The two adjacent sides of a parallelogram are and . Find the unit vector parallel to its diagonal. Also, find its area.
Given, two adjacent sides of a parallelogram are 2 i ^ − 4 j ^ + 5 k ^ and i ^ − 2 j ^ − 3 k ^ .
Consider these adjacent sides as a → and b → .
a → = 2 i ^ − 4 j ^ + 5 k ^ b → = i ^ − 2 j ^ − 3 k ^
Let the diagonal be c → .
c → = a → + b → c → = ( 2 i ^ − 4 j ^ + 5 k ^ ) + ( 1 i ^ − 2 j ^ − 3 k ^ ) c → = ( 2 + 1 ) i ^ − ( 4 + 2 ) j ^ + ( 5 − 3 ) k ^ c → = 3 i ^ − 6 j ^ + 2 k ^
The magnitude of c → is,
| c → | = 3 2 + ( − 6 ) 2 + 2 2 | c → | = 9 + 36 + 4 | c → | = 49 | c → | = 7
Unit vector of c → is,
c ^ = 1 | c → | × c → c ^ = 1 7 ( 3 i ^ − 6 j ^ + 2 k ^ )
The area of parallelogram is | a → × b → | ,
| a → × b → | = | i ^ j ^ k ^ 2 − 4 5 1 − 2 − 3 | | a → × b → | = i ^ [ ( − 3 × − 4 ) − ( − 2 × 5 ) ] − j ^ [ ( 2 × − 3 ) − ( 1 × 5 ) ] + k ^ [ ( − 2 × 2 ) − ( − 4 × 1 ) ] | a → × b → | = i ^ ( 12 + 10 ) − j ^ ( − 11 ) + k ^ ( − 4 + 4 ) | a → × b → | = 22 i ^ + 11 j ^ + 0 k ^
The magnitude of | a → × b → | is,
| a → × b → | = 22 2 + 11 2 | a → × b → | = 11 2 ( 2 2 + 1 ) | a → × b → | = 11 5
Thus, the area of parallelogram is 11 5 square units.
Given, two adjacent sides of a parallelogram are
Consider these adjacent sides as
Let the diagonal be
The magnitude of
Unit vector of
The area of parallelogram is
The magnitude of
Thus, the area of parallelogram is
