Question

Given lines L1 : Ty (1-1)+(21) -2 1 and L2: y 4 8+t2 3 (a) Find the point of intersection of lines Lị and L2. (b) Determine the cosine of the angle between lines L, and L2 at the point of intersection. © Find an equation in form ax +by+cz = d for the plane containing lines L, and Lu. (d) Find the intersection, if any, of the line Ly and the plane P : 3x – 4y + 72 = 48.

Answer 1

محطم 'را lines Given مسر 2 ي }م !را 3 7

Nρώ, (a) If the line that in so from 4-23 رنا۔ ah & L2 are intenseet then we get a point both lines holtz. A) 4+1 aas+tao 3+ 3 = 8+27 2.8-27-541) a-sa 7+37 ay 3+3+=-57 tivi) ( This happen becouse at the point of intensection both of three coordinates are same for L&L2]. Solving (1) & () we get, sad, to-2 the equation liii) [As 1+ 3(-2):5). sal, tand be the solution of (i), (ii), (ii). from (1) this satisfy So Then 2 so the point of intersection is (2,4, 1)? (b) from a we get the dirs. of 4 is ())-(1)--(1)-9).- )( () and and dirs.of La is Torm -3 Cosa - To V14 so the Then if the angle be stored between hi & L2 is a then, .3 -2+2-3 2191 Wheth 114+9 angle between h&L2 at the point of intersection, cosine is at all (e) Let ax aby &cz =d be the equation of plane cont lines h. L2. Then dors. of the normal line of the plane is a,b,c. whose plane containing

to the normal line of so of e intenseet then (V) since " & L2 are on the plane so L1&l₂ are perpendicular the plane . Then -20+5-2=0 a tab+3e zo solving this we have azt, b=1, card. the equation becomes, x+y-x=d (2) Miko we find d. the point A8 intersection of Lid La are on the plane so put (2,4,1) in (a) we tue required equation of plane is x+y-3.5. 4-251 (d) from (1) we get any point on he is J-1 3+3 liv) If Li and plane Therefore, 314.25) -4(3+3)+7(2-5)=48 12- 6S 12-45+14-75=48 gel, Then ( a î 14 - 173 = 48 8 = -2 8 Then from (lu) we get the point So the required point is ( 4 )