Calculating Straight Line Distance from A to B

[wonder1]

Hi all,

for the below questions i used the 3D distance formula

$$\sqrt{((117t-20)^2+((117−20)^2+(90−22)^2+(15−20))}$$

$$d=(97)^2+(68)^2+(−5)^2$$

$$AB=118.566437$$

to give me an answer of 118.566437

i don't understand what it means by finding formulas for x(t), y(t), and z(t)

is it $$97x^2+68y^2−5z^2$$ ?4. The camera needs to move from position A = (20,22,20) to position B = (117,90,15), on a straight line, with constant speed, in 5 seconds. Find formulas for x(t), y(t), and z(t), assuming that the movement starts at time t = 0.

 

[MarkFL]

Hi all,

for the below questions i used the 3D distance formula

$$\sqrt{((117t-20)^2+((117−20)^2+(90−22)^2+(15−20))}$$

$$d=(97)^2+(68)^2+(−5)^2$$

$$AB=118.566437$$

to give me an answer of 118.566437

i don't understand what it means by finding formulas for x(t), y(t), and z(t)

is it $$97x^2+68y^2−5z^2$$ ?4. The camera needs to move from position A = (20,22,20) to position B = (117,90,15), on a straight line, with constant speed, in 5 seconds. Find formulas for x(t), y(t), and z(t), assuming that the movement starts at time t = 0.

Let's look at motion in one dimension...suppose the initial coordinate is $x_0$ and the final coordinate is $x_1$, and we want this change in position to occur in $t$ units of time, with a constant speed $v_x$, which means we can write:

$$\d{x}{t}=v_x$$ where $x\left(t_0\right)=x_0$ and $x\left(t_1\right)=x_1$.

Can you now solve this IVP?

 

[wonder1]

Let's look at motion in one dimension...suppose the initial coordinate is $x_0$ and the final coordinate is $x_1$, and we want this change in position to occur in $t$ units of time, with a constant speed $v_x$, which means we can write:

$$\d{x}{t}=v_x$$ where $x\left(t_0\right)=x_0$ and $x\left(t_1\right)=x_1$.

Can you now solve this IVP?

First distance vector
(117,90,15) - (20,22,20) = (97,68,-5)

Equation

AB(t) = (20+97t, 22+68t, 20-5t)

X=20+97t
Y=22+68t
Z=20-5t

 

[MarkFL]

Solving the IVP I posted, we would obtain:

$$x(t)=\frac{x_1-x_0}{t_1-t_0}\left(t-t_0\right)+x_0$$

Now, applying this to all 3 dimensions, using the given values, we get

$$x(t)=\frac{97}{5}t+20$$

$$y(t)=\frac{68}{5}t+22$$

$$z(t)=-t+20$$