Question

How do you find \[u=\dfrac{1}{2}v-w+2z\] and given that \[v=\ <4,-3,5>,\ w=\ <2,6,-1>\] and \[z=<3,0,4>\]?

Answer 1

Hint: In the given question, we have been asked to find the value of a given expression and we have given three vectors i.e. \[v=\ <4,-3,5>,\ w=\ <2,6,-1>\] and \[z=<3,0,4>\]. In the given each vector we have three components that represent ‘x’, ‘y’ and ‘z’ directions. So for adding the vectors first we need to add only the values of x-direction and then we add the values of y-direction and then adding the values of z-direction and then write it in the given vector form that is in the question. For subtracting the two vectors, we need to subtract the values of x-direction then subtract the values of y-direction and same with the values of z-direction then write it in the vector form. To multiply a scalar unit by vector, we need to multiply each component of the vector by that scalar unit. Later we will simplify the given expression and we will get our required solution.

Complete step by step answer:
We have given three vectors,
\[\overrightarrow{v}=\ <4,-3,5>\ i.e.\ \overrightarrow{v}=\left( \begin{matrix}
   4 \\
   -3 \\
   5 \\
\end{matrix} \right)\]

\[\Rightarrow\overrightarrow{w}=\ <2,6,-1>\ i.e.\ \overrightarrow{w}=\left( \begin{matrix}
   2 \\
   6 \\
   -1 \\
\end{matrix} \right)\]

\[\Rightarrow\overrightarrow{z}=<3,0,4>\ i.e.\ \overrightarrow{z}=\left( \begin{matrix}
   3 \\
   0 \\
   4 \\
\end{matrix} \right)\]

Now,
\[u=\dfrac{1}{2}v-w+2z\]
\[\Rightarrow u=\dfrac{1}{2}\left( \begin{matrix}
   4 \\
   -3 \\
   5 \\
\end{matrix} \right)-\left( \begin{matrix}
   2 \\
   6 \\
   -1 \\
\end{matrix} \right)+2\left( \begin{matrix}
   3 \\
   0 \\
   4 \\
\end{matrix} \right)\]

Simplifying the above, we get
\[u=\left( \begin{matrix}
   2 \\
   \left( \dfrac{-3}{2} \right) \\
   \left( \dfrac{5}{2} \right) \\
\end{matrix} \right)-\left( \begin{matrix}
   2 \\
   6 \\
   -1 \\
\end{matrix} \right)+\left( \begin{matrix}
   6 \\
   0 \\
   8 \\
\end{matrix} \right)\]

Simplifying the above expression by subtracting the first two vectors we get
\[u=\left( \begin{matrix}
   2-2 \\
   \left( \dfrac{-3}{2}-6 \right) \\
   \left( \dfrac{5}{2}-\left( -1 \right) \right) \\
\end{matrix} \right)+\left( \begin{matrix}
   6 \\
   0 \\
   8 \\
\end{matrix} \right)\]

Simplifying the components of the above vector, we get
\[u=\left( \begin{matrix}
   0 \\
   \left( \dfrac{-3}{2}-\dfrac{12}{2} \right) \\
   \left( \dfrac{5}{2}+\dfrac{2}{2} \right) \\
\end{matrix} \right)+\left( \begin{matrix}
   6 \\
   0 \\
   8 \\
\end{matrix} \right)\]

\[\Rightarrow u=\left( \begin{matrix}
   0 \\
   \left( \dfrac{-15}{2} \right) \\
   \left( \dfrac{7}{2} \right) \\
\end{matrix} \right)+\left( \begin{matrix}
   6 \\
   0 \\
   8 \\
\end{matrix} \right)\]

Now, adding the two vectors, we get
\[u=\left( \begin{matrix}
   0+6 \\
   \left( \dfrac{-15}{2}+0 \right) \\
   \left( \dfrac{7}{2}+8 \right) \\
\end{matrix} \right)\]

Simplifying the each component of the above vector by taking LCM, we get
\[u=\left( \begin{matrix}
   6 \\
   \left( \dfrac{-15}{2} \right) \\
   \left( \dfrac{7}{2}+\dfrac{16}{2} \right) \\
\end{matrix} \right)\]

Simplifying the above vector, we get
\[u=\left( \begin{matrix}
   6 \\
   \left( \dfrac{-15}{2} \right) \\
   \left( \dfrac{23}{2} \right) \\
\end{matrix} \right)\]

Therefore, the value of \[u=\dfrac{1}{2}v-w+2z\]is equal to the\[u=\left( \begin{matrix}
   6 \\
   \left( \dfrac{-15}{2} \right) \\
   \left( \dfrac{23}{2} \right) \\
\end{matrix} \right)\].

Note:Students need to remember that addition and subtraction of vectors are different from the normally performing addition and subtraction on numbers, it is because a vector contains or represents the x, y and z direction respectively. So while adding or subtracting the given vectors we can only add or subtract the value of the same direction only. And to multiply a vector by a given scalar unit, we need to multiply each component of the given vector by that given scalar unit.