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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.