Question

Prove that $[\overrightarrow{a}.\overrightarrow{b}.\overrightarrow{c}+\overrightarrow{d}]=[\overrightarrow{a}.\overrightarrow{b}.\overrightarrow{c}]+[\overrightarrow{a}.\overrightarrow{b}.\overrightarrow{d}]$.

Answer 1

Consider, $[\overrightarrow{a}\cdot \overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{d}]$

$=[\overrightarrow{a}\cdot \overrightarrow{b}\cdot (\overrightarrow{c}+\overrightarrow{d}\left)\right]$

$=[\overrightarrow{a}\cdot \overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{a}\cdot \overrightarrow{b}\cdot \overrightarrow{d}]$ ............ [Since, dot is distributive over addition]

$=[\overrightarrow{a}\cdot \overrightarrow{b}\cdot \overrightarrow{c}]+[\overrightarrow{a}\cdot \overrightarrow{b}\cdot \overrightarrow{d}]$

Hence, $[\overrightarrow{a}\cdot \overrightarrow{b}\cdot \overrightarrow{c}+\overrightarrow{d}]$$=[\overrightarrow{a}\cdot \overrightarrow{b}\cdot \overrightarrow{c}]+[\overrightarrow{a}\cdot \overrightarrow{b}\cdot \overrightarrow{d}]$