Question

If $\overrightarrow \alpha = 3\hat i + 4\hat j + 5\hat k$ and $\overrightarrow \beta = 2\hat i + \hat j - 4\hat k$, then express $\overrightarrow \beta$ in the form of $\overrightarrow \beta = \overrightarrow {{\beta _1}} + \overrightarrow {{\beta _2}}$, where, ${\overrightarrow \beta _1}$ is parallel to $\overrightarrow \alpha$ and ${\overrightarrow \beta _2}$ is perpendicular to $\overrightarrow \alpha$.

Answer 1

Let us consider two vectors $\overrightarrow A ,\overrightarrow B$ be two parallel vectors. Then they are scalar multiple of one another.
That is, $\overrightarrow A = k\overrightarrow B$ or $\overrightarrow B = c\overrightarrow A$, where $k$ and $c$ are scalar constants.
Let us consider two vectors $\overrightarrow A ,\overrightarrow B$. They are called perpendicular vector if and only if $\overrightarrow A .\overrightarrow B = 0$.

Complete step-by-step answer:
It is given that,
$\overrightarrow \alpha = 3\hat i + 4\hat j + 5\hat k$ and $\overrightarrow \beta = 2\hat i + \hat j - 4\hat k$
Also, $\overrightarrow \beta$ in the form of $\overrightarrow \beta = \overrightarrow {{\beta _1}} + \overrightarrow {{\beta _2}}$, where, ${\overrightarrow \beta _1}$ is parallel to $\overrightarrow \alpha$ and ${\overrightarrow \beta _2}$ is perpendicular to $\overrightarrow \alpha$.
Since, ${\overrightarrow \beta _1}$ is parallel to $\overrightarrow \alpha$
${\overrightarrow \beta _1}$ can be written as ${\overrightarrow \beta _1} = \lambda \overrightarrow \alpha$, where, $\lambda$is any scalar.
So let us multiply $\lambda$to a vector value of A, ${\overrightarrow \beta _1} = 3\lambda \hat i + 4\lambda \hat j + 5\lambda \hat k$
Again, $\overrightarrow \beta = \overrightarrow {{\beta _1}} + \overrightarrow {{\beta _2}}$
Let us consider the Substitution of the values of given vectors we get,
$2\hat i + \hat j - 4\hat k = 3\lambda \hat i + 4\lambda \hat j + 5\lambda \hat k + {\overrightarrow \beta _2}$
Here we are going to simplify in order to get ${\overrightarrow \beta _2}$,
${\overrightarrow \beta _2} = (2 - 3\lambda )\hat i + (1 - 4\lambda )\hat j - (4 + 5\lambda )\hat k$
As per the given condition, ${\overrightarrow \beta _2}$ is perpendicular to $\overrightarrow \alpha$
Then, ${\overrightarrow \beta _2}.\overrightarrow \alpha = 0$
Substituting the values of given vectors to get the result we get,
$(2 - 3\lambda ).3 + (1 - 4\lambda ).4 - (4 + 5\lambda ).5 = 0$
Now we are going to simplify the above term, we get,
$6 - 9\lambda + 4 - 16\lambda - 20 - 25\lambda = 0$
Simplifying again the above term, we get,
$- 10 - 50\lambda = 0$
Therefore,
$\lambda = \dfrac{{ - 1}}{5}$
Now, to find the values of ${\overrightarrow \beta _1}$ and ${\overrightarrow \beta _2}$ we will substitute the value of $\lambda$.
So, we get as,
${\overrightarrow \beta _1} = \dfrac{{ - 3}}{5}\hat i + \dfrac{{ - 4}}{5}\hat j - \hat k$
And, ${\overrightarrow \beta _2} = (2 + \dfrac{3}{5})\hat i + (1 + \dfrac{4}{5})\hat j - (4 - 1)\hat k$
Solving the above to get the desired result, we get,
${\overrightarrow \beta _2} = \dfrac{{13}}{5}\hat i + \dfrac{9}{5}\hat j - 3\hat k$
Therefore,
$2\hat i + \hat j - 4\hat k = (\dfrac{{ - 3}}{5}\hat i + \dfrac{{ - 4}}{5}\hat j - \hat k) + (\dfrac{{13}}{5}\hat i + \dfrac{9}{5}\hat j - 3\hat k)$

Note: Let us consider two vectors $\overrightarrow A ,\overrightarrow B$. Then the dot product of them is,
$\overrightarrow A .\overrightarrow B = \left| A \right|.\left| B \right|.\cos \theta$
$\theta$ be the angle between them.
In vector, $\hat i,\hat j,\hat k$ are the unit vectors of the axes X, Y, Z respectively.
Then by the conditions of dot product of vectors we get,
$\hat i.\hat i = \left| {\hat i} \right|.\left| {\hat i} \right|.\cos {0^ \circ } = 1$
Similarly, $\hat j.\hat j = 1,\hat k.\hat k = 1$
Again, $\hat i.\hat j = \left| {\hat i} \right|.\left| {\hat j} \right|.\cos {90^ \circ } = 0$
Similarly, $\hat j.\hat k = \hat k.\hat i = 0$