Question and Answers Forum

All Questions      Topic List

Vector Calculus Questions

Previous in All Question      Next in All Question      

Previous in Vector Calculus      Next in Vector Calculus      

Question Number 179053 by Giantyusuf last updated on 24/Oct/22

Answered by Tokugami last updated on 27/Oct/22

OD=OA+AD  AD=OD−OA= ((d),((3d)) ) − ((8),(4) ) = (((d−8)),((3d−4)) )  BD=OD−OB= ((d),((3d)) ) −  (((12)),(6) ) = (((d−12)),((3d−6)) )    ∣AD∣=(√((d−8)^2 +(3d−4)^2 ))  ∣BD∣=(√((d−12)^2 +(3d−6)^2 ))  ((√((d−8)^2 +(3d−4)^2 ))=(√((d−12)^2 +(3d−6)^2 )))^2   (d−8)^2 +(3d−4)^2 =(d−12)^2 +(3d−6)^2   (d^2 −16d+64)+(9d^2 −24d+16)=(d^2 −24d+144)+(9d^2 −36d+36)  10d^2 −40d+80=10d^2 −60d+180  20d=100  d=5

$${OD}={OA}+{AD} \\ $$$${AD}={OD}−{OA}=\begin{pmatrix}{{d}}\\{\mathrm{3}{d}}\end{pmatrix}\:−\begin{pmatrix}{\mathrm{8}}\\{\mathrm{4}}\end{pmatrix}\:=\begin{pmatrix}{{d}−\mathrm{8}}\\{\mathrm{3}{d}−\mathrm{4}}\end{pmatrix} \\ $$$${BD}={OD}−{OB}=\begin{pmatrix}{{d}}\\{\mathrm{3}{d}}\end{pmatrix}\:−\:\begin{pmatrix}{\mathrm{12}}\\{\mathrm{6}}\end{pmatrix}\:=\begin{pmatrix}{{d}−\mathrm{12}}\\{\mathrm{3}{d}−\mathrm{6}}\end{pmatrix} \\ $$$$ \\ $$$$\mid{AD}\mid=\sqrt{\left({d}−\mathrm{8}\right)^{\mathrm{2}} +\left(\mathrm{3}{d}−\mathrm{4}\right)^{\mathrm{2}} } \\ $$$$\mid{BD}\mid=\sqrt{\left({d}−\mathrm{12}\right)^{\mathrm{2}} +\left(\mathrm{3}{d}−\mathrm{6}\right)^{\mathrm{2}} } \\ $$$$\left(\sqrt{\left({d}−\mathrm{8}\right)^{\mathrm{2}} +\left(\mathrm{3}{d}−\mathrm{4}\right)^{\mathrm{2}} }=\sqrt{\left({d}−\mathrm{12}\right)^{\mathrm{2}} +\left(\mathrm{3}{d}−\mathrm{6}\right)^{\mathrm{2}} }\right)^{\mathrm{2}} \\ $$$$\left({d}−\mathrm{8}\right)^{\mathrm{2}} +\left(\mathrm{3}{d}−\mathrm{4}\right)^{\mathrm{2}} =\left({d}−\mathrm{12}\right)^{\mathrm{2}} +\left(\mathrm{3}{d}−\mathrm{6}\right)^{\mathrm{2}} \\ $$$$\left({d}^{\mathrm{2}} −\mathrm{16}{d}+\mathrm{64}\right)+\left(\mathrm{9}{d}^{\mathrm{2}} −\mathrm{24}{d}+\mathrm{16}\right)=\left({d}^{\mathrm{2}} −\mathrm{24}{d}+\mathrm{144}\right)+\left(\mathrm{9}{d}^{\mathrm{2}} −\mathrm{36}{d}+\mathrm{36}\right) \\ $$$$\cancel{\mathrm{10}{d}^{\mathrm{2}} }−\mathrm{40}{d}+\mathrm{80}=\cancel{\mathrm{10}{d}^{\mathrm{2}} }−\mathrm{60}{d}+\mathrm{180} \\ $$$$\mathrm{20}{d}=\mathrm{100} \\ $$$${d}=\mathrm{5} \\ $$$$ \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com