Question

Write the magnitude of the vector \[\bar{a}\] in terms of dot product.

Answer 1

Hint: In this question, we need to consider the vector \[\bar{a}\]and then as we need to find the dot product to find the magnitude. Then again consider the same vector \[\bar{a}\]such that the angle between them is \[{{0}^{\circ }}\]. Now, using the dot product formula which is given by \[\bar{a}\cdot \bar{b}=\left| {\bar{a}} \right|\left| {\bar{b}} \right|\cos \theta \]on substituting the respective vectors and angle then simplify further gives the magnitude of the given vector \[\bar{a}\].

Complete step by step solution:
A vector has both magnitude and direction
Magnitude: The length of the vector AB is called the magnitude of the vector AB which is represented as \[\left| AB \right|\].
DOT PRODUCT of TWO VECTORS:
If a and b are two non-zero vectors, then the scalar product or dot product of a and b is denoted by \[a\cdot b\]and is defined as \[\bar{a}\cdot \bar{b}=\left| {\bar{a}} \right|\left| {\bar{b}} \right|\cos \theta \], where \[\theta \]is the angle between the two vectors a and b and \[0\le \theta \le \pi \]

Angle between the two like vectors is 0.
Now, from the given question we need to find the magnitude of the vector \[\bar{a}\] in terms of dot product.
Let us now consider the two vectors \[\bar{a}\], \[\bar{a}\] and the angle between them is 0 as they are like.
As we already know from the dot product formula that
\[\bar{a}\cdot \bar{b}=\left| {\bar{a}} \right|\left| {\bar{b}} \right|\cos \theta \]
Let us now substitute the respective vectors and the corresponding angle in the above formula
\[\Rightarrow \bar{a}\cdot \bar{a}=\left| {\bar{a}} \right|\left| {\bar{a}} \right|\cos {{0}^{\circ }}\]
Now, this can be further written in the simplified form as
\[\Rightarrow \bar{a}\cdot \bar{a}={{\left| {\bar{a}} \right|}^{2}}\text{ }\left[ \because \cos {{0}^{\circ }}=1 \right]\]
Now, let us apply the square root on both the sides
\[\Rightarrow \sqrt{\bar{a}\cdot \bar{a}}=\sqrt{{{\left| {\bar{a}} \right|}^{2}}}\text{ }\]
Now, on further simplification we get,
\[\therefore \left| {\bar{a}} \right|=\sqrt{\bar{a}\cdot \bar{a}}\]
Note: It is important to note that we need to consider the like vector so that we know the angle between them and there will be no unknown magnitude. Then on further simplification accordingly we get the required result.
It is also to be noted that if we consider any other vector like unit vector or something though we know the magnitude we do not know the angle between the given vector and the vector we considered. So, we cannot simplify it further.