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proof-that-a-b-c-a-c-b-c-or-the-distribiuting-law-of-dot-products-




Question Number 19857 by Nayon last updated on 16/Aug/17
proof that (a_− +b_− ).c_− =a_− .c_− +b_− .c_−   or the distribiuting law of dot products
$${proof}\:{that}\:\left(\underset{−} {{a}}+\underset{−} {{b}}\right).\underset{−} {{c}}=\underset{−} {{a}}.\underset{−} {{c}}+\underset{−} {{b}}.\underset{−} {{c}} \\ $$$${or}\:{the}\:{distribiuting}\:{law}\:{of}\:{dot}\:{products} \\ $$$$ \\ $$
Commented by Nayon last updated on 16/Aug/17
mr w1 pls explain
$${mr}\:{w}\mathrm{1}\:{pls}\:{explain} \\ $$
Commented by mrW1 last updated on 16/Aug/17
(a+b)∙c=∣a+b∣×cos θ×∣c∣  =PQ×∣c∣  =(PQ+QR)×∣c∣  =(∣a∣×cos θ_1 +∣b∣×cos θ_2 )×∣c∣  =∣a∣×cos θ_1 ×∣c∣+∣b∣×cos θ_2 ×∣c∣  =a∙c+b∙c
$$\left(\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\right)\centerdot\boldsymbol{\mathrm{c}}=\mid\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}\mid×\mathrm{cos}\:\theta×\mid\boldsymbol{\mathrm{c}}\mid \\ $$$$=\mathrm{PQ}×\mid\boldsymbol{\mathrm{c}}\mid \\ $$$$=\left(\mathrm{PQ}+\mathrm{QR}\right)×\mid\boldsymbol{\mathrm{c}}\mid \\ $$$$=\left(\mid\boldsymbol{\mathrm{a}}\mid×\mathrm{cos}\:\theta_{\mathrm{1}} +\mid\boldsymbol{\mathrm{b}}\mid×\mathrm{cos}\:\theta_{\mathrm{2}} \right)×\mid\boldsymbol{\mathrm{c}}\mid \\ $$$$=\mid\boldsymbol{\mathrm{a}}\mid×\mathrm{cos}\:\theta_{\mathrm{1}} ×\mid\boldsymbol{\mathrm{c}}\mid+\mid\boldsymbol{\mathrm{b}}\mid×\mathrm{cos}\:\theta_{\mathrm{2}} ×\mid\boldsymbol{\mathrm{c}}\mid \\ $$$$=\boldsymbol{\mathrm{a}}\centerdot\boldsymbol{\mathrm{c}}+\boldsymbol{\mathrm{b}}\centerdot\boldsymbol{\mathrm{c}} \\ $$
Commented by mrW1 last updated on 16/Aug/17

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