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Question Number 8150 by trapti rathaur@ gmail.com last updated on 02/Oct/16

$$showthattheplanex+2y-3z+d=0isperpendiculrtoeachof$$$$theplaneis2x+5y+4z+1=0and4x+7y+3z+2=0.$$

Commented by 123456 last updated on 03/Oct/16

$$\alpha :x+2y-3z+d=0$$$$\beta :2x+5y+4z+1=0$$$$\gamma :4x+7y+3z+2=0$$$$\mathrm{lets}\overrightarrow{r},\overrightarrow{s},\overrightarrow{t}\mathrm{be}\mathrm{the}\mathrm{normal}\mathrm{vector}\mathrm{of}$$$$\alpha ,\beta ,\gamma \mathrm{we}\mathrm{have}$$$$\overrightarrow{r}=(1,2,-3)$$$$\overrightarrow{s}=(2,5,4)$$$$\overrightarrow{t}=(4,7,3)$$$$\mathrm{if}\mathrm{two}\mathrm{plan}\mathrm{are}\mathrm{perp},\mathrm{their}\mathrm{normal}\mathrm{vector}$$$$\mathrm{are}\mathrm{perp}\mathrm{too}$$$$\mathrm{two}\mathrm{vector}\mathrm{are}\mathrm{perp}\mathrm{if}\mathrm{their}\mathrm{dot}\mathrm{product}$$$$\mathrm{equal}\mathrm{to}0$$$$\overrightarrow{r}\cdot \overrightarrow{s}=(1,2,-3)\cdot (2,5,4)$$$$=1\times 2+2\times 5-3\times 4$$$$=2+10-12$$$$=12-12$$$$=0$$$$\overrightarrow{r}\cdot \overrightarrow{t}=(1,2,-3)\cdot (4,7,3)$$$$=1\times 4+2\times 7-3\times 3$$$$=4+14-9$$$$=18-9$$$$=9\ne 0$$$$\overrightarrow{s}\cdot \overrightarrow{t}=(2,5,4)\cdot (4,7,3)$$$$=2\times 4+5\times 7+4\times 3$$$$=8+35+12$$$$=55\ne 0$$

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