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Question Number 98018 by hardylanes last updated on 11/Jun/20

the vector equations of two lines l_1  and l_2  are given by  l_1 :r=5i−j+k+λ(−3i+2j)  l_2 :r=2i+3j+2k+μ(2j+k)  find  thd position vdctor of intersection of thf linds l_1  and l_2    thd cartesian equatkon of the plane π, containing the lines l_(1 ) and l_2   the sine of the anhle between the plane π and the line, l_3 :r=i−5j−2k+s(2i+2j−k)

thevectorequationsoftwolinesl1andl2aregivenbyl1:r=5ij+k+λ(3i+2j)l2:r=2i+3j+2k+μ(2j+k)findthdpositionvdctorofintersectionofthflindsl1andl2thdcartesianequatkonoftheplaneπ,containingthelinesl1andl2thesineoftheanhlebetweentheplaneπandtheline,l3:r=i5j2k+s(2i+2jk)

Answered by Rio Michael last updated on 11/Jun/20

(a) l_1  : r = (5−3λ)i + (−1 + 2λ)j + k = (((5−3λ)),((−1 + 2λ)),(1) )  l_2 : r = 2i + (3 + 2μ)j + (2 + μ)k =  ((2),((3 + 2μ)),((2 + μ)) )  at intersection, l_1  = l_(2 )  ⇒  5−3λ = 2 ⇔ λ = 1  and  2 + μ = 1 ⇔ μ = −1  point of intersection r_0  = 2i + j + k  (b) equation of plane  r.n = r_0 .n  n^�  = d_1  ×d_2  =  determinant ((i,j,k),((−3),2,0),(0,2,1)) = i determinant ((2,0),(2,1))−j determinant (((−3),0),(0,1))+ k determinant (((−3),2),(0,2))= 2i +3j −6k  ⇒   r.(2i + 3j−6k) = (2i + j +k).(2i+3j−6k)           2x + 3y−6z = 1  (c) sin θ = ((n.d)/(∣n∣∣d∣)) = (((2i +3j−6k).(2i + 2j−k))/((√(49)) .(√9))) = ((16)/(21))   ⇒ sin θ = ((16)/(21))

(a)l1:r=(53λ)i+(1+2λ)j+k=(53λ1+2λ1)l2:r=2i+(3+2μ)j+(2+μ)k=(23+2μ2+μ)atintersection,l1=l253λ=2λ=1and2+μ=1μ=1pointofintersectionr0=2i+j+k(b)equationofplaner.n=r0.nn¯=d1×d2=|ijk320021|=i|2021|j|3001|+k|3202|=2i+3j6kr.(2i+3j6k)=(2i+j+k).(2i+3j6k)2x+3y6z=1(c)sinθ=n.dn∣∣d=(2i+3j6k).(2i+2jk)49.9=1621sinθ=1621

Commented by hardylanes last updated on 11/Jun/20

thanks again

Commented by peter frank last updated on 11/Jun/20

good

good

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