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vectors explanation needed
- Thread starter msbgcse
- Start date
msbgcse said:question: vector DC is parallel to vector OB. Calculate the value of k. I know that DC = k3a - 2a + 4b and OB = 6b. I also know that k=2/3. I just don't know why or how, and what is the explanation for this. Please can you help?
I presume a and b are vectors?
Then you have 6b is parallel to (3k-2) a + 4b.
Now 6b is parallel to 4a. The only way it can be parallel to (3k-2)a + 4b is
* if a is parallel to b, OR
* 3k-2 = 0
do you see why? And does this help you proceed?
Hello, msbgcse!
Please give us the original statement of the problem.
Your version is confusing.
I tried to interpret the problem.
\(\displaystyle \text{We have: }\;\begin{Bmatrix}\overrightarrow{DC} &=& (3k\!-\!2)\vec a + 4\vec b \\ \overrightarrow{OB} &=& \qquad\quad 0\vec a + 6\vec b \end{Bmatrix}\)
\(\displaystyle \text{Since }\overrightarrow{DC} \parallel \overrightarrow{OB},\:\text{ their corresponding components are proportional.}\)
. . \(\displaystyle \text{That is: }\:\overrightarrow{DC} \:=\:m\!\cdot\!\overrightarrow{OB}\)
. . \(\displaystyle \text{Therefore: }\;\begin{bmatrix}(3k-2)a \:=\:m(0) & \Rightarrow & k \:=\:\frac{3}{2} \\ \\[-3mm] 4b \:=\:m(6b) & \Rightarrow & m \:=\:\frac{2}{3} \end{bmatrix}\)
Please give us the original statement of the problem.
Your version is confusing.
I tried to interpret the problem.
\(\displaystyle \text{Given: }\:\overrightarrow{DC} \parallel \overrightarrow{OB}\)
\(\displaystyle \text{Calculate the value of }k.\) . . What is k ?
\(\displaystyle \text{I know that: }\:\overrightarrow{DC} \:=\: k3\vec a - 2\vec a + 4\vec b\,\text{ and }\,\overrightarrow{OB} \:=\: 6\vec b\)
. . How do you know this? . Why weren't we told this the beginning?
\(\displaystyle \text{I also know that }k\,=\,\tfrac{2}{3}\)
. . Yay!
\(\displaystyle \text{We have: }\;\begin{Bmatrix}\overrightarrow{DC} &=& (3k\!-\!2)\vec a + 4\vec b \\ \overrightarrow{OB} &=& \qquad\quad 0\vec a + 6\vec b \end{Bmatrix}\)
\(\displaystyle \text{Since }\overrightarrow{DC} \parallel \overrightarrow{OB},\:\text{ their corresponding components are proportional.}\)
. . \(\displaystyle \text{That is: }\:\overrightarrow{DC} \:=\:m\!\cdot\!\overrightarrow{OB}\)
. . \(\displaystyle \text{Therefore: }\;\begin{bmatrix}(3k-2)a \:=\:m(0) & \Rightarrow & k \:=\:\frac{3}{2} \\ \\[-3mm] 4b \:=\:m(6b) & \Rightarrow & m \:=\:\frac{2}{3} \end{bmatrix}\)