Question

If a,b,c are mutually perpendicular unit vectors then show that det{a+b+c}=cube root of 3

Answer 1

a, b and c are three mutual perpendicular unit vector .

|a | = | b | = | c | = 1

a.b = 0
b.c = 0
c.a = 0

so, ( ab + bc + ca) = 0 -----------_(1)

we know, that ,

(a + b + c)² = a² + b² + c² +2( ab + bc + ca)

also we know ,
a² = |a|²

so,

( a + b + c)² = |a|² + |b|² + |c|² +2( ab+ bc +ca) = | a|² + |b|² + |c|² +2 ×0 from eqn (1)

(a + b + c)² = 1² + 1² + 1² = 3

(a + b +c )² = 3

(a + b + c ) = ±√3

so, det{ a + b + c } = | a + b + c | = √3