9\). The choice 'Center \((1, -2)\), radius \(3\)' reverses the sign of the shift, misplacing the center. When identifying shifts in parametric circle equations, the center coordinates are the opposite sign of the constants added to \(\cos(t)\) and \(\sin(t)\).

The component form is found by subtracting the initial point from the terminal point: \(\langle -3-2, 4-(-1) \rangle = \langle -5, 5 \rangle\), with magnitude \(\sqrt{25+25} = 5\sqrt{2}\). The choice \(\langle 5, -5 \rangle\) reverses the subtraction order, giving the negative of the correct vector. Always subtract initial point coordinates from terminal point coordinates, in that specific order, to find a vector's component form.

Setting \(t = 1\) and \(t = -1\) both give \(x = 0\), and checking \(y\) gives \(y = 0\) for both, so the curve passes through \((0,0)\) twice, confirming a self-intersection there. The choice 'No, it never crosses itself' is incorrect because two distinct parameter values produce the identical point, which is the definition of a self-intersection. To find self-intersections in parametric curves, look for two different \(t\)-values that yield the same \((x, y)\) coordinates.

Using \(\vec{u}\cdot\vec{v} = |\vec{u}||\vec{v}|\cos\theta = (1)(1)\cos(120^\circ) = -0.5\) since both vectors are unit vectors. The choice \(0\) is wrong because it would only apply if the angle were \(90^\circ\), not \(120^\circ\). Remembering that the dot product formula depends directly on the cosine of the angle helps avoid confusing perpendicularity with other angle relationships.

Using the Pythagorean identity \(\sec^2(t) - \tan^2(t) = 1\), substituting \(x = \sec(t)\) and \(y = \tan(t)\) gives \(x^2 - y^2 = 1\), a hyperbola. The choice \(x^2+y^2=1\) is wrong because that identity applies to sine and cosine, not secant and tangent. Recognizing which Pythagorean identity matches the given trig functions is essential for correctly eliminating the parameter.

Both vectors have magnitude \(\sqrt{3^2+4^2} = 5\) and \(\sqrt{4^2+(-3)^2}=5\), so besides their dot product being zero (confirming perpendicularity), they also share equal magnitudes. The choice 'they are parallel' is incorrect and contradictory since perpendicular vectors cannot also be parallel unless one is the zero vector. Checking both magnitude and dot product together reveals deeper geometric relationships between vectors beyond a single property.

Substituting into \(y(t) = 2t+1 = 5\) gives \(t=2\), and checking \(x(t) = t^2 = 4\) confirms \(t=2\) works, while \(t=-2\) satisfies \(x(t)=4\) but gives \(y(t) = -3 \ne 5\), so only \(t=2\) is valid. The choice '\(t = 2\)' alone omits the necessary verification step that eliminates the extraneous solution \(t=-2\) from the x-equation. When solving parametric point problems, both equations must be checked simultaneously to rule out extraneous parameter values.

The vector projection formula is \(\text{proj}_{\vec{v}}\vec{u} = \frac{\vec{u}\cdot\vec{v}}{|\vec{v}|^2}\vec{v} = \frac{4}{1}\langle 1,0 \rangle = \langle 4, 0 \rangle\). The choice \(\langle 4, 2 \rangle\) is incorrect because it simply returns the original vector \(\vec{u}\) without projecting it onto \(\vec{v}\)'s direction. Vector projection isolates only the component of \(\vec{u}\) that lies along \(\vec{v}\), discarding any perpendicular component.

Adding the vectors gives \(\langle 170, 40 \rangle\), and the magnitude is \(\sqrt{170^2+40^2} = \sqrt{28900+1600} = \sqrt{30500} \approx 206.5\) km/h. The choice \(170\) km/h only reflects the horizontal component and ignores the vertical wind effect entirely. Ground speed calculations always require finding the magnitude of the full resultant vector, not just one component.

Since \(16 = 4^2\) and \(9 = 3^2\), matching the ellipse equation to \(\left(\frac{x}{4}\right)^2+\left(\frac{y}{3}\right)^2=1\) gives the parametrization \(x = 4\cos(t)\), \(y = 3\sin(t)\). The choice \(x = 3\cos(t)\), \(y = 4\sin(t)\) swaps the semi-axis lengths, which would produce a differently shaped ellipse. Always take the square root of the denominators in the standard ellipse equation to find the correct coefficients for the parametric form.

Parallel vectors have proportional components, so \(\frac{2}{6} = \frac{k}{-3}\), which gives \(k = \frac{2 \times (-3)}{6} = -1\). The choice \(3\) is wrong because it fails to maintain the correct ratio between the corresponding x and y components. When testing for parallel vectors, set up a proportion between corresponding components rather than comparing them individually.

These equations are the standard parametric form of a cycloid, which traces the path of a point on the rim of a circle as it rolls along a straight line. The choice 'a circle' is wrong because a true circle requires equations like \(x = r\cos(t)\) and \(y = r\sin(t)\) without the extra linear term \(t\) present here. Recognizing the characteristic combination of a linear term with trigonometric functions is the key to identifying cycloid equations.

The magnitude of a vector \(\langle a, b \rangle\) is found using \(|\vec{v}| = \sqrt{a^2 + b^2} = \sqrt{36 + 64} = \sqrt{100} = 10\), applying the Pythagorean theorem to the vector's components. The choice "\(14\)" is wrong because it comes from simply adding \(6 + 8\) instead of using the square root of the sum of squares. Students should remember that vector magnitude always requires squaring components, summing, and taking the square root, never simple addition.

Substituting \(t = 3\) into \(x = t+1\) gives \(x = 4\), and into \(y = t^2 - 2\) gives \(y = 9 - 2 = 7\), matching both coordinates of the point simultaneously, which is the requirement for a valid parameter value. The choice "\(t = -3\)" is wrong because although it satisfies \(y = 7\) since \((-3)^2 - 2 = 7\), it gives \(x = -3 + 1 = -2\), which does not match the required \(x = 4\). When solving for a parameter value, students must verify the solution satisfies both equations simultaneously, not just one.

Using the formula \(\vec{u} \cdot \vec{v} = |\vec{u}||\vec{v}|\cos\theta = 5 \cdot 8 \cdot \cos(120^\circ) = 40 \cdot (-0.5) = -20\), the negative cosine of an obtuse angle correctly produces a negative dot product. The choice "\(20\)" is wrong because it drops the negative sign, forgetting that \(\cos(120^\circ) = -0.5\) rather than \(0.5\), which would apply only to an acute angle like \(60^\circ\). Students should remember that the sign of the dot product directly indicates whether the angle between two vectors is acute (positive), obtuse (negative), or exactly \(90^\circ\) (zero).