Menu Close

Question-168361




Question Number 168361 by leicianocosta last updated on 08/Apr/22
Answered by som(math1967) last updated on 09/Apr/22
x=−kz−5   x+5=−kz   ((x+5)/(−k))=z  y =−3z⇒(y/(−3))=z  equn.of r  ((x+5)/(−k))=(y/(−3))=(z/1)    equn. of s     ((x+3)/(−1))=((y−2)/(−2))=((z+1)/1)  r and s perpendicular   (−k)(−1)+(−3)(−2)+(1)(1)=0   ⇒k+7=0 ∴k=−7
$$\boldsymbol{{x}}=−\boldsymbol{{kz}}−\mathrm{5} \\ $$$$\:\boldsymbol{{x}}+\mathrm{5}=−\boldsymbol{{kz}} \\ $$$$\:\frac{\boldsymbol{{x}}+\mathrm{5}}{−\boldsymbol{{k}}}=\boldsymbol{{z}} \\ $$$$\boldsymbol{{y}}\:=−\mathrm{3}\boldsymbol{{z}}\Rightarrow\frac{{y}}{−\mathrm{3}}=\boldsymbol{{z}} \\ $$$$\boldsymbol{{equn}}.\boldsymbol{{of}}\:\boldsymbol{{r}}\:\:\frac{\boldsymbol{{x}}+\mathrm{5}}{−\boldsymbol{{k}}}=\frac{\boldsymbol{{y}}}{−\mathrm{3}}=\frac{\boldsymbol{{z}}}{\mathrm{1}} \\ $$$$\:\:{equn}.\:{of}\:\boldsymbol{{s}} \\ $$$$\:\:\:\frac{\boldsymbol{{x}}+\mathrm{3}}{−\mathrm{1}}=\frac{\boldsymbol{{y}}−\mathrm{2}}{−\mathrm{2}}=\frac{\boldsymbol{{z}}+\mathrm{1}}{\mathrm{1}} \\ $$$$\boldsymbol{{r}}\:\boldsymbol{{and}}\:\boldsymbol{{s}}\:\boldsymbol{{perpendicular}} \\ $$$$\:\left(−\boldsymbol{{k}}\right)\left(−\mathrm{1}\right)+\left(−\mathrm{3}\right)\left(−\mathrm{2}\right)+\left(\mathrm{1}\right)\left(\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\Rightarrow\boldsymbol{{k}}+\mathrm{7}=\mathrm{0}\:\therefore\boldsymbol{{k}}=−\mathrm{7} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *