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Question:

Let →PR = 3^i + ^j - 2^k and →SQ = ^i + ^j + ^k determine diagonals of a parallelogram PQRS and →PT = ^i + 2^j + 3^k be another vector. Then the volume of the parallelepiped determined by the vectors →PT, →PQ and →PS is

10

30

5

20

Solution:

Area of base (PQRS) = 1/2 ||→PR x →SQ|| = 1/2 ||^i ^j ^k
3 1 -2
1 1 1|| = 1/2 |5^i + 10^j + 0^k| = 5|^i - ^j + ^k| = 5√3
Height = proj. of PT on ^i - ^j + ^k = ||1 + 3||/√3 = 2√3
Volume = (5√3)(2√3) = 10 cu. units