Question

Write a unit vector in the direction of the sum of the vectors \[=2\hat{i}+2\hat{j}-5\hat{k}\] and \[=2\hat{i}+\hat{j}-7\hat{k}\].

Answer 1

Hint: In this question, in order find a unit vector in the direction of the sum of the vectors \[=2\hat{i}+2\hat{j}-5\hat{k}\] and \[=2\hat{i}+\hat{j}-7\hat{k}\] we will first evaluate the sum of the vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\]. Then we know that for a vector \[=x\hat{i}+y\hat{j}+z\hat{k}\], the magnitude of the vector \[=x\hat{i}+y\hat{j}+z\hat{k}\] is denoted by \[\left| \right|\] is given by \[\left| \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\] . Also the unit vector of the vector \[\] is given by \[\dfrac{}{\left| \right|}\] which is equals \[\dfrac{}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}\]. Now using this we will have to find the magnitude of the vector \[\overrightarrow{a}+\overrightarrow{b}\] which is denoted by \[\left| \overrightarrow{a}+\overrightarrow{b} \right|\] and then in order to evaluate the unit vector of the sum \[\overrightarrow{a}+\overrightarrow{b}\], we will have to find \[\dfrac{\overrightarrow{a}+\overrightarrow{b}}{\left| \overrightarrow{a}+\overrightarrow{b} \right|}\].

Complete step by step answer:
Let the vector \[\overrightarrow{a}\] is given by \[=2\hat{i}+2\hat{j}-5\hat{k}\] and the vector \[\overrightarrow{b}\] is given by \[=2\hat{i}+\hat{j}-7\hat{k}\].
On plotting this points on the graph we have

Now the sum of both the vectors \[\overrightarrow{a}\] and \[\overrightarrow{b}\] is given by the sum \[\overrightarrow{a}+\overrightarrow{b}\].
That is
\[\begin{align}
  & \overrightarrow{a}+\overrightarrow{b}=\left( 2\hat{i}+2\hat{j}-5\hat{k} \right)+\left( 2\hat{i}+\hat{j}-7\hat{k} \right) \\
 & =\left( 2+2 \right)\hat{i}+\left( 2+1 \right)\hat{j}+\left( -5-7 \right)\hat{k} \\
 & =4\hat{i}+3\hat{j}-12\hat{k} \\
\end{align}\]
Therefore we have \[\overrightarrow{a}+\overrightarrow{b}=4\hat{i}+3\hat{j}-12\hat{k}.............(1)\].
Now since we know that for a vector \[=x\hat{i}+y\hat{j}+z\hat{k}\], the magnitude of the vector \[=x\hat{i}+y\hat{j}+z\hat{k}\] is denoted by \[\left| \right|\] is given by \[\left| \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}\] .
Thus magnitude of the vector \[\overrightarrow{a}+\overrightarrow{b}\] which is denoted by \[\left| \overrightarrow{a}+\overrightarrow{b} \right|\].
Now comparing the sum \[\overrightarrow{a}+\overrightarrow{b}\] with the vector \[=x\hat{i}+y\hat{j}+z\hat{k}\], we will get
\[x=4,y=3\] and \[z=-12\].
Thus on calculating the magnitude of the sum \[\overrightarrow{a}+\overrightarrow{b}\], we will have
\[\begin{align}
  & \left| \overrightarrow{a}+\overrightarrow{b} \right|=\sqrt{{{4}^{2}}+{{3}^{2}}+{{\left( -12 \right)}^{2}}} \\
 & =\sqrt{16+9+144} \\
 & =\sqrt{169}
\end{align}\]
Now we know that \[\sqrt{169}=\pm 13\], but the magnitude of the vector cannot be negative.
Therefore we have
\[\begin{align}
  & \left| \overrightarrow{a}+\overrightarrow{b} \right|=\sqrt{169} \\
 & =13...............(2)
\end{align}\]
Also since we know that the unit vector of the vector \[\] is given by \[\dfrac{}{\left| \right|}\] which is equals \[\dfrac{xi+yj+zk}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}\].
In order to calculate the unit vector of the sum \[\overrightarrow{a}+\overrightarrow{b}\], we will have to find \[\dfrac{\overrightarrow{a}+\overrightarrow{b}}{\left| \overrightarrow{a}+\overrightarrow{b} \right|}\].
Now we will substitute the values in equation (1) and equation (2) in \[\dfrac{\overrightarrow{a}+\overrightarrow{b}}{\left| \overrightarrow{a}+\overrightarrow{b} \right|}\], we will get
\[\begin{align}
  & \dfrac{\overrightarrow{a}+\overrightarrow{b}}{\left| \overrightarrow{a}+\overrightarrow{b} \right|}=\dfrac{4\hat{i}+3\hat{j}-12\hat{k}}{13} \\
 & =\dfrac{4}{13}\hat{i}+\dfrac{3}{13}\hat{j}-\dfrac{12}{13}\hat{k}
\end{align}\].
Therefore the unit vector in the direction of the sum of the vectors \[=2\hat{i}+2\hat{j}-5\hat{k}\] and \[=2\hat{i}+\hat{j}-7\hat{k}\] is given by \[\dfrac{4}{13}\hat{i}+\dfrac{3}{13}\hat{j}-\dfrac{12}{13}\hat{k}\].

Note:
In this problem, please to don consider the magnitude of the vector \[\overrightarrow{a}+\overrightarrow{b}\] as \[-13\] and then substitute in into \[\dfrac{\overrightarrow{a}+\overrightarrow{b}}{\left| \overrightarrow{a}+\overrightarrow{b} \right|}\] to find the unit vector of the sum because the magnitude of the vector cannot be negative.