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

Hint: 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, \[\hat j.\hat k = \hat k.\hat i = 0\]