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Question: In a parallelogram $ABCD$, $|AB| = a$, $|AD| = b$ and $|AC| = c$, then $\overrightarrow{DB}.\overrig...

In a parallelogram ABCDABCD, ∣AB∣=a|AB| = a, ∣AD∣=b|AD| = b and ∣AC∣=c|AC| = c, then DB→.AB→\overrightarrow{DB}.\overrightarrow{AB} has the value:

A

3a2+b2−c22\frac{3a^2 + b^2 - c^2}{2}

B

a2+3b2−c22\frac{a^2 + 3b^2 - c^2}{2}

C

a2−b2+3c22\frac{a^2 - b^2 + 3c^2}{2}

D

a2+3b2+c22\frac{a^2 + 3b^2 + c^2}{2}

Answer

3a2+b2−c22\frac{3a^2 + b^2 - c^2}{2}

Explanation

Solution

Let the parallelogram be ABCDABCD. We are given ∣AB∣=a|AB| = a, ∣AD∣=b|AD| = b, and ∣AC∣=c|AC| = c. We can represent the sides as vectors: AB→\overrightarrow{AB} and AD→\overrightarrow{AD}. The diagonal AC→=AB→+AD→\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}. The other diagonal DB→=AB→−AD→\overrightarrow{DB} = \overrightarrow{AB} - \overrightarrow{AD}.

We are asked to find DB→⋅AB→\overrightarrow{DB} \cdot \overrightarrow{AB}. DB→⋅AB→=(AB→−AD→)⋅AB→\overrightarrow{DB} \cdot \overrightarrow{AB} = (\overrightarrow{AB} - \overrightarrow{AD}) \cdot \overrightarrow{AB} Using the distributive property of the dot product: DB→⋅AB→=AB→⋅AB→−AD→⋅AB→\overrightarrow{DB} \cdot \overrightarrow{AB} = \overrightarrow{AB} \cdot \overrightarrow{AB} - \overrightarrow{AD} \cdot \overrightarrow{AB} We know that AB→⋅AB→=∣AB→∣2=a2\overrightarrow{AB} \cdot \overrightarrow{AB} = |\overrightarrow{AB}|^2 = a^2. So, DB→⋅AB→=a2−AD→⋅AB→\overrightarrow{DB} \cdot \overrightarrow{AB} = a^2 - \overrightarrow{AD} \cdot \overrightarrow{AB} Now we need to find AD→⋅AB→\overrightarrow{AD} \cdot \overrightarrow{AB}. We use the given length of the diagonal ACAC. ∣AC→∣2=c2|\overrightarrow{AC}|^2 = c^2 ∣AB→+AD→∣2=c2|\overrightarrow{AB} + \overrightarrow{AD}|^2 = c^2 (AB→+AD→)⋅(AB→+AD→)=c2(\overrightarrow{AB} + \overrightarrow{AD}) \cdot (\overrightarrow{AB} + \overrightarrow{AD}) = c^2 ∣AB→∣2+∣AD→∣2+2(AB→⋅AD→)=c2|\overrightarrow{AB}|^2 + |\overrightarrow{AD}|^2 + 2 (\overrightarrow{AB} \cdot \overrightarrow{AD}) = c^2 a2+b2+2(AB→⋅AD→)=c2a^2 + b^2 + 2 (\overrightarrow{AB} \cdot \overrightarrow{AD}) = c^2 From this, we can express the dot product: 2(AB→⋅AD→)=c2−a2−b22 (\overrightarrow{AB} \cdot \overrightarrow{AD}) = c^2 - a^2 - b^2 AB→⋅AD→=c2−a2−b22\overrightarrow{AB} \cdot \overrightarrow{AD} = \frac{c^2 - a^2 - b^2}{2} Substitute this back into the expression for DB→⋅AB→\overrightarrow{DB} \cdot \overrightarrow{AB}: DB→⋅AB→=a2−(c2−a2−b22)\overrightarrow{DB} \cdot \overrightarrow{AB} = a^2 - \left( \frac{c^2 - a^2 - b^2}{2} \right) DB→⋅AB→=a2−c22+a22+b22\overrightarrow{DB} \cdot \overrightarrow{AB} = a^2 - \frac{c^2}{2} + \frac{a^2}{2} + \frac{b^2}{2} DB→⋅AB→=2a2+a2+b2−c22\overrightarrow{DB} \cdot \overrightarrow{AB} = \frac{2a^2 + a^2 + b^2 - c^2}{2} DB→⋅AB→=3a2+b2−c22\overrightarrow{DB} \cdot \overrightarrow{AB} = \frac{3a^2 + b^2 - c^2}{2}