Question

Find the projection of \[\vec a\] on \[\vec b\] , if \[\vec a.\vec b = 8\] and \[\vec b = 2\hat i + 6\hat j + 3\hat k\] .

Answer 1

Hint: Any term represented with a one-sided arrowhead on top of it is generally said to be a vector. Vectors possess both direction as well as magnitude. The magnitude of a vector can be found if the vector is expressed in terms of unit vectors in a plane. Magnitude of such a vector is the square root of the sum of squares of the coefficients of unit vectors.

Complete step-by-step answer:
Let us see what we are given from the question;
They have given us two vectors \[\vec a\] and \[\vec b\] .
We are also given a dot product of these vectors \[\vec a\] and \[\vec b\] , it is said to be: \[\vec a.\vec b = 8\] .
Also the vector \[\vec b\] has been represented in this way: \[\vec b = 2\hat i + 6\hat j + 3\hat k\] .
We should remember that $\hat i,\hat j,\hat k$ are unit vectors which are denoted for each direction on the $XYZ$ plane. Vector $\hat i$ is a unit vector along the x-axis, $\hat j$ along the y-axis and $\hat k$ along the z-axis. Unit vectors are those who have their magnitude as $1$.
Now we have an already existing formula for the projection of one vector on the other. Say the vectors under consideration are \[\vec a\] and \[\vec b\] , then the projection of \[\vec a\] on \[\vec b\] is represented as follows:
$ \Rightarrow $Projection of vector \[\vec a\] on the vector \[\vec b\] $ = (\vec a.\vec b) \times \dfrac{1}{{|\vec b|}}$
But to find the value of projection, we need to first find the individual terms that is: \[\vec a.\vec b\] and $\dfrac{1}{{|\vec b|}}$.
From the question, we already have the value of \[\vec a.\vec b = 8\] , so we can keep that term aside.
Next to find the value of $\dfrac{1}{{|\vec b|}}$, we are given the representation of vector $\vec b$ on the XYZ plane.
The representation is \[\vec b = 2\hat i + 6\hat j + 3\hat k\] , and this representation is useful because using this we can easily find the magnitude of the vector $\vec b$.
When we are given a vector say $\vec p$ that is expressed as $x\hat i + y\hat j + z\hat k$, then the formula for its magnitude $|\vec p|$ is given by $\sqrt {{x^2} + {y^2} + {z^2}} $, where $x,y,z$ are the coefficients of $\hat i,\hat j,\hat k$.
In the given vector $\vec b$, the coefficients of $\hat i,\hat j,\hat k$ are $2,6,3$ respectively.
Substituting the values of coefficients of $\hat i,\hat j,\hat k$ into the formula to find the magnitude of the vector $\vec b$
$ \Rightarrow |\vec b| = \sqrt {{2^2} + {6^2} + {3^2}} $
$ \Rightarrow |\vec b| = \sqrt {4 + 36 + 9} $
$ \Rightarrow |\vec b| = \sqrt {49} $
$ \Rightarrow |\vec b| = 7$
Moving on to substituting the values of \[\vec a.\vec b\] and $|\vec b|$ into the formula to find the projection;
$ \Rightarrow $Projection of vector \[\vec a\] on the vector \[\vec b\] $ = (8) \times \dfrac{1}{{(7)}}$
$ \Rightarrow $Projection of vector \[\vec a\] on the vector \[\vec b\] $ = \dfrac{8}{7}$
Therefore the final answer is $ = \dfrac{8}{7}$.
So, the correct answer is “ $ \dfrac{8}{7}$”.

Note: We have understood that when a quantity has both magnitude and direction it is said to be a vector. But if a quantity has only magnitude, and does not indicate direction then it is said to be a scalar quantity. Due to the lack of direction for scalar quantities, it is said to be one dimensional, but vectors having direction can be expressed as multidimensional.