Question

Find the sum of the vectors \[3\hat i + 7\hat j - 4\hat k\] and \[\hat i - 5\hat j - 8\hat k\] and hence find the unit vector along the sum of these vectors.

Answer 1

Hint:The addition of two vectors \[a\hat i + b\hat j + c\hat k\] and \[d\hat i + e\hat j + f\hat k\] is $(a + d)\hat i + (b + e)\hat j + (c + f)\hat k$. To find the unit vector of any vector, divide the vector by its magnitude.

Complete step-by-step answer:
We are given three vectors \[3\hat i + 7\hat j - 4\hat k\] and \[\hat i - 5\hat j - 8\hat k\].
First, we find the sum of the vectors.
To add the vectors, just simply add the components of the same vector. It means $\hat i$ component add to $\hat i$ component, $\hat j$ component add to $\hat j$ component and $\hat k$ component add to $\hat k$ component.
Write the expression for the addition of the vectors.
\[(3 + 1 + 1)\hat i + (7 - 5 - 5)\hat j + ( - 4 - 8 - 8)\hat k\]
Solve the expression.
\[5\hat i - 3\hat j - 20\hat k\]
Hence, the sum of the vectors is \[5\hat i - 3\hat j - 20\hat k\].
Let the addition of the vector be represented by $\vec x$.
Therefore, \[\vec x = 5\hat i - 3\hat j - 20\hat k\]
Now, we have to find the unit vector along the sum of these vectors.
It means we have to find the unit vector along $\vec x$.
We use the formula for calculating the unit vector.
We know that the unit vector of any vector can be evaluated by dividing that vector by its magnitude.
It means we have to divide the vector $\vec x$ by its magnitude that is $|x|$.
The unit vector of $\vec x$ is denoted by $\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{x} $.
Mathematically it can be written as,
$\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{x} = \dfrac{{\vec x}}{{|x|}}$
Now, we have to find the magnitude of $\vec x$.
The magnitude of any vector $\vec p = a\hat i + b\hat j + c\hat k$ is defined as:
$|\vec p| = \sqrt {{a^2} + {b^2} + {c^2}} $
Therefore, the magnitude of the vector \[\vec x = 5\hat i - 3\hat j - 20\hat k\] is:
$|\vec x| = \sqrt {{5^2} + {{( - 3)}^2} + {{( - 20)}^2}} $
Solve the expression and evaluate the magnitude of the $\vec x$.
$
  |\vec x| = \sqrt {25 + 9 + 400} \\
   \Rightarrow |\vec x| = \sqrt {434} \\
 $
Therefore, the unit vector of $\vec x$ is:
 \[\overset{\lower0.5em\hbox{$\smash{\scriptscriptstyle\frown}$}}{x} = \dfrac{{5\hat i - 3\hat j - 20\hat k}}{{\sqrt {434} }}\]

Note:The magnitude of any vector can never be a negative value. For the subtraction of vectors, we follow the same rule as in the addition of the vectors. The magnitude of the unit vector is always $1$.