Question

Show that (a + b) . (b + c) × (c + a) = 2[a b c].

Answer 1

Question.

Show that (a + b) . (b + c) × (c + a) = 2[a b c].



Answer



FORMULA USED.


[a.(b×c)]=[abc]


or as u know

a (c×a)=0

b.(b×c)=o

b.(c×a)=0


are all coplanar so value become 0

Solution

(a + b) . (b + c) × (c + a)


=(a+b).{(b×c)+(b×a)+(c×c)+(c×a)}


=(a+b).{(b×c)+(b×a)+(c×a)}. [ (c×c)=0]



=a.(b×c)+a.(b×a)+a.(c×a)+b.(b×c)+b.(b×a)+b.(c×a)


apply coplanar properties

=>a.(b×c)+b.(c×a)



=[abc]+[abc]


=2[abc]. Ans

Answer 2

Answer:

Step-by-step explanation:

We know that a.( b + c ) = ( a.b.c )

( a+b).(b+c).(c+a) = $\left[\begin{array}{ccc}1&1&0\\0&1&1\\1&0&1\end{array}\right]$$\left[\begin{array}{ccc}a&b&c\end{array}\right]$

= [ 1(1-0)-1(0-1)+0(0-1)]$\left[\begin{array}{ccc}a&b&c\end{array}\right]$

= ( 1+1 )$\left[\begin{array}{ccc}a&b&c\end{array}\right]$

=2$\left[\begin{array}{ccc}a&b&c\end{array}\right]$

= RHS