# 1 0 x 2 y 5 z 1 2 2 z 1 5 z 1 4 z 2 5 z 1 z 1 let

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1=0x=2y5z+1
¿2(2z1)5z+1¿4z25z+1¿z1Let take z = t. The parametric equations arex=−9t+3y=−2t+1z = tSoThe vector equation isw=(3,1,0)+t(−9,2,1)16).n1=(4,1,9)n2=(1,2,3)n3=(2,3,5)(n2×n3)=(1,2,3)×(2,3,5)=(−1,11,1)n1(n2×n3)=(4,1,9)(1,11,1)=−4+11+9=160Sincen1(n2×n3)0, the planes intersect at a point.üFind the point of intersection4x+y9z=0……. (i)x+2y+3z=0…… (ii)2x3y5=0……. (iii)Multiply equation (ii) by 3 and subtract from (i)4x+y9z=03x+6y+9z=0_________________7x+7y=0x=−yNow, x value substitute into the equation (iii)
2x3y+5=02(y)3y+5=05y+5=0y=−1x=−y=−(1)=1Now x and y value put into the equation (i)4x+y9z=04(1)+(1)9Z=09z=−3z=13The point of intersection(1,1,13)ü17).P(−4,2,6)2x3y+z8=0The coordinates of the normal to the plane aren=(2,3,1). Let's take another point R from the point Pon the plane.Now, let's choosex=1y=−12(1)3(1)+z8=02+3+z8=0z=3So the another point on the plane isR(1,1,3).Now, the direction ofPRPR=RP=(1,1,3)(4,2,6)=(5,3,3)Now,
ProjnPR=|PR ∙nn∙n||n|¿|(5,3,3)(2,3,1)(2,3,1)(2,3,1)|(2)2+(3)2+(1)2¿|1614|14¿(1.14) (3.74)¿4.26UnitsThe shortest distance between P and the plane is 4.26 units.üFinal mark = 85%
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