Vector Functions — Question 7

Question 7 – Solution

Strategy. Treat scalar division and vector algebra componentwise; for orthogonality, set the dot product equal to zero.

Step 1: Common domain The vector functions themselves are polynomial and defined for all real $t$. The scalar denominators impose $$t-2\ne 0,\qquad t+1\ne 0.$$ Therefore the common domain is $$\boxed{\mathbb R\setminus\{-1,2\}}.$$

Step 2: Vector combination $$\begin{align*} 2\mathbf a(t)-\mathbf b(t) &=\left\langle 2t,\ 2-2t,\ 4t\right\rangle-\left\langle 1,t,t^2\right\rangle\\ &=\boxed{\left\langle 2t-1,\ 2-3t,\ 4t-t^2\right\rangle}. \end{align*}$$

Step 3: Orthogonality $$\begin{align*} \mathbf a(t)\cdot\mathbf b(t) &=t(1)+(1-t)t+(2t)t^2\\ &=2t-t^2+2t^3\\ &=t(2t^2-t+2). \end{align*}$$ The quadratic has discriminant $(-1)^2-4(2)(2)=-15<0$, so it has no real roots. Consequently $$\boxed{t=0}$$ is the only real solution. Directly, $\mathbf a(0)=\left\langle 0,1,0\right\rangle$ and $\mathbf b(0)=\left\langle 1,0,0\right\rangle$, whose dot product is zero.