Answer
Verified
387.6k+ views
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.
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.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main
Using the following information to help you answer class 12 chemistry CBSE
Trending doubts
Difference Between Plant Cell and Animal Cell
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
How do you solve x2 11x + 28 0 using the quadratic class 10 maths CBSE
Select the correct plural noun from the given singular class 10 english CBSE
What organs are located on the left side of your body class 11 biology CBSE
The sum of three consecutive multiples of 11 is 363 class 7 maths CBSE
What is the z value for a 90 95 and 99 percent confidence class 11 maths CBSE
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
How many squares are there in a chess board A 1296 class 11 maths CBSE