8.6 Vectors in Space

z 2 -2 -2 O y 2 2 -2 x

Theorem Distance Formula in Space If P1 = (x1, y1, z1) and P2 = (x2, y2, z2) are two points in space, the distance d from P1 to P2 is

Find the distance from P1 = (2, 3, 0) to

If v is a vector with initial point at the origin O and terminal point at P = (a, b, c), then we can represent v in terms of the vectors i, j, and k as v = ai + bj + ck

Position Vector P = (a, b, c) v = ai + bj + ck

Theorem Suppose that v is a vector with initial point P1= (x1, y1, z1), not necessarily the origin, and terminal point P2 = (x2, y2, z2). If v = P1P2 then v is equal to the position vector v = (x1- x2)i+(y1 - y2)j+ (z1 - z2)k

P1= (x1, y1, z1), P2 = (x2, y2, z2). v = (x1- x2)i+(y1 - y2)j+ (z1 - z2)k

Find the position vector of the vector v= P1P2 if P1= (0, 2, -1) and P2 = (-2, 3,-1). v = (-2 - 0)i+ (3 - 2)j+[-1-(-1)]k = -2i + j

If v = -2i+ 3j + 4k and w = 3i+ 5j - k find (a) v + w (b) v - w (a) v + w= (-2i+ 3j + 4k) + (3i+ 5j - k) = (-2+3)i + (3+5)j + (4-1)k = i + 8j + 3k (b) v - w= (-2i+ 3j + 4k) - (3i+ 5j - k) = (-2-3)i + (3-5)j + (4-(-1))k = -5i -2j + 5k

For any nonzero vector v, the vector Theorem Unit Vector in Direction of v For any nonzero vector v, the vector is a unit vector that has the same direction as v.

Find the unit vector in the same direction as w = 3i+ 5j - k . First we will find the magnitude of v.

Theorem Properties of Dot Product If u, v, and w are vectors, then Commutative Property Distributive Property

Theorem Angle between Vectors

We find that

Theorem Direction Angles

Find the direction angles of v= -3i+2j-k.

Theorem Property of Direction Cosines

A nonzero vector v in space can be written in terms of its magnitude and direction cosines as