Page 176 - Schaum's Outline of Theory and Problems of Advanced Calculus

CHAP. 7] VECTORS 167

Fig. 7-24

Since A is perpendicular to B, A B is a vector perpendicular to the plane of A and B and having magnitude
AB sin 908 ¼ AB or magnitude of AB. This is equivalent to multiplying vector B by A and rotating the
resultant vector through 908 to the position shown in Fig. 7-25.

Fig. 7-25 Fig. 7-26

Similarly, A C is the vector obtained by multiplying C by A and rotating the resultant vector through
908 to the position shown.
In like manner, A ðB þ CÞ is the vector obtained by multiplying B þ C by A and rotating the resultant
vector through 908 to the position shown.
Since A ðB þ CÞ is the diagonal of the parallelogram with A B and A C as sides, we have
A ðB þ CÞ¼ A B þ A C.

7.18. Prove that A ðB þ CÞ¼ A B þ A C in the general case where A, B, and C are non-
coplanar. See Fig. 7-26.
Resolve B into two component vectors, one perpendicular to A and the other parallel to A, and denote
them by B ? and B k respectively. Then B ¼ B ? þ B k .
If is the angle between A and B,then B ? ¼ B sin . Thus the magnitude of A B ? is AB sin ,the
same as the magnitude of A B. Also, the direction of A B ? is the same as the direction of A B.
Hence A B ? ¼ A B.
Similarly, if C is resolved into two component vectors C k and C ? , parallel and perpendicular respec-
tively to A,then A C ? ¼ A C.
Also, since B þ C ¼ B ? þ B k þ C ? þ C k ¼ðB ? þ C ? ÞþðB k þ C k Þ it follows that

A ðB ? þ C ? Þ¼ A ðB þ CÞ
Now B ? and C ? are vectors perpendicular to A and so by Problem 7.17,

Then A ðB ? þ C ? Þ¼ A B ? þ A C ?
A ðB þ CÞ¼ A B þ A C