Question

Write unit vector in the direction of the sum of vectors $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}+3\hat{k}$.

Answer 1

Hint: We start solving the problem by recalling the definition of the sum of two vectors. We then find the sum of the vectors given in the problem by following the definition. Once we find the sum of the vectors, we divide the resultant vector with its magnitude to find the required unit vector.

Complete step-by-step solution:
Given that we have two vectors $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}+3\hat{k}$. We need to find the unit vector in the direction of the sum of vectors $\vec{a}$ and $\vec{b}$.
Let us first find the sum of the given vectors $\vec{a}$ and $\vec{b}$. We know that sum of two vectors is again a vector. Let us assume the vector we get after the sum be $\vec{c}$.
We have $\vec{c}=\vec{a}+\vec{b}$. We know that the sum of the two vectors $a\hat{i}+b\hat{j}+c\hat{k}$ and $d\hat{i}+e\hat{j}+f\hat{k}$ is defined as $\left( a+d \right)\hat{i}+\left( b+e \right)\hat{j}+\left( c+f \right)\hat{k}$.
So, $\vec{c}=\left( 2\hat{i}-\hat{j}+2\hat{k} \right)+\left( -\hat{i}+\hat{j}+3\hat{k} \right)$.
$\Rightarrow \vec{c}=\left( 2-1 \right)\hat{i}+\left( -1+1 \right)\hat{j}+\left( 2+3 \right)\hat{k}$.
$\Rightarrow \vec{c}=\hat{i}+0\hat{j}+5\hat{k}$.
$\Rightarrow \vec{c}=\hat{i}+5\hat{k}$.
We got the sum of the vectors $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}+3\hat{k}$ as $\vec{c}=\hat{i}+5\hat{k}$.
We need to find the unit vector in the direction of the vector $\vec{c}=\hat{i}+5\hat{k}$. Let us represent this unit vector to be $\hat{d}$.
We know that the unit vector is found by dividing the vector with its magnitude. We know that the unit vector in the direction of $x\hat{i}+y\hat{j}+z\hat{k}$ is defined as $+\dfrac{x\hat{i}+y\hat{j}+z\hat{k}}{\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}}$.
Now, the unit vector in the direction of vector $\vec{c}=\hat{i}+5\hat{k}$ is $\hat{d}=+\dfrac{\hat{i}+5\hat{k}}{\sqrt{{{1}^{2}}+{{5}^{2}}}}$.
$\Rightarrow \hat{d}=\dfrac{\hat{i}+5\hat{k}}{\sqrt{1+25}}$.
$\Rightarrow \hat{d}=\dfrac{\hat{i}+5\hat{k}}{\sqrt{26}}$.
We have found the unit vector in the direction of the vector $\vec{c}=\hat{i}+5\hat{k}$ as $\dfrac{\hat{i}+5\hat{k}}{\sqrt{26}}$.
$\therefore$ The unit vector in the direction of the sum of vectors $\vec{a}=2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{b}=-\hat{i}+\hat{j}+3\hat{k}$ is $\dfrac{\hat{i}+5\hat{k}}{\sqrt{26}}$.

Note: We should know that the magnitude of the unit vector is 1. We also should know that the unit vector is parallel to the vector we divide with magnitude. If we need to find the vector in the opposite direction of the sum of the vectors, we take a negative sign instead of a positive sign. Similarly, we can expect problems to get the vectors with a given magnitude.