Question

Let we have a vector $\vec a = i - 2j + 3k$ if $\vec b$ is a vector such that $\vec a.\vec b = {\left| {\vec b} \right|^2}$ and $\left| {\vec a - \vec b} \right| = \sqrt 7$, then $\left| {\vec b} \right| =$
$\left( a \right)$ 7
$\left( b \right)$ 14
$\left( c \right)\sqrt 7$
$\left( d \right)$ 21

Answer 1

In this particular type of question use the concept of squaring on both sides and expand the square using the property, (${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$) and also use the concept of modulus, the modulus of any vector is given as, ($\left| {ai + bj + ck} \right| = \sqrt {{a^2} + {b^2} + {c^2}}$) so use these properties to reach the solution of the question.

Complete step-by-step answer:
Given data:
$\vec a = i - 2j + 3k$.................. (1)
$\vec a.\vec b = {\left| {\vec b} \right|^2}$................... (2)
$\left| {\vec a - \vec b} \right| = \sqrt 7$............... (3)
Now squaring on both sides in equation (3) we have,
$\Rightarrow {\left| {\vec a - \vec b} \right|^2} = {\left( {\sqrt 7 } \right)^2}$
Now open the square according to property ${\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab$ we have,
$\Rightarrow {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} - 2\vec a.\vec b = 7$.................. (4)
Now first find out ${\left| {\vec a} \right|^2}$
So it is given that, $\vec a = i - 2j + 3k$ so take modulus on both sides we have,
$\Rightarrow \left| {\vec a} \right| = \left| {i - 2j + 3k} \right|$
Now as we all know that the modulus of $\left| {ai + bj + ck} \right| = \sqrt {{a^2} + {b^2} + {c^2}}$ so use this property in the above equation we have,
$\Rightarrow \left| {\vec a} \right| = \sqrt {{1^2} + {{\left( { - 2} \right)}^2} + {3^2}}$
Now simplify this we have,
$\Rightarrow \left| {\vec a} \right| = \sqrt {1 + 4 + 9} = \sqrt {14}$
Now take square on both sides we have,
$\Rightarrow {\left| {\vec a} \right|^2} = {\left( {\sqrt {14} } \right)^2} = 14$............... (5)
Now substitute the value from equation (2) and (5) in equation in equation (4) we have,
$\Rightarrow {\left| {\vec a} \right|^2} + {\left| {\vec b} \right|^2} - 2\vec a.\vec b = 7$
$\Rightarrow 14 + {\left| {\vec b} \right|^2} - 2{\left| {\vec b} \right|^2} = 7$
Now simplify this we have,
$\Rightarrow 14 - {\left| {\vec b} \right|^2} = 7$
$\Rightarrow {\left| {\vec b} \right|^2} = 14 - 7 = 7$
Now take square root on both sides we have,
$\Rightarrow \sqrt {{{\left| {\vec b} \right|}^2}} = \sqrt 7$
$\Rightarrow \left| {\vec b} \right| = \sqrt 7$
So this is the required answer.
Hence option (c) is the correct answer.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the standard identity of squaring the difference of numbers which is stated above, so first apply this property in equation (3) as above and simplify then use the property of modulus as above and again simplify we will get the required answer.