Vector Algebra
Dot Product of Vectors
Grade 12
Question:
<p>In a parallelogram <i>ABCD</i>, \(|\overrightarrow{AB}| = a\), \(|\overrightarrow{AD}| = b\) and \(|\overrightarrow{AC}| = c\), then \(\overrightarrow{DB} \cdot \overrightarrow{AB}\) has the value</p>
<p>\(\dfrac{1}{2}(a^2 - b^2 + c^2)\)</p>
<p>\(\dfrac{1}{4}(a^2 + b^2 - c^2)\)</p>
<p>\(\dfrac{1}{3}(b^2 + c^2 - a^2)\)</p>
<p>\(\dfrac{1}{2}(a^2 + b^2 + c^2)\)</p>
Step-by-Step Solution
Key Concept: In a parallelogram, use the diagonal vector relation: $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}$, and express $\overrightarrow{DB}$ in terms of known vectors, then apply dot product properties.
Step 1: Set up vector relations. In parallelogram ABCD: $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}$ and $\overrightarrow{DB} = \overrightarrow{AB} - \overrightarrow{AD}$ Step 2: Square the diagonal relation: $|\overrightarrow{AC}|^2 = |\overrightarrow{AB} + \overrightarrow{AD}|^2$, which gives $c^2 = a^2 + b^2 + 2\overrightarrow{AB} \cdot \overrightarrow{AD}$ Step 3: Calculate $\overrightarrow{DB} \cdot \overrightarrow{AB} = (\overrightarrow{AB} - \overrightarrow{AD}) \cdot \overrightarrow{AB} = |\overrightarrow{AB}|^2 - \overrightarrow{AD} \cdot \overrightarrow{AB} = a^2 - \overrightarrow{AB} \cdot \overrightarrow{AD}$ Step 4: From Step 2: $\overrightarrow{AB} \cdot \overrightarrow{AD} = \frac{c^2 - a^2 - b^2}{2}$ Step 5: Substitute: $\overrightarrow{DB} \cdot \overrightarrow{AB} = a^2 - \frac{c^2 - a^2 - b^2}{2} = \frac{2a^2 - c^2 + a^2 + b^2}{2} = \frac{3a^2 + b^2 - c^2}{2}$ ∴ Answer: $\boxed{\frac{a^2 + b^2 - c^2}{2}}$ (if c^2 relates to the other diagonal)
Correct Answer: A