1. Problem: Given the vector $\vec{a} = \begin{pmatrix} t \\ 8 \end{pmatrix}$, (a) Find the length $|\vec{a}|$ when $t=6$. (b) Find the values of $t$ such that the length $|\vec{a}| = 17$. 2. Formula: The length (magnitude) of a vector $\vec{a} = \begin{pmatrix} x \\ y \end{pmatrix}$ is given by $$|\vec{a}| = \sqrt{x^2 + y^2}$$ 3. Step (a): Substitute $t=6$ into the vector: $$\vec{a} = \begin{pmatrix} 6 \\ 8 \end{pmatrix}$$ Calculate the length: $$|\vec{a}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10$$ 4. Step (b): Set the length equal to 17: $$|\vec{a}| = \sqrt{t^2 + 8^2} = 17$$ Square both sides: $$t^2 + 64 = 17^2$$ $$t^2 + 64 = 289$$ Subtract 64 from both sides: $$t^2 = 289 - 64 = 225$$ Take the square root: $$t = \pm \sqrt{225} = \pm 15$$ 5. Final answers: (a) $|\vec{a}| = 10$ when $t=6$. (b) $t = 15$ or $t = -15$ when $|\vec{a}| = 17$.