第12章 12.18

(1)

a⃗=e1⃗+3e2⃗\vec{a} = \vec{e_1} + 3\vec{e_2}

b⃗=2e1⃗+te2⃗\vec{b} = 2\vec{e_1} + t\vec{e_2}

a⃗⋅b⃗=0\vec{a} \cdot \vec{b} = 0

(e1⃗+3e2⃗)(2e1⃗+te2⃗)=0(\vec{e_1} + 3\vec{e_2})(2\vec{e_1} + t\vec{e_2}) = 0

2∣e1⃗∣2+3t∣e2⃗∣2=02|\vec{e_1}|^2 + 3t|\vec{e_2}|^2 = 0

2+3t=02 + 3t = 0

t=−23t = -\frac{2}{3}

(2)

a⃗=(1,1)b⃗=(1,−3)\vec{a} = (1,1) \qquad \vec{b} = (1,-\sqrt{3})

(ta⃗+b⃗)⋅a⃗=0(t\vec{a} + \vec{b}) \cdot \vec{a} = 0

(1+t,−3+t)⋅(1,1)=0(1+t, -\sqrt{3}+t) \cdot (1,1) = 0

1+t−3+t=01 + t - \sqrt{3} + t = 0

t=3−12t = \frac{\sqrt{3} - 1}{2}

解説: r31bn1z