Answer
Verified
496.2k+ views
Hint – In this question we have been given a vector and we have to find a vector which is parallel to the given vector. Parallel means that it must be in the same direction as that of the given vector, so use the concept that a unit vector is always in direction to the vector and is of a magnitude 1. So multiplying it with a scalar quantity alters its length.
Complete step-by-step answer:
Given vector
$5\hat i - \hat j + 2\hat k$.
Let
$\vec a = 5\hat i - \hat j + 2\hat k$
Now we have to find out the vector which is parallel to the given vector and has magnitude 8 units.
As we know when we divide the vector with its modulus then the vector converts into unit vector and when we multiply by 8 with this unit vector the vector converts into the vector which has magnitude 8 units.
Let the parallel vector to the given vector be $\vec x$
$ \Rightarrow \vec x = 8\left( {\dfrac{{\vec a}}{{\left| {\vec a} \right|}}} \right)$……………. (1)
So, first find out the modulus of vector a.
As we know $\vec p = x\hat i + y\hat j + z\hat k$
Then $\left| {\vec p} \right| = \sqrt {{x^2} + {y^2} + {z^2}} $ so, use this property in above equation we have,
$ \Rightarrow \left| {\vec a} \right| = \sqrt {{5^2} + {{\left( { - 1} \right)}^2} + {2^2}} = \sqrt {25 + 1 + 4} = \sqrt {30} $
Now from equation (1) we have,
$ \Rightarrow \vec x = 8\left( {\dfrac{{\vec a}}{{\left| {\vec a} \right|}}} \right) = 8\left( {\dfrac{{5\hat i - \hat j + 2\hat k}}{{\sqrt {30} }}} \right) = \dfrac{{40\hat i - 8\hat j + 16\hat k}}{{\sqrt {30} }}$
$ \Rightarrow \vec x = \dfrac{{40}}{{\sqrt {30} }}\hat i - \dfrac{8}{{\sqrt {30} }}\hat j + \dfrac{{16}}{{\sqrt {30} }}\hat k$
So, this is the required vector which is parallel to the given vector and has magnitude 8 units.
Note – Whenever we face such type of problems the key concept is simply to take out the unit vector of the given vector as it will be in direction to the given vector then multiply it with the scalar constant to obtain the desired magnitude vector.
Complete step-by-step answer:
Given vector
$5\hat i - \hat j + 2\hat k$.
Let
$\vec a = 5\hat i - \hat j + 2\hat k$
Now we have to find out the vector which is parallel to the given vector and has magnitude 8 units.
As we know when we divide the vector with its modulus then the vector converts into unit vector and when we multiply by 8 with this unit vector the vector converts into the vector which has magnitude 8 units.
Let the parallel vector to the given vector be $\vec x$
$ \Rightarrow \vec x = 8\left( {\dfrac{{\vec a}}{{\left| {\vec a} \right|}}} \right)$……………. (1)
So, first find out the modulus of vector a.
As we know $\vec p = x\hat i + y\hat j + z\hat k$
Then $\left| {\vec p} \right| = \sqrt {{x^2} + {y^2} + {z^2}} $ so, use this property in above equation we have,
$ \Rightarrow \left| {\vec a} \right| = \sqrt {{5^2} + {{\left( { - 1} \right)}^2} + {2^2}} = \sqrt {25 + 1 + 4} = \sqrt {30} $
Now from equation (1) we have,
$ \Rightarrow \vec x = 8\left( {\dfrac{{\vec a}}{{\left| {\vec a} \right|}}} \right) = 8\left( {\dfrac{{5\hat i - \hat j + 2\hat k}}{{\sqrt {30} }}} \right) = \dfrac{{40\hat i - 8\hat j + 16\hat k}}{{\sqrt {30} }}$
$ \Rightarrow \vec x = \dfrac{{40}}{{\sqrt {30} }}\hat i - \dfrac{8}{{\sqrt {30} }}\hat j + \dfrac{{16}}{{\sqrt {30} }}\hat k$
So, this is the required vector which is parallel to the given vector and has magnitude 8 units.
Note – Whenever we face such type of problems the key concept is simply to take out the unit vector of the given vector as it will be in direction to the given vector then multiply it with the scalar constant to obtain the desired magnitude vector.
Recently Updated Pages
A particle is undergoing a horizontal circle of radius class 11 physics CBSE
A particle is thrown vertically upwards with a velocity class 11 physics CBSE
A particle is rotated in a vertical circle by connecting class 11 physics CBSE
A particle is projected with a velocity v such that class 11 physics CBSE
A particle is projected with a velocity u making an class 11 physics CBSE
A particle is projected vertically upwards and it reaches class 11 physics CBSE
Trending doubts
Fill the blanks with the suitable prepositions 1 The class 9 english CBSE
Which are the Top 10 Largest Countries of the World?
How do you graph the function fx 4x class 9 maths CBSE
Who was the leader of the Bolshevik Party A Leon Trotsky class 9 social science CBSE
The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths
Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE
Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE
Which is the largest saltwater lake in India A Chilika class 8 social science CBSE
Ghatikas during the period of Satavahanas were aHospitals class 6 social science CBSE