Question and Answers Forum

All Questions      Topic List

Vector Questions

Previous in All Question      Next in All Question      

Previous in Vector      Next in Vector      

Question Number 133199 by rexford last updated on 19/Feb/21

Answered by mr W last updated on 20/Feb/21

AB^(→) =(−1,5,−3) AC^(→) =(−4,3,3) AD^(→) =(1,7,λ+1) 1×(15+9)−7×(−3−12)+(λ+1)(−3+20)=0 24+7×15+17(λ+1)=0 ⇒λ=−((146)/(17))

$$\overset{\rightarrow} {\boldsymbol{{AB}}}=\left(−\mathrm{1},\mathrm{5},−\mathrm{3}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{AC}}}=\left(−\mathrm{4},\mathrm{3},\mathrm{3}\right) \\ $$$$\overset{\rightarrow} {\boldsymbol{{AD}}}=\left(\mathrm{1},\mathrm{7},\lambda+\mathrm{1}\right) \\ $$$$\mathrm{1}×\left(\mathrm{15}+\mathrm{9}\right)−\mathrm{7}×\left(−\mathrm{3}−\mathrm{12}\right)+\left(\lambda+\mathrm{1}\right)\left(−\mathrm{3}+\mathrm{20}\right)=\mathrm{0} \\ $$$$\mathrm{24}+\mathrm{7}×\mathrm{15}+\mathrm{17}\left(\lambda+\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=−\frac{\mathrm{146}}{\mathrm{17}} \\ $$

Commented by liberty last updated on 20/Feb/21

typo. 24+105+17=105+41=146

$$\mathrm{typo}.\:\mathrm{24}+\mathrm{105}+\mathrm{17}=\mathrm{105}+\mathrm{41}=\mathrm{146} \\ $$

Commented by mr W last updated on 20/Feb/21

yes, thanks!

$${yes},\:{thanks}! \\ $$

Commented by rexford last updated on 20/Feb/21

Thanks

$${Thanks} \\ $$

Answered by john_santu last updated on 20/Feb/21

let n^→ = a^→ ×b^→ be a normal vector plane then n^→ = determinant (((3 −2 −1)),((2 3 −4)))= 11i^� +10j^� +13k^� since the points A,B,C and D so n^→ .CD = 0 ; where CD = (5,4,λ−2) ⇔ 55+40+13λ−26=0 ⇔ λ = ((26)/(99))

$$\mathrm{let}\:\overset{\rightarrow} {\mathrm{n}}\:=\:\overset{\rightarrow} {{a}}×\overset{\rightarrow} {{b}}\:\mathrm{be}\:\mathrm{a}\:\mathrm{normal}\:\mathrm{vector}\:\mathrm{plane} \\ $$$$\mathrm{then}\:\overset{\rightarrow} {\mathrm{n}}\:=\:\begin{vmatrix}{\mathrm{3}\:\:\:\:−\mathrm{2}\:\:\:\:\:−\mathrm{1}}\\{\mathrm{2}\:\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:−\mathrm{4}}\end{vmatrix}=\:\mathrm{11}\hat {\mathrm{i}}+\mathrm{10}\hat {\mathrm{j}}+\mathrm{13}\hat {\mathrm{k}} \\ $$$$\mathrm{since}\:\mathrm{the}\:\mathrm{points}\:\mathrm{A},\mathrm{B},\mathrm{C}\:\mathrm{and}\:\mathrm{D}\: \\ $$$$\mathrm{so}\:\overset{\rightarrow} {\mathrm{n}}.\boldsymbol{\mathrm{CD}}\:=\:\mathrm{0}\:;\:\mathrm{where}\:\boldsymbol{\mathrm{CD}}\:=\:\left(\mathrm{5},\mathrm{4},\lambda−\mathrm{2}\right) \\ $$$$\Leftrightarrow\:\mathrm{55}+\mathrm{40}+\mathrm{13}\lambda−\mathrm{26}=\mathrm{0} \\ $$$$\Leftrightarrow\:\lambda\:=\:\frac{\mathrm{26}}{\mathrm{99}}\: \\ $$

Answered by EDWIN88 last updated on 20/Feb/21

Answered by liberty last updated on 20/Feb/21

normal vector plane n^→ =AB×AC = determinant (((−1 5 −3 )),((−4 3 3))) n^→ = 24i^� +15j^� +17k^� put AD = i^� +7j^� +(λ+1)k^� so we get equation of plane is n^→ .AD = 0 ⇒24+105+17λ+17 = 0 ⇒λ=−((146)/(17))

$$\mathrm{normal}\:\mathrm{vector}\:\mathrm{plane}\: \\ $$$$\overset{\rightarrow} {\mathrm{n}}\:=\mathrm{AB}×\mathrm{AC}\:=\begin{vmatrix}{−\mathrm{1}\:\:\:\:\:\:\:\mathrm{5}\:\:\:\:\:\:−\mathrm{3}\:\:\:\:}\\{−\mathrm{4}\:\:\:\:\:\:\mathrm{3}\:\:\:\:\:\:\:\:\:\mathrm{3}}\end{vmatrix} \\ $$$$\overset{\rightarrow} {\mathrm{n}}=\:\mathrm{24}\hat {\mathrm{i}}+\mathrm{15}\hat {\mathrm{j}}+\mathrm{17}\hat {\mathrm{k}} \\ $$$$\mathrm{put}\:\mathrm{AD}\:=\:\hat {\mathrm{i}}+\mathrm{7}\hat {\mathrm{j}}+\left(\lambda+\mathrm{1}\right)\hat {\mathrm{k}} \\ $$$$\mathrm{so}\:\mathrm{we}\:\mathrm{get}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{plane}\:\mathrm{is} \\ $$$$\overset{\rightarrow} {\mathrm{n}}.\mathrm{AD}\:=\:\mathrm{0} \\ $$$$\Rightarrow\mathrm{24}+\mathrm{105}+\mathrm{17}\lambda+\mathrm{17}\:=\:\mathrm{0} \\ $$$$\Rightarrow\lambda=−\frac{\mathrm{146}}{\mathrm{17}} \\ $$

Commented by rexford last updated on 20/Feb/21

Thank you very much

$${Thank}\:{you}\:{very}\:{much} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com