Question

12 abcd is a trapezium. find \\( \overrightarrow { a c } \\) and then \\( \overrightarrow { d c } \\) in terms of \\( \mathbf { u } \\) and \\( \mathbf { v } \\).

Explanation:

Step1: Find $\overrightarrow{AC}$

By the triangle law of vector addition, $\overrightarrow{AC}=\overrightarrow{AB}+\overrightarrow{BC}$.
We know that $\overrightarrow{AB}=-\mathbf{v}$ (opposite direction of $\mathbf{v}$) and $\overrightarrow{BC} = 9\mathbf{u}$.
So, $\overrightarrow{AC}=9\mathbf{u}+(-\mathbf{v})=9\mathbf{u}-\mathbf{v}$.

Step2: Find $\overrightarrow{DC}$

Since $\overrightarrow{DC}=\overrightarrow{AC}-\overrightarrow{AD}$.
We know that $\overrightarrow{AD} = 4\mathbf{u}$ and $\overrightarrow{AC}=9\mathbf{u}-\mathbf{v}$.
Substitute the values: $\overrightarrow{DC}=(9\mathbf{u}-\mathbf{v})-4\mathbf{u}$.
Simplify the expression: $\overrightarrow{DC}=(9 - 4)\mathbf{u}-\mathbf{v}=5\mathbf{u}-\mathbf{v}$.

Answer:

$\overrightarrow{AC}=9\mathbf{u}-\mathbf{v}$; $\overrightarrow{DC}=5\mathbf{u}-\mathbf{v}$