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Question Number 2499 by Syaka last updated on 21/Nov/15

$$ABCD.EFGHiscube,XismidpointEF$$$$ifAB=6cm,howdistanceAXtoBD??$$

Commented by prakash jain last updated on 21/Nov/15

$$\mathrm{What}\mathrm{is}\mathrm{AM}?\mathrm{Is}\mathrm{X}=\mathrm{M}?$$

Commented by Syaka last updated on 21/Nov/15

$$sorrySir,Imistakewhenwritethisquestion$$$$andnow{I}^{\prime }vecorrected$$

Answered by prakash jain last updated on 21/Nov/15

$$\mathrm{Let}\mathrm{A}(0,0,0)$$$$\mathrm{A}=(0,0,0)\mathrm{B}=(6,0,0)\mathrm{C}=(6,6,0)\mathrm{D}=(0,6,0)$$$$\mathrm{E}=(0,0,6)\mathrm{F}=(6,0,6)\mathrm{G}=(6,6,6)\mathrm{H}=(0,6,6)$$$$\mathrm{X}=(3,0,6)$$$$\mathrm{direction}\mathrm{vector}{\overrightarrow{d}}_1=\overrightarrow{\mathrm{BD}}=(-6,6,0)$$$$\mathrm{BD}:\frac{x-6}{-6}=\frac{y}{6}=t,z=0$$$$x=6-6t,y=6t,z=0\left(\mathrm{some}pointQ\mathrm{on}\mathrm{line}\mathrm{BD}\right)$$$$\mathrm{direction}\mathrm{vector}{\overrightarrow{d}}_2=\overrightarrow{\mathrm{AX}}=(3,0,6)$$$$\mathrm{AX}:\frac{x}{3}=\frac{z}{6}=s,y=0$$$$x=3s,z=6s,y=0\left(\mathrm{some}\mathrm{point}\mathrm{R}\mathrm{on}\mathrm{line}\mathrm{AX}\right)$$$$\overrightarrow{\mathrm{QR}}=(3s-6+6t,-6t,6s)$$$$\mathrm{To}\mathrm{calculate}\mathrm{the}\mathrm{distance}\mathrm{we}\mathrm{need}\mathrm{to}\mathrm{solve}$$$$\mathrm{for}s\mathrm{and}t\mathrm{such}\overrightarrow{\mathrm{QR}}\mathrm{\perp }\overrightarrow{\mathrm{BD}}\mathrm{and}\overrightarrow{\mathrm{QR}}\mathrm{\perp }\overrightarrow{\mathrm{AX}}.$$$$\overrightarrow{\mathrm{QR}}\bullet \overrightarrow{\mathrm{BD}}=0\mathrm{and}\overrightarrow{\mathrm{QR}}\bullet \overrightarrow{\mathrm{AX}}=0$$$$\mathrm{Once}s\mathrm{and}t\mathrm{are}\mathrm{found}\mathrm{we}\mathrm{can}\mathrm{find}\mathrm{coordinates}$$$$\mathrm{of}\mathrm{Q}\mathrm{and}\mathrm{R}\mathrm{and}\mathrm{calculate}\mathrm{the}\mathrm{distance}.$$$$\overrightarrow{\mathrm{QR}}\bullet \overrightarrow{\mathrm{BD}}=0\Rightarrow -18s+36-36t-36t=0$$$$s+4t=2....\left(1\right)$$$$$$$$\overrightarrow{\mathrm{QR}}\bullet \overrightarrow{\mathrm{AX}}=0\Rightarrow 9s-18+18t+36s=0$$$$5s+2t=2....\left(2\right)$$$$\mathrm{From}\left(1\right)\mathrm{and}\left(2\right)t=\frac{4}{9},s=\frac{2}{9}$$$$x=6-\frac{24}{9}=\frac{10}{3},y=\frac{8}{3},z=0\left(pointQ\mathrm{on}\mathrm{line}\mathrm{BD}\right)$$$$x=\frac{2}{3},z=\frac{4}{3},y=0\left(\mathrm{point}\mathrm{R}\mathrm{on}\mathrm{line}\mathrm{AX}\right)$$$$\overrightarrow{\mathrm{QR}}\mathrm{is}\mathrm{\perp }\mathrm{to}\mathrm{both}\mathrm{BD}\mathrm{and}\mathrm{AX}\mathrm{and}\mathrm{hence}\mathrm{its}$$$$\mathrm{length}\mathrm{gives}\mathrm{the}\mathrm{distance}.$$$$\mathrm{QR}=\sqrt{{\left(\frac{8}{3}\right)}^2+{\left(\frac{8}{3}\right)}^2+{\left(\frac{4}{3}\right)}^2}=\frac{1}{3}\sqrt{128+16}=4$$

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