Question

Prove the arithmetic properties of the Cross Product 1. 2. a. Line L1 is parallel to the vector u Si+j, line L2 is parallel to the vector u-3i +4j and both lines pass through point P(-1,-2). Determine the parametric equations for line L1 and Lz b. Given line L:x(t)-2t+8,y(t)-10-3t. Does L and Ls has common 3. a. Find the equation of the plane A that pass through point P(3,-2,0) with b. Given A2 be the plane -2x3y +5z 4, find a parametric equation 4. Find the reflection vector R when an incoming vector V-(5,-1) is reflected by intersection point? If yes, find its point of intersection. normal vector N (1,2,-5). representation of the line of intersection of plane A and A2 the ellipse (2 sin(t), cos(t)) at the point where t-0.85 on the ellipse. 5. a. Given curves p(t) 3i (2/t)j (t3/8)k and q(s) (2s 1)i (s2) Vs)k. Find the point and angle of intersection for the given curves. If q(s) is the position of an object at time s, calculate the velocity and speed of the object ats 2 b. 6. The acceleration of an object is a(t) 6i (24t) (2t-3)k and its velocity at time t- 2 is v(2) (17,-46,-1/4). Find a formula for r(t) as its position at time t ifr(1) 1,-2,-2) 7. Determine the curvature of the curve r() (2t)i -(cos(t))i (3 sin(t))k at point t 0 and t m/6. If necessary, use π 3.14.

please answer question 4-7

Answer 1

Solution:

$\text{4.}$

r (t) 2 cos t and y'(t)--sin t

So a tangent vector to a ellipse at t 0.85 is T-(2 cos 0.85.-sin 0.85

(1.31997,-0.75128) and then a normal vector is N (075 128. 1.31997) T

Finallv

2.43643 2.3067424393 ProjN (0.75128,1.31997) (0 0.7933, 1.3938

So, R V - ProjV-(5,-1) (0.7933, 1.3938)4.2067,-2.3938)

$\text{5.}$

a)

If p(t) and q(s) intersects then at the point of intersections the x.y and z

$\text{coordinates of both curves will be equal,}$

$2s+1=3=>s=1$

s2 =-=> t = 2

t3 8 Vs

$\text{Hence the point of intersection is, }\left ( 3,1,-1 \right )$

$\text{angle of intersection is the angle between the tangent vectors at, }\left ( 3,1,-1 \right )$

2 3t2

$p'(2)=\left \langle 0,-\frac{1}{2},-\frac{3}{2} \right \rangle$

$q'(s)=\left \langle 1,2s,-\frac{1}{2\sqrt{s}} \right \rangle$

$q'(1)=\left \langle 1,2,-\frac{1}{2} \right \rangle$

$\text{Hence the angle is,}$

$\cos \theta =\frac{p'(2)\cdot q'(1)}{\left \| p'(2) \right \|\left \| q'(1) \right \|}=-\frac{1}{\sqrt{210}}$

$\theta =-\cos^{-1} \left ( \frac{1}{\sqrt{210}} \right )$

$\text{b)}$

$q(s)=\left \langle 2s+1,s^2,-\sqrt{s} \right \rangle$

$v(s)=q'(s)=\left \langle 1,2s,-\frac{1}{2\sqrt{s}} \right \rangle$

$v(2)=\left \langle 1,4,-\frac{1}{2\sqrt{2}} \right \rangle$

$\text{speed}=\left \| q'(s) \right \|=\sqrt{1+4s^2+\frac{1}{4s}}$

$\text{speed}(2)=\sqrt{1+4s^2+\frac{1}{4s}}=\sqrt{\frac{137}{8}}$

$\text{6.}$

$a(t)=\left \langle 6,-24t,\frac{2}{t^3} \right \rangle$

$v'(t)=\left \langle 6,-24t,\frac{2}{t^3} \right \rangle$

$v(t)=\left \langle 6t+A,-12t^2+B,-\frac{1}{t^2}+C \right \rangle$

$\text{Since }v(2)=\left \langle 17,-46,-\frac{1}{4} \right \rangle$

$\text{therefore,}v(2)=\left \langle 12+A,-48+B,-\frac{1}{4}+C \right \rangle=\left \langle 17,-46,-\frac{1}{4} \right \rangle$

$=>A=5,B=2,C=0$

Hence

$r'(t)=v(t)=\left \langle 6t+5,-12t^2+2,-\frac{1}{t^2} \right \rangle$

$r(t)=\left \langle 3t^2+5t+A,-4t^3+2t+B,\frac{1}{t}+C \right \rangle$

$\text{we have, }r(1)=\left \langle 1,-2,-2 \right \rangle$

therefore, r (1) = 〈84 A.-2 + B. 1-C) = 〈1,-2,-2)

$=>A=-7,B=0,C=-3$

Hence

$r(t)=\left \langle 3t^2+5t-7,-4t^3+2t,\frac{1}{t}-3 \right \rangle$