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

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2 2 2 7 This is the parametrized form of the two-point formula for a line in the plane determined by a pair of points. 3. LINES IN SPACE We will use these ideas to prove the following. 2. In any triangle, the three lines joining a vertex to the midpoint of the opposite side meet at a single point. Proof. Label the vertices of the triangle A, B and C, and their position → − → → vectors − a , b and − c , respectively. 23). 2 that the position vectors of the midpoints of the sides are −−→′ 1 − → → OA = ( b + − c) 2 −−→′ 1 − → OB = (→ c +− a) 2 −−→′ 1 − → − OC = (→ a + b ), 2 and so the line ℓA through A and A′ can be parametrized (using r as the parameter) by → → → r− r − r− − → → → p A (r) = (1 − r)− a + (b +− c ) = (1 − r)− a + b + → c.

The dot product has the following algebraic properties: 1. It is commutative: − → → → → v ·− w =− w ·− v 2. It distributes over vector sums11 : − → → → → → → → u · (− v +− w) = − u ·− v +− u ·− w 3. it respects scalar multiples: → → → → → → (r − v )·− w = r(− v ·− w) = − v · (r − w ). 17) yields a number of geometric properties: 10 11 Also the scalar product, direct product, or inner product → In this formula, − u is an arbitrary vector, not necessarily of unit length. 4. 3. The dot product has the following geometric properties: − → → → 1.

Suppose P1 (x1 , y1 , z1 ) and P2 (x2 , y2 , z2 ) are distinct points. The line through P1 and P2 is given by the parametrization7 → → → − p 1 + t− p2 p (t) = (1 − t)− with coordinates x = (1 − t)x1 + tx2 y = (1 − t)y1 + ty2 z = (1 − t)z1 + tz2 . The line segment P1 P2 consists of the points for which 0 ≤ t ≤ 1. The → value of t gives the portion of P1 P2 represented by the segment P1 − p (t); in particular, the midpoint of P1 P2 has position vector 1 − → (→ p1+− p 2) = 2 1 1 1 (x1 + x2 ), (y1 + y2 ), (z1 + z2 ) .