Question

If $\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b = \widehat {4i} - \widehat {2j} + \widehat {3k},\overrightarrow c = \widehat i - \widehat {2j} + \widehat k$ . Find a vector of magnitude 6 units which is parallel to the vector $2\overrightarrow a - \overrightarrow b - 3\overrightarrow c $.

Answer 1

Hint: We are given three vectors and let $\overrightarrow r = 2\overrightarrow a - \overrightarrow b - 3\overrightarrow c $and substituting the given vectors we get the new vector r and in order to find the unit vector we use the formula $\widehat r = \dfrac{{\overrightarrow r }}{{\left| {\overrightarrow r } \right|}}$and the magnitude formula is given as $\sqrt {{{\left( {{\text{coefficient of i}}} \right)}^2} + {{\left( {{\text{coefficient of j}}} \right)}^2} + {{\left( {{\text{coefficient of k}}} \right)}^2}} $and after obtaining the unit vector the required vector is obtained by multiplying 6 with the unit vector.

Complete step by step solution:
Let $\overrightarrow r = 2\overrightarrow a - \overrightarrow b - 3\overrightarrow c $
We are given that $\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b = \widehat {4i} - \widehat {2j} + \widehat {3k},\overrightarrow c = \widehat i - \widehat {2j} + \widehat k$
Substituting this in we get
 $
   \Rightarrow \overrightarrow r = 2\left( {\widehat i + \widehat j + \widehat k} \right) - \left( {\widehat {4i} - \widehat {2j} + \widehat {3k}} \right) - 3\left( {\widehat i - \widehat {2j} + \widehat k} \right) \\
   \Rightarrow \overrightarrow r = 2\widehat i + \widehat {2j} + 2\widehat k - \widehat {4i} + \widehat {2j} - \widehat {3k} - 3\widehat i + 6\widehat j - 3\widehat k \\
   \Rightarrow \overrightarrow r = - 5\widehat i + 10\widehat j - \widehat {4k} \\
 $
Now the unit vector is given by the formula
$ \Rightarrow \widehat r = \dfrac{{\overrightarrow r }}{{\left| {\overrightarrow r } \right|}}$ ……….(1)
Where $\overrightarrow r $is the given vector and $\left| {\overrightarrow r } \right|$is the magnitude of $\overrightarrow r $
The magnitude of a vector is given by $\sqrt {{{\left( {{\text{coefficient of i}}} \right)}^2} + {{\left( {{\text{coefficient of j}}} \right)}^2} + {{\left( {{\text{coefficient of k}}} \right)}^2}} $
Therefore the modulus of is given as
\[
   \Rightarrow \left| {\overrightarrow r } \right| = \sqrt {{{\left( { - 5} \right)}^2} + {{\left( {10} \right)}^2} + {{\left( { - 4} \right)}^2}} \\
   \Rightarrow \left| {\overrightarrow r } \right| = \sqrt {25 + 100 + 16} = \sqrt {141} \\
   \Rightarrow \left| {\overrightarrow r } \right| = \sqrt {141} \\
 \]
Using this in (1) we get
$ \Rightarrow \widehat r = \dfrac{{ - 5\widehat i + 10\widehat j - \widehat {4k}}}{{\sqrt {141} }}$
Since we are given the magnitude of the required vector is 6
The vector parallel to r with magnitude 6 is given by $6\times \widehat r$
$
   \Rightarrow 6\times \dfrac{{ - 5\widehat i + 10\widehat j - \widehat {4k}}}{{\sqrt {141} }} \\
   \Rightarrow \dfrac{{ - 30\widehat i + 60\widehat j - 24\widehat k}}{{\sqrt {141} }} \\
 $

Therefore the required vector is $\dfrac{{ - 30\widehat i + 60\widehat j - 24\widehat k}}{{\sqrt {141} }}$.

Note :
Vectors are parallel if they have the same direction. Both components of one vector must be in the same ratio to the corresponding components of the parallel vector.
A unit vector is a vector of length 1, sometimes also called a direction vector.