At what point do the curves r1(t)=t,4−t,35+t2 and r2(s)=7−s,s−3,s2 intersect? (x,y,z)= Find angle of intersection, θ, correct to the nearest degree.

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curwyrm · Accepted Answer

Consider the following curves:(r_1(t)=)r2(s)=&amp;lt;7−s,s−3,s2&amp;gt;Find the point of intersection of the curves and angle of intersection θ.The parametric equations corresponding to the curve r1(t) are as follows:x=ty=4−tz=35+t2The parametric equations corresponding to the curve r2(s) are as follows:x=7−sy=s−3z=s2At the point of intersection of two curves, the corresponding coordinates of parametric lines are same.t=7−s⇒s+t=74−t=s−3⇒s+t=735+t2=s2⇒s2−t2=35That is,s2−t2=35(s−t)(s+t)=35(s−t)7=35s−t=5Solve the equations s−t=5 and s+t=7, to find the values of s and t.s−t+s+t=5+7⇒2s=12⇒s=6The corresponding value of t is as follows:t=7−s⇒t=7−6⇒t=1Thus, the values are t=1 and s=6Substitute t=1 in r1(t), to find the point of intersection.r1(t)=&amp;lt;1,4−1,35+12&amp;gt;=&amp;lt;1,3,36&amp;gt;r1(t)=&amp;lt;1,3,36&amp;gt;The angle between the normal vectors is given as cos⁡θ=r1′(t)⋅r2′(s)|r1′(t)||r2′(s)|The direction vector of the line r1(t) is ()The direction vector of the line r2(s) is &amp;lt;−s,s,s2&amp;gt;At t=1,r1′(t)=&amp;lt;1,−1,2t&amp;gt;r1′(1)=&amp;lt;1,−1,2&amp;gt;At s=6,r2′(s)=&amp;lt;−1,1,2s&amp;gt;r2′(6)=&amp;lt;−1,1,12&amp;gt;cos⁡θ=r1′(t)⋅r2′(s)|r1′(t)||r2′(s)|=r1′(1)⋅r2′(6)|r1′(1)||r2′(6)|=&amp;lt;1,−1,2&amp;gt;⋅&amp;lt;−1,1,12&amp;gt;1+1+4⋅1+1+144=−1−1+246⋅146=226⋅146=11219Thus, θ=cos−1⁡(11219)

At what point do the curves r_1(t)=t,4-t,35+t^2 and r_2(s)=7-s,s-3,s^2 intersect? (x,y,z)= Find angle of intersection, \theta, correct to the nearest degree.

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