Calculus 3 with the Math Sorcerer

Learn to plot points in space with three dimensions using the x, y, z axes, including projections onto the xy plane and coordinates (x, y, z) with signs.

Compute the magnitude of a vector in space by squaring its components, summing them, and taking the square root; for components 3, -2, 1, this yields sqrt(14).

Compute the magnitude of a three-dimensional vector by summing the squares of its components and taking the square root; an example yields 5 for (3, 0, 4).

Compute the magnitude of the space vector V = (-2, 3, 4) by squaring its components and summing to 29. Take the square root to obtain sqrt(29).

Normalize the given vector to a unit vector, then multiply by six to obtain a vector with magnitude six in the same direction.

Find a vector with a given magnitude in the same direction as another by forming a unit vector and scaling it, as shown with (1,2,3) and magnitude 3.

Compute a unit vector in the direction of v by dividing v by its magnitude sqrt(29), yielding (3, 2, 4)/sqrt(29). Obtain the opposite by multiplying by -1.

Explore the dot product, also known as the inner or scalar product, and how multiplying corresponding components yields a single number, revealing orthogonality, magnitude, and angles.

Learn the dot product of vectors in the plane, defined as multiplying corresponding components and summing, illustrated with 4,6 and 2,8 to yield 56.

Compute the dot product of two planar vectors by multiplying corresponding components and adding them; for u=(3,6) and v=(-2,4), the result is 18.

Compute the dot product of two space vectors by multiplying corresponding coordinates and summing the results, illustrated with vectors [1, 2, 4] and [-3, -2, 0], yielding -7.

Compute the dot product of two vectors in space by multiplying corresponding components, handling a missing component as zero, and summing to obtain -1.

Compute the dot product of the two three-dimensional vectors; if the result is zero, they are orthogonal. The example yields zero, confirming orthogonality.

Compute the angle between two vectors in the plane by applying the dot product over the product of magnitudes; the result is cos theta = sqrt(2)/2, so theta = pi/4.

Calculate the angle between two vectors in space using the dot product and magnitudes, applying theta = arccos((u·v)/(|u||v|)) and computing from the given example.

Compute the angle between space vectors using the dot product, confirming orthogonality when u·v = 0 and theta = pi/2 (90 degrees) through component form and the cosine formula.

Apply the dot product formula using the magnitudes and the angle between the vectors. With |u|=6 and |v|=3, theta=pi/6, the dot product is 9 sqrt(3).

Compute the dot product from given magnitudes 50 and 20 and angle 5π/6 using u·v = |u||v| cosθ, giving -500√3.

Learn to construct two opposite-direction vectors that are orthogonal to a given vector by solving dot products to zero, often by placing a zero component, and verify with examples.

Explore projections by decomposing u into its projection onto v and the orthogonal component, using the projection formula (u·v)/(||v||^2) v and a worked example with u=(6,7) and v=(1,4).

Apply the projection formula (u·v)/|v|^2 times v to find u projected onto v, then subtract from u to get the orthogonal component; example uses u=(2,3) and v=(5,1).

Project vector u onto v and compute the orthogonal component using W1 and W2. Apply the projection formula u·v over ||v||^2 times v in a 3d example.

Derive direction angles and direction cosines for a vector in 3-D, normalize it, express V in unit-vector components, and verify cos^2 alpha plus cos^2 beta plus cos^2 gamma equals 1.

Derive the direction cosines and direction angles of a vector from its component form and magnitude; divide components by the magnitude and verify the squares sum to one.

Express the vector (1, 8, 4) in component form, find its magnitude 9, then direction cosines are 1/9, 8/9, 4/9, and direction angles are arccos of these values.

Find the direction angles of the vector from components (5, 9, 7) by computing magnitude sqrt(155) and the cosines 5/sqrt(155), 9/sqrt(155), 7/sqrt(155), then determine alpha, beta, gamma in degrees.

Compute the cross product of two vectors by expanding the determinant with i, j, k and 2x2 minors.

Compute the cross product of two vectors using a determinant with the first row and the standard plus-minus-plus pattern, producing a vector orthogonal to the original ones.

Compute the cross product of two vectors using a determinant, express the result in component form, and verify the resulting vector is perpendicular to both originals.

Use the cross product to obtain a vector orthogonal to the given vectors, then normalize it to a unit vector by dividing by its norm sqrt(59), yielding (1,3,7)/sqrt(59).

Find a unit vector orthogonal to both u and v by crossing them and normalizing; the cross product is (-14, -10, 26) divided by sqrt(972).

Learn to calculate the distance between two points in space using the 3d distance formula, with a step-by-step example that yields the result sqrt(6).

Find the distance between two points in space using the distance formula, subtract coordinates, square the differences, sum them, and take the square root (example yields sqrt(62)).

Plug the center (4, -2, -3) and radius 2 into the standard sphere equation to get (x - 4)^2 + (y + 2)^2 + (z + 3)^2 = 4.

derive the sphere equation from its center and radius using the standard form (x - x0)^2 + (y - y0)^2 + (z - z0)^2 = r^2; for center (3, -1, 0) and radius 3, the equation is (x - 3)^2 + (y + 1)^2 + z^2 = 9.

determine the center at (-1, 3, 5/2) and the radius sqrt(45)/2, yielding the sphere equation (x+1)^2+(y-3)^2+(z-5/2)^2=45/4.

Complete the square to write the equation of the sphere in standard form, and determine the center (3, 1/2, -1) and radius 4.

derive a line in space from a point and a parallel vector; obtain parametric form x1 + a t, y1 + b t, z1 + c t, and symmetric form.

Compute the line through a point parallel to a vector using parametric form; when the z-component is zero, symmetric equations cannot be formed.

Derive the parametric line through (3,5,6) parallel to (1,-2,4): x = 3 + t, y = 5 - 2t, z = 6 + 4t, and the symmetric form (x-3)/1 = (y-5)/(-2) = (z-6)/4.

Find the line through (2,7,1) parallel to (-2,-4,6) using parametric equations x=2-2t, y=7-4t, z=1+6t and symmetric form (x-2)/-2=(y-7)/-4=(z-1)/6.

determine the line through the point (1,0,1) parallel to the given line using parametric form x=1+3t, y=-2t, z=1+t and the symmetric form (x-1)/3=(y-0)/-2=(z-1)/1.

Identify two points on the line, form the direction vector from their difference, and express the line parametrically.

Derive the plane equation from a point on the plane and a normal vector, using n · (x - x1, y - y1, z - z1) = 0.

Learn to find the equation of a plane through a point with a given normal vector using the form (x−x0)A+(y−y0)B+(z−z0)C=0, resulting in y=5 for the example.

Find the equation of a plane from a point and a normal vector using the standard form a(x−x1)+b(y−y1)+c(z−z1)=0, with the normal components a,b,c and the point (x1,y1,z1).

Form two vectors from the three points, cross them to obtain the normal vector, and apply the point-normal form to obtain 7x - 28y - 2z = 0.

Find plane through two points parallel to z-axis by crossing inter-point vector with z-axis to obtain normal, then apply plane equation with a known point.

Derive the plane equation with a point and a line's direction as the plane normal; using (8,3,3) and (6,1,-8) yields 6x + y - 8z - 27 = 0.

Derive the plane equation from a point and a line on the plane by using the line’s direction vector, finding a second point, cross product for the normal, then substitute.

Find the plane through two points perpendicular to another plane using two parallel vectors, cross product for the normal, and a point to form the plane equation.