Question

$\left( a.i \right)i+\left( a.j \right)j+\left( a.k \right)k$ is equal to?
$A)a$
$B)2a$
$C)3a$
$D)0$

Answer 1

Hint: This question may need to have the knowledge of vectors. Vector is an object which has both magnitude and direction. So on representing a vector $\overrightarrow{V}$ in terms of the direction vectors which are $i,j,k$ in $x,y$ and $z$ direction respectively is given as $\overrightarrow{V}=ui+vj+wk$ . Here $u,v,w$ represents the magnitude of the vector $\overrightarrow{V}$. In this question we will consider $a$ vector as $\vec{a}=xi+yj+zk$ and will then as given in the question.

Complete step-by-step solution:
The question ask us to find the value of expression $\left( a.i \right)i+\left( a.j \right)j+\left( a.k \right)k$ where $a$ is considered to be a vector and $i,j,k$ represents the direction of the vector. Now we will write the vector $\vec{a}$ in terms of direction, which is $i,j,k$ so that the question solving becomes easier. So the vector $\vec{a}$ becomes $\vec{a}=xi+yj+zk$. Now we will substitute the value of the vector $\vec{a}$ which is in terms of the direction vector in the above equation. On doing that we get:
$\Rightarrow \left( a.i \right)i+\left( a.j \right)j+\left( a.k \right)k$
$\Rightarrow \left( \left( xi+yj+zk \right).i \right)i+\left( \left( xi+yj+zk \right).j \right)j+\left( \left( xi+yj+zk \right).k \right)k$
On expanding the above expression we get:
$\Rightarrow \left( xi.i+yj.i+zk.i \right)i+\left( xi.j+yj.j+zk.j \right)j+\left( xi.k+yj.k+zk.k \right)k$
Now we know that the dot product of the same direction is $1$, but the dot product of the rest of the combinations are zero. So on expanding the above expression we get:
$\Rightarrow \left( x+0+0 \right)i+\left( 0+y+0 \right)j+\left( 0+0+z \right)k$
On calculating further we get:
$\Rightarrow xi+yj+zk$
The above expression is the same as the vector $\vec{a}$.
$\therefore $ : $\left( a.i \right)i+\left( a.j \right)j+\left( a.k \right)k$ is equal to $A)a$

Note: Do remember dot product of two perpendicular vectors results to zero. Now $i,j,k$ are unit vectors in x, y and z direction which means that all the three vectors are perpendicular. Since the dot product of the two vectors are linked with the $\cos \alpha $ where $\alpha $ is the angle between the two vectors as the formulas says $\vec{a}.\vec{b}=\left| {\vec{a}} \right|\left| {\vec{b}} \right|\cos \alpha $ . In the case of the perpendicular vector $\alpha ={{90}^{\circ }}$ which means the value of $\cos \alpha $ will be $0$. So the dot product of $j.k,i.j,i.k$ etc is zero, while that of $i.i,j.j,k.k$ are $1$.