Question

Express \[2\hat i - \hat j + 3\hat k\] as the sum of a vector parallel and a vector perpendicular to \[2\hat i + 4\hat j - 2\hat k\].

Answer 1

Hint: Let us consider two vectors \[\overrightarrow A ,\overrightarrow B \]as 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 constant.
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 solution:
Let us consider, \[\overrightarrow A = 2\hat i - \hat j + 3\hat k\] and \[\overrightarrow B = 2\hat i + 4\hat j - 2\hat k\]
According to the problem we have to represent \[\overrightarrow A = 2\hat i - \hat j + 3\hat k\] as the sum of two vectors which are vector parallel and a vector perpendicular to\[\overrightarrow B = 2\hat i + 4\hat j - 2\hat k\].
Let us consider, \[\overrightarrow A = \overrightarrow C + \overrightarrow D \] where, \[\overrightarrow C \] is perpendicular to \[\overrightarrow B \]and \[\overrightarrow D \]is parallel to\[\overrightarrow B \].
So, from the given hint we have, \[\overrightarrow {C.} \overrightarrow B = 0\] and \[\overrightarrow D = \alpha \overrightarrow B \] for some constant\[\alpha \].
That is \[\overrightarrow A \]can be written as \[\overrightarrow A = \overrightarrow C + \alpha \overrightarrow B \]… (1)
Let us multiply \[\overrightarrow B \]on both sides of the equation we get,
\[\overrightarrow B \overrightarrow {.A} = \overrightarrow B .\overrightarrow C + \alpha \overrightarrow B .\overrightarrow B \]… (2)
Here \[\overrightarrow B \overrightarrow {.A} = (2\hat i - \hat j + 3\hat k).(2\hat i + 4\hat j - 2\hat k) = 4 - 4 - 6\]
\[\overrightarrow B .\overrightarrow B = (2\hat i + 4\hat j - 2\hat k).(2\hat i + 4\hat j - 2\hat k) = 4 + 16 + 4\]
Also \[\overrightarrow {C.} \overrightarrow B = 0\]
Substitute the values we have found in equation (2) we get,
\[4 - 4 - 6 = 0 + \alpha (16 + 4 + 4)\]
Let us solve the above equation to find \[\alpha \]we get,
\[\alpha = \dfrac{{ - 6}}{{24}} = \dfrac{{ - 1}}{4}\]
Let us consider, \[\overrightarrow C = x\hat i + y\hat j + z\hat k\]
Now substitute the values of \[\alpha = \dfrac{{ - 1}}{4}\] in (1) we get,
\[2\hat i - \hat j + 3\hat k = \dfrac{{ - 1}}{4}(2\hat i + 4\hat j - 2\hat k) + (x\hat i + y\hat j + z\hat k)\]
Let us take the common terms on right hand side,
\[2\hat i - \hat j + 3\hat k = (x - \dfrac{1}{2})\hat i + (y - 1)\hat j + (z + \dfrac{1}{2})\hat k\]
Let us equate the coefficient of the same components of the vectors in the above equation we have,
\[2 = - \dfrac{1}{2} + x\]; \[ - 1 = - 1 + y\]; \[3 = \dfrac{1}{2} + z\]
Let us solve the three equations we have,
\[x = \dfrac{5}{2};y = 0;z = \dfrac{5}{2}\]
So, \[\overrightarrow C = \dfrac{5}{2}\hat i + \dfrac{5}{2}\hat k\]

$\therefore$The sum of required two vectors is \[2\hat i - \hat j + 3\hat k = \dfrac{{ - 1}}{4}(2\hat i + 4\hat j - 2\hat k) + (\dfrac{5}{2}\hat i + \dfrac{5}{2}\hat k)\]

Note:
Let us consider two vectors \[\overrightarrow A ,\overrightarrow B \]. Then the dot product of them is \[\overrightarrow A .\overrightarrow B = \left| A \right|.\left| B \right|.\cos \theta \]. Here \[\theta \] is the angle between them. In vector \[\hat i,\hat j,\hat k\] are the unit vectors of the axes X, Y, Z respectively.
Also, we must be careful that in dot product the resultant is scalar whereas in cross-product the resultant is a vector. Here in dot product, we must multiply the coefficients of each vector but in cross-product, we form a determinant and solve it.