Find the magnitude and direction of the vector JK given points J(1,0) & K(-2,3)

Mathematics

1 answer:

Stella [2.4K]3 years ago

3 0

Draw points J and K in the coordinate axis, and join them so that the top of the vector is point K.

the magnitude of the vector is just the length of the segment JK, so we can either write it as |JK| or |KJ|, it does not make any change.

Let O be the point in the x axis such that KO is perpendicular to OJ, as shown in the picture.

From the Pythagorean theorem:

$|KJ|= \sqrt{ |OK|^{2} + |OJ|^{2} } = \sqrt{3^{2} + 3^{2} } = \sqrt{2*3^{2}}=3 \sqrt{2}$ units

since |OK|=|OJ|, the right triangle KOJ is isosceles, so the measure of KJO is 45°, which means that the angle of vector JK to the positive x axis is 180°-45°=135°

Answer: magnitude= $3 \sqrt{2}$ units, direction : 135°

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Consider F and C below. F(x, y, z) = y2 sin(z) i + 2xy sin(z) j + xy2 cos(z) k C: r(t) = t2 i + sin(t) j + t k, 0 ≤ t ≤ π (a) Fi

Liono4ka [1.6K]

Answer:

a) $f (x,y,z)= xy^2\sin(z)$

b) $\int_C F \cdot dr =0$

Step-by-step explanation:

Recall that given a function f(x,y,z) then $\nabla f = (\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z})$ . To find f, we will assume it exists and then we will find its form by integration.

First assume that F = $\nabla f$ . This implies that

$\frac{\partial f}{\partial x} = y^2\sin(z)$ if we integrate with respect to x we get that

$f(x,y,z) = xy^2\sin(z) + g(y,z)$ for some function g(y,z). If we take the derivative of this equation with respect to y, we get

$\frac{\partial f}{\partial y} = 2xy\sin(z) + \frac{\partial g}{\partial y}$

This must be equal to the second component of F. Then

$2xy\sin(z) + \frac{\partial g}{\partial y}=2xy\sin(z)$

This implies that $\frac{\partial g}{\partial y}=0$ , which means that g depends on z only. So $f(x,y,z) = xy^2\sin(z) + g(z)$

Taking the derivative with respect to z and making it equal to the third component of F, we get

$xy^2\cos(z)+\frac{dg}{dz} = xy^2\cos(z)$

which implies that $\frac{dg}{dz}=0$ which means that g(z) = K, where K is a constant. So

$f (x,y,z)= xy^2\sin(z)$

b) To evaluate $\int_C F \cdot dr$ we can evaluate it by using f. We can calculate the value of f at the initial and final point of C and the subtract them as follows.