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Question-19912




Question Number 19912 by ajfour last updated on 17/Aug/17
Commented by ajfour last updated on 18/Aug/17
Find tangent of the acute angle  between AB and SC and the  shortest distance between them.  The side of the equilateral triangle  is even a. SA is perpendicular to  the plane of △ABC.
$${Find}\:{tangent}\:{of}\:{the}\:{acute}\:{angle} \\ $$$${between}\:{AB}\:{and}\:{SC}\:{and}\:{the} \\ $$$${shortest}\:{distance}\:{between}\:{them}. \\ $$$${The}\:{side}\:{of}\:{the}\:{equilateral}\:{triangle} \\ $$$${is}\:{even}\:\boldsymbol{{a}}.\:{SA}\:{is}\:{perpendicular}\:{to} \\ $$$${the}\:{plane}\:{of}\:\bigtriangleup{ABC}. \\ $$
Answered by ajfour last updated on 19/Aug/17
Commented by ajfour last updated on 19/Aug/17
AB^(→) =(a/2)i^� +((a(√3))/2)j^�   ;  AB=a  SC^(→) =−(a/2)i^� +((a(√3))/2)j^� −ak^�  ;  SC=a(√2)  let angle between AB and SC   be α .  (AB)(SC)cos α=AB^(→) .SC^(→)   ⇒  a^2 (√2)cos α=−(a^2 /4)+((3a^2 )/4)=(a^2 /2)       cos α=(1/(2(√2)))   ⇒  tan α=(√7)       α=tan^(−1) (√7^ )   AB and SC are a pair of skew  lines; let shortest distance be d.      d=((AC^(→) .(AB^(→) ×SC^(→) ))/(∣AB^(→) ×SC^(→) )))  =(((−(a/2)i^� +((√3)/2)j^� ).(−((√3)/2)i^� +(1/2)j^� +((√3)/2)k^� ))/((((√7)/2))))    d=(2/( (√7)))(((a(√3))/4)+((a(√3))/4)) =(2/( (√7)))(((a(√3))/2))          d=a(√(3/7)) .
$$\overset{\rightarrow} {{AB}}=\frac{{a}}{\mathrm{2}}\hat {{i}}+\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}\hat {{j}}\:\:;\:\:{AB}={a} \\ $$$$\overset{\rightarrow} {{SC}}=−\frac{{a}}{\mathrm{2}}\hat {{i}}+\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}\hat {{j}}−{a}\hat {{k}}\:;\:\:{SC}={a}\sqrt{\mathrm{2}} \\ $$$${let}\:{angle}\:{between}\:{AB}\:{and}\:{SC}\: \\ $$$${be}\:\alpha\:. \\ $$$$\left({AB}\right)\left({SC}\right)\mathrm{cos}\:\alpha=\overset{\rightarrow} {{AB}}.\overset{\rightarrow} {{SC}} \\ $$$$\Rightarrow\:\:{a}^{\mathrm{2}} \sqrt{\mathrm{2}}\mathrm{cos}\:\alpha=−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}+\frac{\mathrm{3}{a}^{\mathrm{2}} }{\mathrm{4}}=\frac{{a}^{\mathrm{2}} }{\mathrm{2}} \\ $$$$\:\:\:\:\:\mathrm{cos}\:\alpha=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}\:\:\:\Rightarrow\:\:\mathrm{tan}\:\alpha=\sqrt{\mathrm{7}}\: \\ $$$$\:\:\:\:\alpha=\mathrm{tan}^{−\mathrm{1}} \sqrt{\mathrm{7}^{} }\: \\ $$$${AB}\:{and}\:{SC}\:{are}\:{a}\:{pair}\:{of}\:{skew} \\ $$$${lines};\:{let}\:{shortest}\:{distance}\:{be}\:\boldsymbol{{d}}. \\ $$$$\:\:\:\:\boldsymbol{{d}}=\frac{\overset{\rightarrow} {{AC}}.\left(\overset{\rightarrow} {{AB}}×\overset{\rightarrow} {{SC}}\right)}{\left.\mid\overset{\rightarrow} {{AB}}×\overset{\rightarrow} {{SC}}\right)} \\ $$$$=\frac{\left(−\frac{{a}}{\mathrm{2}}\hat {{i}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\hat {{j}}\right).\left(−\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\hat {{i}}+\frac{\mathrm{1}}{\mathrm{2}}\hat {{j}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\hat {{k}}\right)}{\left(\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}\right)} \\ $$$$\:\:{d}=\frac{\mathrm{2}}{\:\sqrt{\mathrm{7}}}\left(\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{4}}+\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{4}}\right)\:=\frac{\mathrm{2}}{\:\sqrt{\mathrm{7}}}\left(\frac{{a}\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:\:\:\:\:{d}={a}\sqrt{\frac{\mathrm{3}}{\mathrm{7}}}\:. \\ $$$$ \\ $$

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