Question

The unit vector parallel to the resultant of vectors $\overrightarrow{A}=4\widehat{i}+3\widehat{j}+6\widehat{k}$ and $\overrightarrow{B}=-\widehat{i}+3\widehat{j}-8\widehat{k}$ is:
A. $\dfrac{1}{7}\left( 3\widehat{i}+6\widehat{j}-2\widehat{k} \right)$
B. $-\dfrac{1}{7}\left( 3\widehat{i}+6\widehat{j}-2\widehat{k} \right)$
C. $\dfrac{1}{49}\left( 3\widehat{i}+6\widehat{j}-2\widehat{k} \right)$
D. $\dfrac{1}{49}\left( 3\widehat{i}+6\widehat{j}-2\widehat{k} \right)$

Answer 1

Hint: The unit vector along the resultant of the given vectors can be found by first calculating the resultant vector of the two. Then find the magnitude of the resultant vector. Use the formula for the unit vector along a known vector.

Formula used:
$\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$
$\widehat{R}=\dfrac{\overrightarrow{R}}{\left| \overrightarrow{R} \right|}$
$\left| \overrightarrow{R} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$

Complete step by step answer:
To find the unit vector parallel to the resultant of vectors A and B, let us first the resultant of the vectors A and B. The resultant of the vectors will be $\overrightarrow{R}=\overrightarrow{A}+\overrightarrow{B}$.
But it is given that $\overrightarrow{A}=4\widehat{i}+3\widehat{j}+6\widehat{k}$ and $\overrightarrow{B}=-\widehat{i}+3\widehat{j}-8\widehat{k}$.
Therefore,
$\Rightarrow \overrightarrow{R}=\left( 4\widehat{i}+3\widehat{j}+6\widehat{k} \right)+\left( -\widehat{i}+3\widehat{j}-8\widehat{k} \right)$.
This means that the resultant of the vectors A and B can be written as $\overrightarrow{R}=\left( (4-1)\widehat{i}+(3+3)\widehat{j}+(6-8)\widehat{k} \right)=\left( 3\widehat{i}+6\widehat{j}-2\widehat{k} \right)$.
Hence, we found the resultant of the two given vectors.

Now, let us calculate the unit vector along (parallel to) this resultant vector.
The unit vector along any vector R is equal to the vector R divided by the magnitude of the vector R.
i.e. $\widehat{R}=\dfrac{\overrightarrow{R}}{\left| \overrightarrow{R} \right|}$ …. (i),
where $\widehat{R}$ is the unit vector along vector R and $\left| \overrightarrow{R} \right|$ is its magnitude.

Therefore, let us calculate the magnitude of vector R.
The magnitude of a vector $\overrightarrow{R}=x\widehat{i}+y\widehat{j}+z\widehat{k}$ is given as $\left| \overrightarrow{R} \right|=\sqrt{{{x}^{2}}+{{y}^{2}}+{{z}^{2}}}$.
Hence, $\left| \overrightarrow{R} \right|=\sqrt{{{3}^{2}}+{{6}^{2}}+{{(-2)}^{2}}}=\sqrt{9+36+4}=\sqrt{49}$.
$\Rightarrow \left| \overrightarrow{R} \right|=\pm 7$.
However, the magnitude of a vector is always positive. Therefore, $\left| \overrightarrow{R} \right|=-7$ is discarded.
$\Rightarrow \left| \overrightarrow{R} \right|=7$.

Now, substitute the values of $\left| \overrightarrow{R} \right|$ and $\overrightarrow{R}$ in (i).
$\therefore \widehat{R}=\dfrac{3\widehat{i}+6\widehat{j}-2\widehat{k}}{7}$
Therefore, we found the unit vector along the resultant of the vectors A and B.

Hence, the correct option is A.

Note: When any two vectors are added, only the similar components are added. This means that the x component of one vector is added to the x component of the other vector, the y component of the first vector is added to the y component of the second vector and the z component of the first vector is added to the z component of the second vector. The same goes with subtraction.