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Question Number 28640 by sk9236421@gmail.com last updated on 28/Jan/18

Each of the angle between vectors a, b and c is equal to 60°. If ∣a∣=4, ∣b∣=2 and ∣c∣=6, then the modulus of a+b+c is

$$\mathrm{Each}\:\mathrm{of}\:\mathrm{the}\:\mathrm{angle}\:\mathrm{between}\:\mathrm{vectors}\:\boldsymbol{\mathrm{a}}, \\ $$$$\boldsymbol{\mathrm{b}}\:\mathrm{and}\:\boldsymbol{\mathrm{c}}\:\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:\mathrm{60}°.\:\mathrm{If}\:\mid\boldsymbol{\mathrm{a}}\mid=\mathrm{4},\:\mid\boldsymbol{\mathrm{b}}\mid=\mathrm{2} \\ $$$$\mathrm{and}\:\mid\boldsymbol{\mathrm{c}}\mid=\mathrm{6},\:\mathrm{then}\:\mathrm{the}\:\mathrm{modulus}\:\mathrm{of} \\ $$$$\boldsymbol{\mathrm{a}}+\boldsymbol{\mathrm{b}}+\boldsymbol{\mathrm{c}}\:\:\mathrm{is} \\ $$

Commented by ajfour last updated on 28/Jan/18

Commented by ajfour last updated on 28/Jan/18

Answered by ajfour last updated on 28/Jan/18

let b^� =6i^� and a^� =2i^� +2(√3)j^� c^� =xi^� +yj^� +zk^� ; ∣c^� ∣=2 b^� .c^� =∣b^� ∣∣c^� ∣cos 60° ⇒ 6x =6×2×(1/2) ⇒ x=1 a^� .c^� =∣a^� ∣∣c^� ∣cos 60° ⇒ 2x+2(√3)y =4×2×(1/2) ⇒ 2+2(√3)y=4 ⇒ y=(1/(√3)) as ∣c^� ∣=2 ⇒ x^2 +y^2 +z^2 =4 so 1+(1/3)+z^2 =4 ⇒ z=±((2(√2))/(√3)) ∣a^� +b^� +c^� ∣=∣6i^� +2i^� +2(√3)j^� +i^� +(1/(√3))j^� ±((2(√2))/(√3))k^� ∣ =(√(81+((49)/3)+(8/3))) =(√(81+19)) ∣a^� +b^� +c^� ∣= 10 .

$${let}\:\:\bar {{b}}=\mathrm{6}\hat {{i}}\:\:\:{and} \\ $$$$\:\:\:\:\:\:\bar {{a}}=\mathrm{2}\hat {{i}}+\mathrm{2}\sqrt{\mathrm{3}}\hat {{j}} \\ $$$$\:\:\bar {{c}}\:={x}\hat {{i}}+{y}\hat {{j}}+{z}\hat {{k}}\:\:\:;\:\:\mid\bar {{c}}\mid=\mathrm{2} \\ $$$$\bar {{b}}.\bar {{c}}\:=\mid\bar {{b}}\mid\mid\bar {{c}}\mid\mathrm{cos}\:\mathrm{60}° \\ $$$$\Rightarrow\:\:\:\mathrm{6}{x}\:=\mathrm{6}×\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}}\: \\ $$$$\Rightarrow\:\:\:{x}=\mathrm{1} \\ $$$$\bar {{a}}.\bar {{c}}\:=\mid\bar {{a}}\mid\mid\bar {{c}}\mid\mathrm{cos}\:\mathrm{60}° \\ $$$$\Rightarrow\:\:\:\mathrm{2}{x}+\mathrm{2}\sqrt{\mathrm{3}}{y}\:=\mathrm{4}×\mathrm{2}×\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\:\:\:\:\Rightarrow\:\:\:\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}{y}=\mathrm{4} \\ $$$$\Rightarrow\:\:\:\:{y}=\frac{\mathrm{1}}{\sqrt{\mathrm{3}}} \\ $$$${as}\:\:\mid\bar {{c}}\mid=\mathrm{2}\:\:\Rightarrow\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{4} \\ $$$${so}\:\:\:\:\:\mathrm{1}+\frac{\mathrm{1}}{\mathrm{3}}+{z}^{\mathrm{2}} =\mathrm{4} \\ $$$$\Rightarrow\:\:\:{z}=\pm\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}} \\ $$$$\mid\bar {{a}}+\bar {{b}}+\bar {{c}}\mid=\mid\mathrm{6}\hat {{i}}+\mathrm{2}\hat {{i}}+\mathrm{2}\sqrt{\mathrm{3}}\hat {{j}}+\hat {{i}}+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\hat {{j}}\pm\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\sqrt{\mathrm{3}}}\hat {{k}}\mid \\ $$$$\:\:=\sqrt{\mathrm{81}+\frac{\mathrm{49}}{\mathrm{3}}+\frac{\mathrm{8}}{\mathrm{3}}}\:=\sqrt{\mathrm{81}+\mathrm{19}} \\ $$$$\mid\bar {{a}}+\bar {{b}}+\bar {{c}}\mid=\:\mathrm{10}\:.\:\: \\ $$

Answered by ajfour last updated on 28/Jan/18

∣a^� +b^� +c^� ∣^2 =a^2 +b^2 +c^2 +2(a^� .b^� +b^� .c^� +c^� .a^� ) =16+4+36+(12+8+24) =100 ∣a^� +b^� +c^� ∣=10 .

$$\mid\bar {{a}}+\bar {{b}}+\bar {{c}}\mid^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left(\bar {{a}}.\bar {{b}}+\bar {{b}}.\bar {{c}}+\bar {{c}}.\bar {{a}}\right) \\ $$$$\:\:\:=\mathrm{16}+\mathrm{4}+\mathrm{36}+\left(\mathrm{12}+\mathrm{8}+\mathrm{24}\right)\:=\mathrm{100} \\ $$$$\mid\bar {{a}}+\bar {{b}}+\bar {{c}}\mid=\mathrm{10}\:. \\ $$

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