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Question Number 154948 by mathdanisur last updated on 23/Sep/21

If  a^(→)  = (-1;0;-3;4)        b^(→)  = (1;-4;0;-3)        c^(→)  = ((∣a^(→) ∣)/(∣b^(→) ∣)) ∙ (2a^(→)  + b^(→) )         d^(→)  = ((∣b^(→) ∣)/(∣a^(→) ∣)) ∙ (a^(→)  + b^(→) )         k^(→)  = -3∙(c^(→)  ∙ ∣c^(→) ∣ ∙ d^(→) )  Find  ∣k^(→) ∣ = ?

$$\mathrm{If}\:\:\overset{\rightarrow} {\mathrm{a}}\:=\:\left(-\mathrm{1};\mathrm{0};-\mathrm{3};\mathrm{4}\right) \\ $$$$\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{b}}\:=\:\left(\mathrm{1};-\mathrm{4};\mathrm{0};-\mathrm{3}\right) \\ $$$$\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{c}}\:=\:\frac{\mid\overset{\rightarrow} {\mathrm{a}}\mid}{\mid\overset{\rightarrow} {\mathrm{b}}\mid}\:\centerdot\:\left(\mathrm{2}\overset{\rightarrow} {\mathrm{a}}\:+\:\overset{\rightarrow} {\mathrm{b}}\right) \\ $$$$\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{d}}\:=\:\frac{\mid\overset{\rightarrow} {\mathrm{b}}\mid}{\mid\overset{\rightarrow} {\mathrm{a}}\mid}\:\centerdot\:\left(\overset{\rightarrow} {\mathrm{a}}\:+\:\overset{\rightarrow} {\mathrm{b}}\right) \\ $$$$\:\:\:\:\:\:\:\overset{\rightarrow} {\mathrm{k}}\:=\:-\mathrm{3}\centerdot\left(\overset{\rightarrow} {\mathrm{c}}\:\centerdot\:\mid\overset{\rightarrow} {\mathrm{c}}\mid\:\centerdot\:\overset{\rightarrow} {\mathrm{d}}\right) \\ $$$$\mathrm{Find}\:\:\mid\overset{\rightarrow} {\mathrm{k}}\mid\:=\:? \\ $$

Commented by JDamian last updated on 23/Sep/21

how is defined the product of two vectors?

Commented by prakash jain last updated on 23/Sep/21

c^→ ∙∣c^→ ∣∙d^→  is scaler.  if ∙ is dot product

$$\overset{\rightarrow} {{c}}\centerdot\mid\overset{\rightarrow} {{c}}\mid\centerdot\overset{\rightarrow} {{d}}\:\mathrm{is}\:\mathrm{scaler}. \\ $$$$\mathrm{if}\:\centerdot\:\mathrm{is}\:\mathrm{dot}\:\mathrm{product} \\ $$

Commented by mathdanisur last updated on 23/Sep/21

Sorry Ser,  ∣k^(→) ∣ = -3(c^(→)  - ∣c^(→) ∣ ∙ d^(→) )

$$\mathrm{Sorry}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\:\mid\overset{\rightarrow} {\mathrm{k}}\mid\:=\:-\mathrm{3}\left(\overset{\rightarrow} {\mathrm{c}}\:-\:\mid\overset{\rightarrow} {\mathrm{c}}\mid\:\centerdot\:\overset{\rightarrow} {\mathrm{d}}\right) \\ $$

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