Question

In parallelogram ABCD, |AB| = a, |AD| = b, |AC| = c then DA.AB has the value

• A. $\frac{1}{2}(a^2+b^2+c^2)$
• B. $\frac{1}{2}(a^2-b^2+c^2)$
• C. $\frac{1}{2}(a^2+b^2-c^2)$
• D. $\frac{1}{2}(b^2+c^2-a^2)$

Answer 1

Answer: (C)

$\frac{1}{2}(a^2+b^2-c^2)$

Answer 2

Let

$$\vec{AB} = \vec{a},\quad \vec{AD} = \vec{d}.$$

Then the diagonal $\vec{AC}$ is given by

$$\vec{AC} = \vec{AB} + \vec{AD} = \vec{a} + \vec{d}.$$

Taking the square of the magnitude of $\vec{AC}$, we have

$$|\vec{AC}|^2 = |\vec{a}+\vec{d}|^2 = |\vec{a}|^2 + |\vec{d}|^2 + 2\,\vec{a}\cdot\vec{d}.$$

Given $|\vec{AB}| = a$, $|\vec{AD}| = b$, and $|\vec{AC}| = c$, this becomes

$$c^2 = a^2 + b^2 + 2\,\vec{a}\cdot\vec{d}.$$

However, note that the problem asks for $\vec{DA}\cdot \vec{AB}$. Since

$$\vec{DA} = -\vec{AD},$$

it follows that

$$\vec{DA}\cdot\vec{AB} = -\vec{AD}\cdot\vec{AB} = -\vec{a}\cdot\vec{d}.$$

Thus, from the equation above,

$$\vec{a}\cdot\vec{d} = \frac{c^2 - a^2 - b^2}{2}.$$

Therefore,

$$\vec{DA}\cdot\vec{AB} = -\frac{c^2 - a^2 - b^2}{2} = \frac{a^2 + b^2 - c^2}{2}.$$