Zadatak vek-003

a) $p\cdot q=|p|\,|q|\cos\dfrac{\pi}{4}=1\cdot\sqrt{2}\cdot\dfrac{\sqrt{2}}{2}=1$. Tada $|p+q|^2=|p|^2+2(p\cdot q)+|q|^2=1+2+2=5$, znači $|p+q|=\sqrt{5}$. Uslov $(p+tq)\cdot q=0$ daje $p\cdot q+t|q|^2=1+2t=0$, odakle $t=-\dfrac12$.

b) $a\cdot b=3\alpha-\alpha-4=2\alpha-4=0$, pa je $\alpha=2$.

v) $|u|^2=(\alpha-1)^2+9$, $|v|^2=(\alpha+1)^2+5$. Iz $|u|^2=|v|^2$:

$(\alpha-1)^2+9=(\alpha+1)^2+5\;\Rightarrow\;4-4\alpha=0,$

odakle $\alpha=1$.