Question

If the resolved part of the force vector $5\widehat i + 4\widehat j + 2\widehat k$ along and perpendicular to the vector$3\widehat i + 4\widehat j + - 5\widehat k$ are $\alpha$ and $\beta$ respectively. Then the value of $\alpha$ is
a.$\dfrac{{21}}{{50}}\left( {3\widehat i + 4\widehat j - 5\widehat k} \right)$
b.$\dfrac{{21}}{{50}}\left( {3\widehat i - 4\widehat j + 5\widehat k} \right)$
c.$\dfrac{{11}}{{50}}\left( {2\widehat i - 4\widehat j + 3\widehat k} \right)$
d.$\dfrac{1}{{50}}\left( {187\widehat i + 116\widehat j + 205\widehat k} \right)$

Answer 1

Here we need to find the value of $\alpha$, which is equal to the resolved part of vector $5\widehat i + 4\widehat j + 2\widehat k$ along the vector $3\widehat i + 4\widehat j + - 5\widehat k$. We will resolve the vector $5\widehat i + 4\widehat j + 2\widehat k$ along another vector$3\widehat i + 4\widehat j + - 5\widehat k$. We will use the formula to resolve the vector$5\widehat i + 4\widehat j + 2\widehat k$ along another vector$3\widehat i + 4\widehat j + - 5\widehat k$. This will give us the value of $\alpha$.

Formula used:
We will use the formula to resolve the part of any vector $a$ along another vector $b$ is $\dfrac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b }}{{{{\left| b \right|}^2}}}$.

Complete step-by-step answer:
It is given that the resolved part of the vector $5\widehat i + 4\widehat j + 2\widehat k$ along the vector $3\widehat i + 4\widehat j + - 5\widehat k$ is equal to $\alpha$and resolved part of the vector $5\widehat i + 4\widehat j + 2\widehat k$perpendicular to the vector$3\widehat i + 4\widehat j + - 5\widehat k$ is equal to$\beta$.
Let us assume $\overrightarrow a = 5\widehat i + 4\widehat j + 2\widehat k$ and $\overrightarrow b = 3\widehat i + 4\widehat j + - 5\widehat k$.
We will put the value of the vector$\overrightarrow a$ and $\overrightarrow b$ in the formula $\Rightarrow$ $\dfrac{{\left( {\overrightarrow a .\overrightarrow b } \right)\overrightarrow b }}{{{{\left| b \right|}^2}}}$.
$\Rightarrow$ $\alpha = \dfrac{{\left[ {\left( {5\widehat i + 4\widehat j + 2\widehat k} \right).\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)} \right]\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)}}{{{{\left| {\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)} \right|}^2}}}$
We find the dot product of the two vectors in the numerator.
$\Rightarrow$ $\alpha = \dfrac{{\left( {5.3 + 4.4 - 2.5} \right)\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)}}{{{{\left| {\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)} \right|}^2}}}$
Multiplying and adding the terms inside the bracket, we get
$\Rightarrow$ $\alpha = \dfrac{{21\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)}}{{{{\left| {\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)} \right|}^2}}}$
Now, we will find the magnitude of the vector in the denominator.
$\Rightarrow$ $\alpha = \dfrac{{21\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)}}{{{{\left| {\sqrt {\left( {{3^2} + {4^2} + {{\left( { - 5} \right)}^2}} \right)} } \right|}^2}}}$
Simplifying the expression, we get
$\Rightarrow$ $\alpha = \dfrac{{21\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)}}{{50}}$
Rewriting the value of $\alpha$, we get
$\Rightarrow$ $\alpha = \dfrac{{21}}{{50}}\left( {3\widehat i + 4\widehat j + - 5\widehat k} \right)$
Thus, the correct option is option (a).

Note: We have calculated the resolved part of a vector along another vector. Vector resolution is defined as a process of breaking one vector into two smaller vectors where both the vectors will be perpendicular to each other. In other words, we can say that the vector resolution is a process of finding two components of a given vector.