115分科測驗數學乙非選第17題詳解

1. 求 $\overset{\large\rightharpoonup}{AB}$： $$\overset{\large\rightharpoonup}{AB} = B - A = (4-1,\,2-1) = (3,1).$$ 2. 設 $C = (x,y)$。 因為 $u = (1,2)$ 與 $\overset{\large\rightharpoonup}{CA}$ 垂直， 所以 $u \cdot \overset{\large\rightharpoonup}{CA} = 0$： $$(1,2) \cdot (1-x,\,1-y) = (1-x) + 2(1-y) = 0 \implies x + 2y = 3.$$ 3. 因為 $u$ 與 $\overset{\large\rightharpoonup}{CB}$ 的夾角餘弦為 $\frac{1}{\sqrt{10}}$， 所以 $$\frac{u \cdot \overset{\large\rightharpoonup}{CB}}{|u|\,|\overset{\large\rightharpoonup}{CB}|} = \frac{(1,2)\cdot(4-x,\,2-y)}{\sqrt{5}\,|\overset{\large\rightharpoonup}{CB}|} = \frac{1}{\sqrt{10}}.$$ 利用 $x+2y = 3$，分子化簡為 $$(4-x) + 2(2-y) = 8 - (x+2y) = 8 - 3 = 5.$$ 因此 $$\frac{5}{\sqrt{5}\,|\overset{\large\rightharpoonup}{CB}|} = \frac{1}{\sqrt{10}} \implies |\overset{\large\rightharpoonup}{CB}| = 5\sqrt{2}.$$ 即 $$(x-4)^2 + (y-2)^2 = 50.$$ 4. 將 $x = 3 - 2y$ 代入： $$(3-2y-4)^2 + (y-2)^2 = 50 \implies (-2y-1)^2 + (y-2)^2 = 50 \implies 4y^2+4y+1 + y^2-4y+4 = 50 \implies 5y^2 + 5 = 50 \implies y^2 = 9 \implies y = \pm 3.$$ 因為 $C$ 在第二象限（$x<0,\,y>0$），所以 $y=3$， 代入 $x = 3-2\times3 = -3$，得 $C = (-3,3)$。 5. 求 $\cos\angle BAC$： $$\overset{\large\rightharpoonup}{AC} = C - A = (-3-1,\,3-1) = (-4,2).$$ $$\cos\angle BAC = \frac{\overset{\large\rightharpoonup}{AB} \cdot \overset{\large\rightharpoonup}{AC}} {|\overset{\large\rightharpoonup}{AB}|\,|\overset{\large\rightharpoonup}{AC}|} = \frac{3\times(-4)+1\times2}{\sqrt{10}\,\sqrt{20}} = \frac{-12+2}{\sqrt{200}} = \frac{-10}{10\sqrt{2}} = -\frac{\sqrt{2}}{2}.$$