Question

Show that a x (b+c) + bx (c+ a) + c x(a+b) = 0​

Answer 1

Step-by-step explanation:

Use distributive laws to get A⃗ ×(B⃗ +C⃗ )=A⃗ ×B⃗ + A⃗ ×C⃗

B⃗ ×(C⃗ +A⃗ )=B⃗ ×C⃗ + B⃗ ×A⃗

C⃗ ×(A⃗ +B⃗ )=C⃗ ×A⃗ + C⃗ ×B⃗

Now just add them and rearrange to get this :

(A×B+B×A)+(B×C+C×B)+(C×A+A×C)

But A⃗ ×B⃗ =−(B⃗ ×A⃗ )

B⃗ ×C⃗ =−(C⃗ ×B⃗ )

C⃗ ×A⃗ =−(A⃗ ×C⃗ )

So you get 0 + 0 + 0 = 0

Tell me if you need further help.

Expand all the brackets like this.

You get 6 term in total.

For a term of kind A⃗ ×B⃗ there is a term of the form B⃗ ×A⃗

Which cancel out when added. There are 3 such pairs who are cyclic permutations of one another.

I hope you get the solution.No

Answer 2

$\huge{★Question ↴}$

●Show that a x (b+c) + bx (c+ a) + c x(a+b) = 0

$\huge{★Answer ↴}$

a x (b+c) + bx (c+ a) + c x(a+b) = 0

abx+acx+bcx+abx+acx+bcx=0

x(ab+ac+bc+ab+ac+bc) =0

x=0

Hope its help❤