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