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

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.