Question

The projection of \[\overrightarrow a = 3\widehat i - \widehat j + 5\widehat k\]on \[\overrightarrow b = 2\widehat i + 3\widehat j + \widehat k\]is
(A) $\dfrac{8}{{\sqrt {135} }}$
(B)$\dfrac{8}{{\sqrt {139} }}$
(C) $\dfrac{8}{{\sqrt {14} }}$
(D) $\sqrt {14} $

Answer 1

Hint :- If the vector a is projected on vector b then,
The scalar projection formula
\[\Pr oj(\overrightarrow a on\overrightarrow b ) = \dfrac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow b } \right|}}\]

Complete step by step solution
Given
\[\overrightarrow a = 3\widehat i - \widehat j + 5\widehat k\]
\[\overrightarrow b = 2\widehat i + 3\widehat j + \widehat k\]
\[\overrightarrow a = {a_1}\widehat i + {b_1}\widehat j + {c_1}\widehat k\]
\[\overrightarrow b = {a_2}\widehat i + {b_2}\widehat j + {c_2}\widehat k\]
We know that,
\[\overrightarrow a .\overrightarrow b = {a_1}{a_2} + {b_1}{b_2} + {c_1}{c_2}\]
\[\begin{gathered}
  \left| {\overrightarrow b } \right| = \sqrt {{{\left( {{a_2}} \right)}^2} + {{\left( {{b_2}} \right)}^2} + {{\left( {{c_2}} \right)}^2}} \\
  \overrightarrow a .\overrightarrow b and\left| {\overrightarrow b } \right| \\
\end{gathered} \]
\[\Pr oj(\overrightarrow a on\overrightarrow b ) = \dfrac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow b } \right|}}\]
\[\Pr oj(\overrightarrow a on\overrightarrow b ) = \dfrac{{(3\widehat i - \widehat j + 5\widehat k).(2\widehat i + 3\widehat j + \widehat k)}}{{\left| {2\widehat i + 3\widehat j + \widehat k} \right|}}\]

$ = \dfrac{{(3)(2) - (1)(3) + (5)(1)}}{{\sqrt {{{(2)}^2} + {{(3)}^2} + {{(1)}^2}} }}$
By Formula of
\[\overrightarrow a .\overrightarrow b and\left| {\overrightarrow b } \right|\]
$ = \dfrac{{6 - 3 + 5}}{{\sqrt {14} }}$
$ = \dfrac{8}{{\sqrt {14} }}$
So option (C) $\dfrac{8}{{\sqrt {14} }}$is the right answer.


Note –If we have to find Projection $\overrightarrow b $ on \[\overrightarrow a \]
Then
\[\Pr oj(\overrightarrow b on\overrightarrow a ) = \dfrac{{\overrightarrow a .\overrightarrow b }}{{\left| {\overrightarrow a } \right|}}\]