4 s 1 1 2 4 5 5 3 simplified equations 3 t s 2 4 t s

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4s[1]−1 + ? = ?[2]−4? = 5 + 5?[3]Simplified equations:
3t+s=−2[4]ts=1[5]5? + 4? = −5[6]Using equations [4] and [5] to solve forsandt.3t+s=−2[4]ts=1[5]Adding [4] and [5] to find a solution fort:4t=−2+14? = −1? = −14Substitutingt=14into equation [4]:3(14)+s=−234+s=−2? = −234? =114
The determined values ofsandtmust satisfy the third equation or equation [6] in orderfor the system of equations to have a solution:LS:5s+4t= 5(114)+4(14)¿5(114)1¿3141¿354RS:= 5As, LS≠RS, there is no solution forsandtthat satisfies the linear system. Therefore,these lines never intersect. The lines are skew.15. Let,P1: 4x+3y2z+7=0[1]P2:x2y+5z1=0[2]The normal vectors on these planes are:n1=(4,3,2)n2=(1,2,5)Confirming that the planes intersect:
413225As the ratios of the corresponding components of the normal vectors are not equivalent,the normal vectors are not collinear. Therefore, the planes are neither coincident norparallel and intersect in a line.Determining the line of intersection by using elimination:Multiplying the equation of plane 2 [2] by (-4):(x2y+5z1=0)×(4)−4𝑥+8𝑦 − 20𝑧 +4 = 0[3]Adding [1] and [3] to eliminatex:4x+3y2z+7=0+−4x+8y20z+4=011y22z+11=011𝑦 = 22𝑧 − 11𝑦 = 2𝑧−1Substitutingy=2z1into [1] to find the value ofxin terms ofz.4x+3y2z+7=04𝑥 + 3(2z1)2z+7=04𝑥 + 6𝑧 − 2𝑧 – 3 + 7 = 04𝑥 + 4𝑧 + 4=04𝑥 = −4𝑧 − 4𝑥 = −𝑧 −1
Let,z=tThe parametric equations of the vector line are:x=−1t𝑦 = −1+2?𝑧 = ?Let, the line of intersection be represented byu:The vector equation is:u=(1,1,0)+t(−1,2,1)16. The system is:4x+y9z=0[1]𝑥 + 2𝑦 + 3𝑧 = 0[2]2𝑥 − 3𝑦 – 5 =0[3]The normal vectors:n1=(4,1,9)n2=(1,2,3)n3=(2,3,5)If the normals are not coplanar then the equations intersect at one point.If,n1(n2×n3)0Then, the normal vectors are not coplanar.
n1(n2×n3)¿(4,1,9)(1,2,3)×(2,3,5)=(4,1,9)(1,11,7)¿4×1+1×11±9×7=4+11+63≠ 0Therefore, the three equations intersect at one point.Multiplying equation [2] by 3:(𝑥 + 2𝑦 + 3𝑧 = 0)×33x+6y+9z=0[4]Adding equations [1] and [4] to eliminatez:4x+y9z=0[1]+3x+6y+9z=0[4]7x+7y=0[5]Multiplying equation [3] by 3 ½:(2𝑥 − 3𝑦 – 5 = 0)×3.57x10.5y=17.5[6]Subtracting [6] from [5] to eliminatexand solve fory:7x+7y=0[5]7x10.5y=17.5[6]17.5y=−17.5𝑦 = −1Theyvalue of the co-ordinate of point of intersection of the three planes is -1.Substituting the value𝑦 = −1into equation[3]:2x3y –5=02𝑥3(1)5=0
2x+35=02𝑥=2𝑥=1Substituting the values𝑦 = −1 andx=1into equation [2]x+2y+3z=01+2(1)+3z=01−2+3𝑧=0−1+3𝑧=03𝑧=1𝑧=13Therefore, the point of intersection of the systems of the equations is (1,-1,13¿.17. The plane is:2x3y+z8=0The normal vector on this plane is:n=(2,3,1)Let,Rrepresent another point on the plane.Let, thexandyvalues of the coordinates on the pointRbe 1 and -1 respectively.Substituting the valuesx =1 andy =-1 into the equation of the plane to solve forz:2x3y+z8=02(1)3(1)+z=8
𝑧=8−3−2𝑧=3Another point on the plane isR(1, -1, 3).The given point isP( -4, 2, 6)Finding the direction ofPR:PR=RP=(1,1,3)(4,2,6)=(1+4,12,36)=(5,3,3)The magnitude of the projection ofPRonn:|PRnnn||n|=|(5,3,3)(2,3,1)(2,3,1)(2,
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Spring
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Tags
Vector Space, Dot Product, 2 W, 8, 8 13

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