MAT1503 Assignment 4 - Solutions
Question 1
1.1) Plane through (0,0,0), parallel to -x+3y-2z=6.
Normal n = <-1,3,-2>. Equation: -x+3y-2z=0.
Answer: -x+3y-2z=0.
1.2) Distance from P(-1,-2,0) to plane 3x-y+4z=-2.
Rewrite: 3x - y + 4z + 2 = 0.
d = |3(-1) -(-2)+0+2| / sqrt(9+1+16) = 1 / sqrt(26).
Answer: 1/√26.
, Question 2
2.1) v1·v2 = (-1)(1)+(1)(-1)+(0)(3)+(-1)(-2)=0 → ■ → θ=90°.
2.2) r=<0,-1,-2,3/4>, ||r||=√(1+4+9/16)=√89/4.
Direction cosines: (0,-4/√89,-8/√89,3/√89).
2.3) r'(t)=<1,1/t²,2t>. r'(1)=<1,1,2>.
(V·r)'=V'·r+V·r'. With V(1)=<-1,1,-2>, V'(1)=<1,-2,2>, r(1)=<1,-1,-1>.
= (1)(1)+(-2)(-1)+(2)(-1) + (-1)(1)+(1)(1)+(-2)(2).
= (1+2-2)+(-1+1-4) = 1-4=-3.
2.4) Work=F cosθ·s=5 cos45°·30=150/√2≈106.07 ft·lb.
2.5) a·v=1008·3500+699.99·4250=6,502,957.5 ZAR.
Interpretation: total revenue.
10% reduction: 0.9a.
2.6) Horizontal=30 cos30°=15√3≈25.98 N.
2.7) v=<-1,3,-5>.
Condition: v·x=0, ||x||=1.
→ Circle (radius 1, center 0).
Parametric: x(θ)=e1 cosθ+e2 sinθ.
2.8) u=<-3,2k,-k>, v=<2,5/2,-k>.
u·v=-6+5k+k²=0 ⇒ k=1 or -6.
Check u+v ■ u-v: requires ||u||²=||v||² ⇒ k²=5/16 ≠1,-6.
So not ■.
2.9) a ■ <2,-1,1>, <1,2,-2>.
u×v=<0,5,5>.
Unit vector=±(0,1/√2,1/√2).
Question 1
1.1) Plane through (0,0,0), parallel to -x+3y-2z=6.
Normal n = <-1,3,-2>. Equation: -x+3y-2z=0.
Answer: -x+3y-2z=0.
1.2) Distance from P(-1,-2,0) to plane 3x-y+4z=-2.
Rewrite: 3x - y + 4z + 2 = 0.
d = |3(-1) -(-2)+0+2| / sqrt(9+1+16) = 1 / sqrt(26).
Answer: 1/√26.
, Question 2
2.1) v1·v2 = (-1)(1)+(1)(-1)+(0)(3)+(-1)(-2)=0 → ■ → θ=90°.
2.2) r=<0,-1,-2,3/4>, ||r||=√(1+4+9/16)=√89/4.
Direction cosines: (0,-4/√89,-8/√89,3/√89).
2.3) r'(t)=<1,1/t²,2t>. r'(1)=<1,1,2>.
(V·r)'=V'·r+V·r'. With V(1)=<-1,1,-2>, V'(1)=<1,-2,2>, r(1)=<1,-1,-1>.
= (1)(1)+(-2)(-1)+(2)(-1) + (-1)(1)+(1)(1)+(-2)(2).
= (1+2-2)+(-1+1-4) = 1-4=-3.
2.4) Work=F cosθ·s=5 cos45°·30=150/√2≈106.07 ft·lb.
2.5) a·v=1008·3500+699.99·4250=6,502,957.5 ZAR.
Interpretation: total revenue.
10% reduction: 0.9a.
2.6) Horizontal=30 cos30°=15√3≈25.98 N.
2.7) v=<-1,3,-5>.
Condition: v·x=0, ||x||=1.
→ Circle (radius 1, center 0).
Parametric: x(θ)=e1 cosθ+e2 sinθ.
2.8) u=<-3,2k,-k>, v=<2,5/2,-k>.
u·v=-6+5k+k²=0 ⇒ k=1 or -6.
Check u+v ■ u-v: requires ||u||²=||v||² ⇒ k²=5/16 ≠1,-6.
So not ■.
2.9) a ■ <2,-1,1>, <1,2,-2>.
u×v=<0,5,5>.
Unit vector=±(0,1/√2,1/√2).
