Vector Functions — Question 8

Question 8 – Solution

Strategy. A directed segment is an affine blend $(1-t)A+tB$. A new interval requires a linear map back to $[0,1]$.

Step 1: Segment formula First compute $$B-A=\left\langle 8,-4,-4\right\rangle.$$ Thus $$\boxed{\mathbf r(t)=A+t(B-A)=\left\langle -2+8t,\ 1-4t,\ 4-4t\right\rangle}, \quad 0\le t\le 1.$$ Indeed, $\mathbf r(0)=A$ and $\mathbf r(1)=B$.

Step 2: Division point At $t=5/8$, $$\mathbf r\left(\frac 58\right) =\left\langle -2+5,\ 1-\frac 52,\ 4-\frac 52\right\rangle =\boxed{\left\langle 3,-\tfrac 32,\tfrac 32\right\rangle}.$$

Step 3: New parameter interval The linear map taking $u=-3$ to $t=0$ and $u=5$ to $t=1$ is $$t=\frac{u+3}{8}.$$ Substitution gives $$\boxed{\mathbf R(u)=\left\langle u+1,\ -\tfrac 12u-\tfrac 12,\ \tfrac 52-\tfrac 12u\right\rangle}, \quad -3\le u\le 5.$$ Checking $u=-3$ and $u=5$ recovers $A$ and $B$ in the required order.