We are given two diagrams containing vectors. We need to find which options are equal to the resultant vectors $\vec{KM}$ and $\vec{DA}$.

1. Problem Description

We are given two diagrams containing vectors. We need to find which options are equal to the resultant vectors $\vec{KM}$ and $\vec{DA}$.

2. Solution Steps

For the first problem, we want to find a vector equivalent to $\vec{KM}$. From the diagram, we can write $\vec{KM} = \vec{KL} + \vec{LM}$. We have $\vec{KL} = a$ and $\vec{LM} = b$. Thus, $\vec{KM} = a + b$. We also have $\vec{KN} = - \vec{NK}$.

We can express $\vec{NM}$ as $3a$. Then $\vec{NL} = \vec{NM} + \vec{ML} = 3a - b$. $\vec{NK} = \vec{NM} + \vec{MK} = 3a - (a+b) = 2a - b$. So $\vec{KN} = -(2a - b) = -2a+b$.

Consider $\vec{MK} = -\vec{KM} = -a - b$. $\vec{NL}$ is not $\vec{KM}$. $\vec{NK}$ is not $\vec{KM}$. $\vec{KN}$ is not $\vec{KM}$.

We also have $\vec{KM} = \vec{KN} + \vec{NM}$, where $\vec{NM} = 3a$. Then $\vec{KM} = \vec{KN} + 3a$.

So $\vec{KN} = \vec{KM} - 3a = a + b - 3a = -2a + b$.

We have $\vec{NL} = \vec{NK} + \vec{KL} = 2a - b + a = 3a - b$.

Let's recalculate $\vec{NM} = \vec{NL} + \vec{LM} = \vec{NL} + b = 3a$.

So $\vec{NL} = 3a - b$.

Now $\vec{NL} = \vec{NK} + \vec{KL}$, so $\vec{NK} = \vec{NL} - \vec{KL} = 3a - b - a = 2a - b$.

For the second problem, we want to find a vector equivalent to $\vec{DA}$. We can write

$\vec{DA} = \vec{DE} + \vec{EA}$.

We have $\vec{DE} = -a$ and $\vec{EA} = -3b -b = -4b$.

Thus $\vec{DA} = -a - 4b$.

However, we can also find $\vec{FE} = 3b$, $\vec{AF} = -b$, $\vec{AB} = a$, $\vec{BC} = 2b$.

$\vec{CD} = -2b$, $\vec{DE} = -a$.

$\vec{BC} = 2b$

$\vec{FC} = \vec{FE} + \vec{EC} = 3b + \vec{EA} + \vec{AC}$.

$\vec{AC} = \vec{AB} + \vec{BC} = a + 2b$.

$\vec{EA} = -(\vec{AF} + \vec{FE}) = -(-b+3b) = -2b$.

So $\vec{DA}$ is not $\vec{FE}$, $\vec{BC}$, $\vec{FC}$.

$\vec{DA} = \vec{DE} + \vec{EA} = -a - 3b - \vec{AF} = -a - 3b + b = -a - 2b.$

$\vec{ED} = a$

$\vec{BC} = 2b$

The correct equation is $\vec{DA} = \vec{DE} + \vec{EA}$.

The vector $\vec{DE} = a$. $\vec{ED} = -a$.

$\vec{EF} = -3b$

$\vec{FA} = b$

$\vec{AB} = a$

$\vec{BC} = 2b$

$\vec{CD} = \vec{CE} + \vec{ED}$

$\vec{CE} = -(\vec{EF} + \vec{FC}) = -(-3b + \vec{FC})$.

Since we need to find a vector expression equal to $\vec{DA}$, we can't use these expressions. The vector $\vec{DA}$ is depicted in the image.

$\vec{EF} = -3b$.

Then $\vec{FE} = 3b$.

Since $\vec{FC} = \vec{FE} + \vec{EC}$.

We have $\vec{DA}$

3. Final Answer

4

1. d) $\vec{KN}$

4

2. d) $\vec{DA}$