Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# The projection of the vector $\overrightarrow{a}=\widehat{i}-2\widehat{j}+\widehat{k}$ on the vector $\overrightarrow{b}=4\widehat{i}-4\widehat{j}+7\widehat{k}$ isA.$\dfrac{9}{19}$B. $\dfrac{19}{9}$C. $9$D. $\sqrt{19}$

Last updated date: 17th Apr 2024
Total views: 35.1k
Views today: 0.35k
Verified
35.1k+ views
Hint: The formula for finding the projection of vector $\overrightarrow{u}$ on vector $\overrightarrow{v}$ is that the length of the projection = $\overrightarrow{u}.\widehat{v}=\overrightarrow{u}.\overrightarrow{\dfrac{v}{|\overrightarrow{v}|}}$. Proceed using this formula.

Complete step by step answer: -
Let’s remember what the formula for finding the projection of a vector on another vector was.
Let us have one vector named $\overrightarrow{u}$ and another vector called $\overrightarrow{v}$. Then, if it is asked to find the projection of $\overrightarrow{u}$ on $\overrightarrow{v}$, then the formula for finding the projection = $\overrightarrow{u}.\widehat{v}=\overrightarrow{u}.\dfrac{\overrightarrow{v}}{|\overrightarrow{v}|}$
In other words, the projection of one vector on another is basically its dot product with the other vector’s unit vector.
The unit vector of a particular vector is found by dividing that vector by its magnitude. The magnitude of the vector is found by taking the under root of the sum of the squares of the $x$, $y$ and $z$ coordinates.
$\widehat{v}=\dfrac{\overrightarrow{v}}{|\overrightarrow{v}|}$
Thus, to find the projection of $\overrightarrow{u}$ on $\overrightarrow{v}$, all we have to do is multiply $\overrightarrow{u}$ with $\overrightarrow{v}$’s unit vector, $\widehat{v}$.
In the question given to us, the vector whose projection has to be found out, or $\overrightarrow{u}$= $\overrightarrow{a}=\widehat{i}-2\widehat{j}+\widehat{k}$ and the vector on which the projection has to be found out, or $\overrightarrow{v}$= $\overrightarrow{b}=4\widehat{i}-4\widehat{j}+7\widehat{k}$.
Thus, we’ll first find out the unit vector of $\overrightarrow{b}$, since we have to find the dot product of $\widehat{b}$ with $\overrightarrow{a}$ to find the projection.
Therefore, $\widehat{b}=\dfrac{\overrightarrow{b}}{|\overrightarrow{b}|}=\dfrac{4\widehat{i}-4\widehat{j}+7\widehat{k}}{\sqrt{{{4}^{2}}+{{4}^{2}}+{{7}^{2}}}}=\dfrac{4\widehat{i}-4\widehat{j}+7\widehat{k}}{\sqrt{81}}=\dfrac{4\widehat{i}-4\widehat{j}+7\widehat{k}}{9}$
Now, to find out the projection, all we have to do is find $\overrightarrow{a}.\widehat{b}$.
Therefore, $\overrightarrow{a}.\widehat{b}=(\widehat{i}-2\widehat{j}+\widehat{k}).\dfrac{(4\widehat{i}-4\widehat{j}+7\widehat{k})}{9}=\dfrac{4+8+7}{9}=\dfrac{19}{9}$
Hence, our required projection = $\dfrac{19}{9}$.
Therefore,  the correct answer is option (b).

Note: It should be important that the student understands why the projection of one vector on another is found like this. The dot product of two vectors $\overrightarrow{u}$ and $\overrightarrow{v}$ = $\overrightarrow{u}.\overrightarrow{v}=|\overrightarrow{u}|\times |\overrightarrow{v}|\times \cos \theta$, where $\theta$ is the angle made between the vectors $\overrightarrow{u}$ and $\overrightarrow{v}$. Now, the projection of a vector $\overrightarrow{u}=|\overrightarrow{u}|\cos \theta$, where $\theta$ is the angle made between the vectors $\overrightarrow{u}$ and $\overrightarrow{v}$, if $\overrightarrow{v}$ is the vector we have to find the projection on. Hence, the projection of $\overrightarrow{u}$ on $\overrightarrow{v}$ = $\overrightarrow{u}.\dfrac{\overrightarrow{v}}{|\overrightarrow{v}|}$, and this is a scalar quantity, because the projection is supposed to be a length.