Question

If $\overrightarrow A = 2\widehat i + 7\widehat j + 3\widehat k$ and $\overrightarrow B = 3\widehat i + 2\widehat j + 5\widehat k$, then find the projection of $\overrightarrow a $ and $\overrightarrow b $.

Answer 1

Hint:
The problem has a vector $\overrightarrow A $ projected on vector $\overrightarrow B $. Thus, one vector component of $\overrightarrow A $ is parallel to the vector $\overrightarrow B $. We can easily calculate the projection of $\overrightarrow A $ and $\overrightarrow B $ by using one formula
\[{\text{Pro}}{{\text{j}}_{(\overrightarrow {B)} }}^{(\overrightarrow {A)} } = \,\dfrac{1}{{|\overrightarrow B |\,}}\,(\overrightarrow A \,.\,\overrightarrow B )\,\,or\,\,\dfrac{{(\overrightarrow A \,.\,\overrightarrow B )\,}}{{|\overrightarrow B |\,}}\]

Stepwise solution:

Given:
$\overrightarrow A = 2\widehat i + 7\widehat j + 3\widehat k$
$\overrightarrow B = 3\widehat i + 2\widehat j + 5\widehat k$
If it is given in the question that $\overrightarrow A $ projects on $\overrightarrow B $ , so that the vector component of $\overrightarrow A $ is in the direction of vector component $\overrightarrow B $. The orthogonal projection of A onto a straight line parallel to B, which is defined as \[A = A\widehat B\] where A is scalar, called a scalar projection of A onto B, and B is the unit vector in the direction of B.
Thus, the projection of $\overrightarrow A $ on $\overrightarrow B $ can be written as:
\[\Pr o{j_{(\overrightarrow {B)} }}^{(\overrightarrow {A)} } = \,\dfrac{1}{{|\overrightarrow B |\,}}\,(\overrightarrow A \,.\,\overrightarrow B )\,\] eq. (1)
Where, \[(\overrightarrow A \,.\,\overrightarrow B )\,\]is the dot product of two vectors $\overrightarrow A $and $\overrightarrow B $ and $|\overrightarrow B |$ is the magnitude of vector $\overrightarrow B $.
$\overrightarrow A = 2\widehat i + 7\widehat j + 3\widehat k$ and $\overrightarrow B = 3\widehat i + 2\widehat j + 5\widehat k$
So, dot product of $\overrightarrow A $ and $\overrightarrow B $ gives:
$(\overrightarrow A .\overrightarrow B )\, = \,(2\widehat i + 7\widehat j + 3\widehat k).\,(3\widehat i + 2\widehat j + 5\widehat k)$
$ \Rightarrow (2 \times 3) + (7 \times 2) + (3 \times 5)$ eq. (2)
$ \Rightarrow (\overrightarrow A .\,\overrightarrow B )\, = \,35$
Here, we take one product of the numbers in the same co ordinate and then add them all to obtain $(\overrightarrow A .\,\overrightarrow B )\,$
Also, remember that the dot product of $\widehat i$, $\widehat j$ and $\widehat k$ with itself is 1.
Now,
$|\overrightarrow B |\, = \,\sqrt {{{(\widehat i)}^2}\, + \,{{(\widehat j\,)}^2} + \,{{(\widehat k)}^2}} $
$|\overrightarrow B |\, = \,\sqrt {{{(3)}^2}\, + \,{{(2\,)}^2} + \,{{(5)}^2}} $
$|\overrightarrow B |\, = \,\sqrt {38} $
Now putting the value in eq. (1), we get
\[\Pr o{j_{(\overrightarrow {B)} }}^{(\overrightarrow {A)} } = \,\dfrac{{35}}{{\sqrt {38} }} = \,5.67\]

Note:
Students must have an idea of certain vector properties before attempting this question. The scalar products of two vectors a and b, denoted by \[\overrightarrow a .\,\overrightarrow b \] is defined as \[\overrightarrow a .\,\overrightarrow b = |\,\overrightarrow a |\,|\overrightarrow b |\,\cos \,\theta \] where $\theta $ is the angle between vector and vector b, $0 \leqslant \theta \leqslant \pi $. If either vector (a) equal to zero or vector (b) equal to zero, then $\theta $ is not defined, and in this case, we define \[\overrightarrow a .\,\overrightarrow b = 0\].