By Nitecki Z.

Show description

Read or Download Calculus in 3D. Geometry, vectors, and multivariate calculus PDF

Best calculus books

Additional resources for Calculus in 3D. Geometry, vectors, and multivariate calculus

Sample text

D) Show that a collection of three position vectors in R3 is linearly independent precisely if the corresponding points determine a plane in space that does not pass through the origin. (e) Show that any collection of four or more vectors in R3 is linearly dependent. 3. 3 Lines in Space Parametrization of Lines A line in the plane is the locus of a “linear” equation in the rectangular coordinates x and y Ax + By = C where A, B and C are real constants with at least one of A and B nonzero. 14) where the slope m is the tangent of the angle the line makes with the horizontal and the y-intercept b is the ordinate (signed height) of its intersection with the y-axis.

V corresponds to P if the arrow OP from the origin to → → P is a representation of − v : that is, − v is the vector representing that → 3 displacement of R which moves the origin to P ; we refer to − v as the position vector of P . We shall make extensive use of the correspondence between vectors and points, often denoting a point by its position vector → − p ∈ R3 . Furthermore, using rectangular coordinates we can formulate a numerical specification of vectors in which addition and multiplication by scalars is −−→ → very easy to calculate: if − v = OP and P has rectangular coordinates → (x, y, z), we identify the vector − v with the triple of numbers (x, y, z) and → − write v = (x, y, z).

P C (t) = − 2 2 To find the intersection of ℓA with ℓB , we need to solve the vector equation − → → p A (r) = − p B (s) − → − −c is which, written in terms of → a , b and → − → s→ − r→ r→ s− → (1 − r)− a + b + − c = → a + (1 − s) b + − c. 2 2 2 2 Assuming the triangle △ABC is nondegenerate—which is to say, none of → − → − → a , b or − c can be expressed as a linear combination of the others—this equality can only hold if corresponding coefficients on the two sides are equal. This leads to three equations in two unknowns s (1 − r) = 2 r = (1 − s) 2 r s = .

Download PDF sample

Rated 4.83 of 5 – based on 10 votes