Question-186701

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Question Number 186701 by Mingma last updated on 08/Feb/23

Commented by aba last updated on 08/Feb/23

$$\frac{\mathrm{4}}{\mathrm{5}} \\ $$

Commented by Mingma last updated on 09/Feb/23

can you show workings?

Answered by som(math1967) last updated on 09/Feb/23

(a^2 +b^2 +c^2 )(x^2 +y^2 +z^2 )−(ax+by+cz)^2 =25×16−20^2 a^2 x^2 +a^2 y^2 +a^2 z^2 +b^2 x^2 +b^2 y^2 +b^2 z^2 c^2 x^2 +c^2 y^2 +c^2 z^2 −a^2 x^2 −b^2 y^2 −c^2 z^2 −2abxy−2bcyz−2cazx=0 (ay−bx)^2 +(bz−cy)^2 +(az−cx)^2 =0 ⇒(ay−bx)=0⇒(x/y)=(a/b)⇒(x/a)=(y/b) ⇒bz−cy=0⇒(y/z)=(b/c)⇒(y/b)=(z/c) ∴(x/a)=(y/b)=(z/c)=k (let) x=ak,y=bk,z=ck x^2 +y^2 +z^2 =25 k^2 (a^2 +b^2 +c^2 )=25 k=(5/4) ((a+b+c)/(x+y+z))=(((a+b+c))/(k(a+b+c)))=(1/k)=(4/5)

$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)−\left({ax}+{by}+{cz}\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:=\mathrm{25}×\mathrm{16}−\mathrm{20}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} {x}^{\mathrm{2}} +{a}^{\mathrm{2}} {y}^{\mathrm{2}} +{a}^{\mathrm{2}} {z}^{\mathrm{2}} +{b}^{\mathrm{2}} {x}^{\mathrm{2}} +{b}^{\mathrm{2}} {y}^{\mathrm{2}} +{b}^{\mathrm{2}} {z}^{\mathrm{2}} \\ $$$${c}^{\mathrm{2}} {x}^{\mathrm{2}} +{c}^{\mathrm{2}} {y}^{\mathrm{2}} +{c}^{\mathrm{2}} {z}^{\mathrm{2}} −{a}^{\mathrm{2}} {x}^{\mathrm{2}} −{b}^{\mathrm{2}} {y}^{\mathrm{2}} −{c}^{\mathrm{2}} {z}^{\mathrm{2}} \\ $$$$−\mathrm{2}{abxy}−\mathrm{2}{bcyz}−\mathrm{2}{cazx}=\mathrm{0} \\ $$$$\left({ay}−{bx}\right)^{\mathrm{2}} +\left({bz}−{cy}\right)^{\mathrm{2}} +\left({az}−{cx}\right)^{\mathrm{2}} =\mathrm{0} \\ $$$$\Rightarrow\left({ay}−{bx}\right)=\mathrm{0}\Rightarrow\frac{{x}}{{y}}=\frac{{a}}{{b}}\Rightarrow\frac{{x}}{{a}}=\frac{{y}}{{b}} \\ $$$$\Rightarrow{bz}−{cy}=\mathrm{0}\Rightarrow\frac{{y}}{{z}}=\frac{{b}}{{c}}\Rightarrow\frac{{y}}{{b}}=\frac{{z}}{{c}} \\ $$$$\therefore\frac{{x}}{{a}}=\frac{{y}}{{b}}=\frac{{z}}{{c}}={k}\:\left({let}\right) \\ $$$${x}={ak},{y}={bk},{z}={ck} \\ $$$$\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{25} \\ $$$${k}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)=\mathrm{25} \\ $$$$\:{k}=\frac{\mathrm{5}}{\mathrm{4}} \\ $$$$\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\frac{\left({a}+{b}+{c}\right)}{{k}\left({a}+{b}+{c}\right)}=\frac{\mathrm{1}}{{k}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$

Answered by mr W last updated on 09/Feb/23

vector method p=ai+bj+ck q=xi+yj+zk ∣p∣=(√(a^2 +b^2 +c^2 ))=(√(16))=4 ∣q∣=(√(x^2 +y^2 +z^2 ))=(√(25))=5 p∙q=ax+by+cz=20 p∙q=∣p∣∣q∣ cos θ=4×5 cos θ=20 cos θ 20 cos θ=20 ⇒cos θ=1 ⇒θ=0° ⇒p//q ⇒(a/x)=(b/y)=(c/z)=λ, say λ=((∣p∣)/(∣q∣))=(4/5) ((a+b+c)/(x+y+z))=((λx+λy+λz)/(x+y+z))=λ=(4/5) ✓

$${vector}\:{method} \\ $$$$\boldsymbol{{p}}={ai}+{bj}+{ck} \\ $$$$\boldsymbol{{q}}={xi}+{yj}+{zk} \\ $$$$\mid\boldsymbol{{p}}\mid=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }=\sqrt{\mathrm{16}}=\mathrm{4} \\ $$$$\mid\boldsymbol{{q}}\mid=\sqrt{{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} }=\sqrt{\mathrm{25}}=\mathrm{5} \\ $$$$\boldsymbol{{p}}\centerdot\boldsymbol{{q}}={ax}+{by}+{cz}=\mathrm{20} \\ $$$$\boldsymbol{{p}}\centerdot\boldsymbol{{q}}=\mid\boldsymbol{{p}}\mid\mid\boldsymbol{{q}}\mid\:\mathrm{cos}\:\theta=\mathrm{4}×\mathrm{5}\:\mathrm{cos}\:\theta=\mathrm{20}\:\mathrm{cos}\:\theta \\ $$$$\mathrm{20}\:\mathrm{cos}\:\theta=\mathrm{20}\:\Rightarrow\mathrm{cos}\:\theta=\mathrm{1}\:\Rightarrow\theta=\mathrm{0}°\:\Rightarrow\boldsymbol{{p}}//\boldsymbol{{q}} \\ $$$$\Rightarrow\frac{{a}}{{x}}=\frac{{b}}{{y}}=\frac{{c}}{{z}}=\lambda,\:{say} \\ $$$$\lambda=\frac{\mid\boldsymbol{{p}}\mid}{\mid\boldsymbol{{q}}\mid}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\frac{{a}+{b}+{c}}{{x}+{y}+{z}}=\frac{\lambda{x}+\lambda{y}+\lambda{z}}{{x}+{y}+{z}}=\lambda=\frac{\mathrm{4}}{\mathrm{5}}\:\checkmark \\ $$

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