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How do you find $\left\| v \right\|$ given $v = i - j$?

Answer
VerifiedVerified
551.1k+ views
Hint: Here we will find the value of $\left\| v \right\|$ by using normal dot product property. It can be easily solved and we get solutions very early.

Complete step by step answer:
In this section, we will now concentrate on the vector operation called the dot product. The dot product of two vectors will produce a scalar instead of a vector.
The dot product is equal to the sum of the product of the horizontal components and the product of the vertical components.
If $v = {a_1}i + {b_1}j$ and $w = {a_2}i + {b_2}j$ are vectors that their dot product is given by,
$v \times w = {a_1}{a_2} + {b_1}{b_2}$
Here we are going to find $\left\| v \right\|$ that is magnitude of vector $v$,
Definition: Magnitude of a vector:
The magnitude of a vector is the length of the vector. The magnitude of the vector $v$ is denoted as $\left\| v \right\|$.
For a two dimensional vector $v = \left( {{a_1},{a_2}} \right)$ the formula for its magnitude is ${\left\| v \right\|^2} = \left( {{a_1}i + {a_2}j} \right).\left( {{a_1}i + {a_2}j} \right) = {a_1}^2 + {a_2}^2$
Here we gave the $v = i - j$
Now we are going to find ${\left\| v \right\|^2} = v \times v$
$ \Rightarrow {\left\| v \right\|^2} = \left( {i - j} \right) \times \left( {i - j} \right)$
Using the dot product in above, we get,
$ \Rightarrow {\left\| v \right\|^2} = \left( {{1^2} + {{( - 1)}^2}} \right)$
Squaring the term we get,
$ \Rightarrow {\left\| v \right\|^2} = \left( {1 + 1} \right)$
Adding the term we get,
$ \Rightarrow {\left\| v \right\|^2} = 2$
Our targeting term is $\left\| v \right\|$ so taking square root on both sides we get,
$ \Rightarrow \left\| v \right\| = \sqrt 2 $

Therefore, the magnitude of the given vector $v$ is $\left\| v \right\| = \sqrt 2 $

Note: A Euclidean vector represents the position of a point in $P$ in a Euclidean space. Geometrically it can be described as an arrow from the origin of the space to that point. The Euclidean norm of a vector is just a special case of Euclidean distance. The distance between its tail and its tip. Two similar notations are used for the Euclidean norm of a vector $x$ .
$\left\| x \right\|$
$\left| x \right|$
A disadvantage of the second notation is that it can also be used to denote the absolute value of scalars and the determinants of matrices, which introduces an element of ambiguity.