
Unit vector parallel to the resultant of vectors $\vec A = 4\hat i - 3\hat j$ and $\vec B = 8\hat i + 8\hat j$
A. $\dfrac{{24\hat i + 5\hat j}}{{13}}$
B. $\dfrac{{12\hat i + 5\hat j}}{{13}}$
C. $\dfrac{{6\hat i + 5\hat j}}{{13}}$
D. None of these
Answer
497.7k+ views
Hint: In order to solve this question, we will first find the resultant of two given vectors where the resultant of two vectors is just the sum of two vectors by its component wise. A unit vector is one which has a magnitude of unit one, and a unit vector parallel to a vector is simply the vector having the same components but having magnitude of unit one.
Complete step by step answer:
According to the question, we have given that two vectors are as $\vec A = 4\hat i - 3\hat j$ and $\vec B = 8\hat i + 8\hat j$. The resultant of these two vectors is represented by $\vec R = \vec A + \vec B$. On putting the values we get,
$\vec R = (4\hat i - 3\hat j) + (8\hat i + 8\hat j)$
$\Rightarrow \vec R = 12\hat i + 5\hat j$
so, the resultant of given two vectors $\vec A = 4\hat i - 3\hat j$ and $\vec B = 8\hat i + 8\hat j$ is $\vec R = 12\hat i + 5\hat j$
Now, we have to find a unit vector parallel to the vector $\vec R = 12\hat i + 5\hat j$ so, a unit vector is simple calculated by using the relation as
$\hat R = \dfrac{{\vec R}}{{\left| R \right|}}$
where $\hat R$ is the unit vector and $\left| R \right| = \sqrt {{x^2} + {y^2}} $ is the magnitude of vector $\vec R = x\hat i + y\hat j$ and in our solution we have,
$\Rightarrow \vec R = 12\hat i + 5\hat j$
where $x = 12{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} y = 5$
On putting the values, we have the magnitude of vector $R$ as,
$\left| R \right| = \sqrt {{{12}^2} + {5^2}} $
$\Rightarrow \left| R \right| = \sqrt {169} $
$\Rightarrow \left| R \right| = 13$
And now we have, $\vec R = 12\hat i + 5\hat j$ and $\left| R \right| = 13$
Putting these values in relation of unit vector $R$ as,
$\hat R = \dfrac{{\vec R}}{{\left| R \right|}}$
We get,
$\therefore \hat R = \dfrac{{12\hat i + 5\hat j}}{{13}}$
So, the unit vector $R$ parallel to the resultant of two vectors is $\hat R = \dfrac{{12\hat i + 5\hat j}}{{13}}$.
Hence, the correct option is B.
Note: It should be remembered that, In three dimensional space a vector is written in the form of $\vec R = x\hat i + y\hat j + z\hat k$ where I, j, k are the unit vectors in the direction of three axes namely X, Y, Z. and x, y, z are the components of the vector R in these three dimensions respectively.
Complete step by step answer:
According to the question, we have given that two vectors are as $\vec A = 4\hat i - 3\hat j$ and $\vec B = 8\hat i + 8\hat j$. The resultant of these two vectors is represented by $\vec R = \vec A + \vec B$. On putting the values we get,
$\vec R = (4\hat i - 3\hat j) + (8\hat i + 8\hat j)$
$\Rightarrow \vec R = 12\hat i + 5\hat j$
so, the resultant of given two vectors $\vec A = 4\hat i - 3\hat j$ and $\vec B = 8\hat i + 8\hat j$ is $\vec R = 12\hat i + 5\hat j$
Now, we have to find a unit vector parallel to the vector $\vec R = 12\hat i + 5\hat j$ so, a unit vector is simple calculated by using the relation as
$\hat R = \dfrac{{\vec R}}{{\left| R \right|}}$
where $\hat R$ is the unit vector and $\left| R \right| = \sqrt {{x^2} + {y^2}} $ is the magnitude of vector $\vec R = x\hat i + y\hat j$ and in our solution we have,
$\Rightarrow \vec R = 12\hat i + 5\hat j$
where $x = 12{\kern 1pt} {\kern 1pt} {\kern 1pt} {\kern 1pt} y = 5$
On putting the values, we have the magnitude of vector $R$ as,
$\left| R \right| = \sqrt {{{12}^2} + {5^2}} $
$\Rightarrow \left| R \right| = \sqrt {169} $
$\Rightarrow \left| R \right| = 13$
And now we have, $\vec R = 12\hat i + 5\hat j$ and $\left| R \right| = 13$
Putting these values in relation of unit vector $R$ as,
$\hat R = \dfrac{{\vec R}}{{\left| R \right|}}$
We get,
$\therefore \hat R = \dfrac{{12\hat i + 5\hat j}}{{13}}$
So, the unit vector $R$ parallel to the resultant of two vectors is $\hat R = \dfrac{{12\hat i + 5\hat j}}{{13}}$.
Hence, the correct option is B.
Note: It should be remembered that, In three dimensional space a vector is written in the form of $\vec R = x\hat i + y\hat j + z\hat k$ where I, j, k are the unit vectors in the direction of three axes namely X, Y, Z. and x, y, z are the components of the vector R in these three dimensions respectively.
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