1. ## Vector question.

A,B,C and D are points of the plane
1)show that :vector 2DA+vector 3DB = zero vector
then AD vector = 3/5 AB vector

2. Originally Posted by lebanon
A,B,C and D are points of the plane
1)show that :vector 2DA+vector 3DB = zero vector
then AD vector = 3/5 AB vector

There some information is missing.

3. Originally Posted by lebanon
A,B,C and D are points of the plane
1)show that :vector 2DA+vector 3DB = zero vector
then AD vector = 3/5 AB vector
Why even mention "C"? We can always set up a coordinate system so that one point, say D, is the origin. Write $A= (x_1, y_1)$, $B= (x_2, y_2)$ in such a coordinate system. The we can represent vector DA as $$ and vector DB as $$. Then $2DA= <2x_1, 2y_1>$, $3DB= <3x_2, 3y_2>$ and 2DA+ 3DB= 0 becomes $<2x_1+ 3x_2, 2y_1+3y_2>= <0, 0>$ or $2x_1= -3x_2$ and $2y_1= -3y_2$.

In that same system, $AB= $ while $AD= <-x_1, -y_1>$.

The result you are asking for, that $<-x_1, -y_1>= <(3/5)(x_2- x_1), (3/5)(y_2- y_1)>$ is the same as $-x_1= (3/5)(x_2- x_1)= (3/5)x_2- (3/5)x_1$ and $-y_1= (3/5)(y_2- y_1)$. That is, $-2x_1= 3x-2$ and $-2y_1= 3y_2$.

4. Hello, lebanon!

A strange problem . . . Point $\,C$ is unnecessary.

$\,A,B,D\text{ are points on the plane.}$

$\text{1) Show that if: }\:2\overrightarrow{DA} + 3\overrightarrow{DB} \:=\: \vec 0$

. . . $\text{then: }\;\overrightarrow{AD} \:=\:\frac{3}{5}\overrightarrow{AB}$

Since $2\overrightarrow{DA} + 3\overrightarrow{DB} \:=\:0\quad\Rightarrow\quad \overrightarrow{DA} \:=\:-\frac{3}{2}\overrightarrow{DB} \quad\Rightarrow \quad \overrightarrow{AD} \:=\:\frac{3}{2}\overrightarrow{DB}$
. . the diagram must look like this:

. . $\begin{array}{ccccc}
\bullet & ---- & \bullet & --- & \bullet} \\
A && D && B \\
& \frac{3}{2}\overrightarrow{DB} && \overrightarrow{DB}
\end{array}$

Since $\overrightarrow{AB} \:=\:\frac{3}{2}\overrightarrow{DB} + \overrightarrow{DB} \:=\:\frac{5}{2}\overrightarrow{DB}$

. . then: . $\dfrac{\overrightarrow{AD}}{\overrightarrow{AB}} \;=\;\dfrac{\frac{3}{2}\overrightarrow{DB}}{\frac{ 5}{2}\overrightarrow{DB}} \quad\Rightarrow\quad \dfrac{\overrightarrow{AD}}{\overrightarrow{AB}} \;=\;\dfrac{3}{5}$

Therefore: . $\overrightarrow{AD} \;=\;\frac{3}{5}\overrightarrow{AB}$