Vectors

Problem: Find the angle between the planes with the given equations:

2x - y + z = 5 and x + y - z = 1

Solution:
Angle between two planes (h_1,k_1,l_1,d_1) and (h_2,k_2,l_l,d_2) is given as

Cos∅=(h_1 h_2+k_1 k_2+l_1 l_2)/(√(h_1^2+k_1^2+l_1^2 )×√(h_2^2+k_2^2+l_2^2 ))

=((2)×(1)+(-1)×(1)+(1)×(-1))/(√((2)^2+(-1)^2+(1)^2 )×√((1)^2+(1)^2+(-1)^2 ))

=0

Or,Cos∅=0

Or,∅=〖Cos 〗^(-1) 0= 90°=π/2

Therefore the angle between the two given planes is π/2

Problem: Find the values of r' (t) and r'' (t) for the given values of t:

r (t) = i cos t + j sin t; t = π/4

Solution:
r (t)= i cos⁡t+ j sin⁡t

r^' (t)=d{r(t)}/dt=∂(Cos t)/∂t i+∂(Sin t)/∂t j=-Sin t i+Cos t j

At t =π/4,r^' (t)=-Sin (π/4) i+Cos (π/4) j =-1/√2 i+1/√2 j

r^'' (t)=d{r^'(t) }/dt=-∂(Sin t)/∂t i+∂(Cos t)/∂t j=-Cos t i-Sin t j

At t =π/4,r^'' (t)=-Cos (π/4) i-Sin (π/4) j =-1/√2 i-1/√2 j

Problem: The acceleration vector a (t), the initial position r₀ = r (0), and the initial velocity v₀ = v (0) of a particle moving in xyz- space are given. Find its position vector r (t) at time t.

a(t) = 6ti - 5j + 12t²k; r₀ = 3i + 4j; v₀ = 4j - 5k

Solution:

Velocity,v(t)= ∫▒〖a (t)dt=∫▒(〗 6ti - 5j + 12t²k) dt

=6×t^2/2 i-5t j+12×t^3/3 k+v_0

=3t^2 i-5t j+4t^3 k+v_0

Given: Initial Velocity v_0=v(0)=3(0)^2 i-5(0)j+4(0)^3 k=0

Thus, the velocity vector is given as:

v(t)=3t^2 i-5t j+4t^3 k+0=3t^2 i-5t j+4t^3 k

Position Vector,r(t)=∫▒〖v (t)dt=∫▒(〗 3t^2 i-5t j+4t^3 k ) dt

=(3t^3)/3 i-(5t^2)/2 j+(4t^4)/4 k+r_0

=t^3 i-(5t^2)/2 j+t^4 k+r_0

Given: Initial Position r_0=r(0)=0^3 i-(5×0^2)/2 j+0^4 k=0
Thus, the position vector, at time t is given as:

r(t)=t^3 i-(5t^2)/2 j+t^4 k+0=t^3 i-(5t^2)/2 j+t^4 k

Problem: Find the curvature of the given plane curve at the indicated point.

x = t - 1, y = t² + 3t + 2, where t = 2

Solution:

x(t) = t - 1
x^' (t)=1
x^'' (t)=0

y(t) = t^2+ 3t + 2
y^' (t)=2t+3
y^'' (t)=2

Curvature is given as

K(t)=|〖x^' (t)y〗^'' (t)-y^' (t) x^'' (t)|/[(x^' (t))^2+(y^' (t))^2 ]^(3/2)

K(t)=|1×2-(2t+3)×0|/[1^2+(2t+3)^2 ]^(3/2)

K(t=2)=|1×2-(2×2+3)×0|/[1^2+(2×2+3)^2 ]^(3/2) =2/353.55=0.00566

Problem: Find the unit tangent and normal vectors at the indicated point.

x = t^3, y = t^2 at (-1,1)

The point (-1 ,1) corresponds to t = -1.

Solution:

The implicit function is given as:
r(t)=( t^3,t^2)

r^' (t)=(3t^2,2t)

r^' (-1)=(3,-2)

The unit tangent vector is given as:

T(t)=r^' (t)=3i-2j

Curvature is given as follows:
|dT/dt|=|(d(3t^2,2t))/dt|=(6t,2)

Curvature,κ=|dT/dt|_(t=-1)=|(-6,2)|=√((-6)^2+2^2 )=6.3246

The unit normal vector is given as:

N(t)=1/κ×|dT/dt|_(t=-1)=(-6i+2j)/6.3246

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