Begin2.DVI

Example 6-11. Derive the law of cosines for the triangle illustrated.

Figure 6-11. Triangle for law of cosines.

Solution Let C
=A
−B
so that the dot product of C
with itself gives

C
·C
= (A
−B
)·(A
−B
) = A
·A
+B
·B
− 2 A
·B

or

C^2 =A^2 +B^2 − 2 AB cos θ,

where A=|A
|, B=|B
|, C=|C
|represent the magnitudes of the vector sides.

Example 6-12. Find the vector equation of the line which passes through the

two points P 1 (x 1 , y 1 , z 1 )and P 2 (x 2 , y 2 , z 2 ).

Solution Let

r 1 =x 1 ˆe 1 +x 2 ˆe 2 +x 3 ˆe 3

and
r 2 =x 2 ˆe 1 +y 2 ˆe 2 +z 2 ˆe 3

denote position vectors to the points P 1 and P 2

respectively and let
r =xˆe 1 +yˆe 2 +zeˆ 3 denote the

position vector of any other variable point on the

line. Observe that the vector
r 2 −
r 1 is parallel

to the line through the points P 1 and P 2 .By vector addition the (x, y, z )position on

the line is given by

r =
r 1 +λ(
r 2 −
r 1 ) −∞ < λ < ∞ (6 .41)