Question

If ${{\vec a , \vec b}}$ and ${{\vec c}}$ are those mutually perpendicular vectors, then the projection of the vector $\left( {{{l }}\dfrac{{{{\bar a}}}}{{{{|a|}}}}{{ + m }}\dfrac{{{{\bar b}}}}{{{{|b|}}}}{{ + n }}\dfrac{{\left( {{{\bar a \times \bar b}}} \right)}}{{{{|a \times b|}}}}} \right)$ along the bisector of vectors ${{\vec a}}$ and ${{\vec a}}$ may be given as?
a) $\dfrac{{{{{l}}^{{2}}}{{ + }}{{{m}}^{{2}}}}}{{\sqrt {{{{l}}^{{2}}}{{ + }}{{{m}}^{{2}}}{{ + }}{{{n}}^{{2}}}} }}$
b) $\sqrt {{{{l}}^{{2}}}{{ + }}{{{m}}^{{2}}}{{ + }}{{{n}}^{{2}}}}$
c) $\dfrac{{\sqrt {{{{l}}^{{2}}}{{ + }}{{{m}}^{{2}}}} }}{{\sqrt {{{{l}}^{{2}}}{{ + }}{{{m}}^{{2}}}{{ + }}{{{n}}^{{2}}}} }}$
d) $\dfrac{{{{l + m}}}}{{\sqrt {{2}} }}$

Answer 1

In this question we have to find the projection of the given vector along the given bisector of vectors ${{\vec a}}$. For that we are going to solve using mutually perpendicular vector methods.
In this type of problem you need to solve the equation when in the question mentioned that ${\vec a , \vec b}$ ${\text{and}}$ ${\vec c}$ are those mutually perpendicular vectors respectively. By using mutually perpendicular vector, we know that ${{\vec a \times \vec b = \vec c}}$, ${{\vec b \times \vec c = \vec a}}$, ${{\vec c \times \vec a = \vec b}}$, ${{\vec a }}{{. \vec b = \vec b }}{{. \vec c = \vec c }}{{. \vec a = 0}}$ and ${{| \hat a }}{{{|}}^{{2}}} = 1$. Using mutually perpendicular we solve the equation.
The first method to find the values for a given expression, we need to do the values of the given expression say as ${{y}}$.
Next substitute the values of mutually perpendicular vectors.

Formula used: Using algebraic formula,
${{| \hat a + \hat b }}{{{|}}^{{2}}}{{ = | \hat a }}{{{|}}^{{2}}}{{ + | \hat b }}{{{|}}^{{2}}} + {{2 \hat a }}{{. \hat b}}$
${{\hat a = \hat b = {\text{unit vector}}}}$

Complete step-by-step solution:
We know for mutually perpendicular vectors,
${{\vec a \times \vec b = \vec c}}$
${{\vec b \times \vec c = \vec a}}$
${{\vec c \times \vec a = \vec b}}$
${{\vec a }}{{. \vec b = \vec b }}{{. \vec c = \vec c }}{{. \vec a = 0}}$
${{| \hat a }}{{{|}}^{{2}}} = 1$
Now, A vector parallel to bisector of angle between vectors ${{\vec a}}$ and ${{\vec b}}$ is given as ${{\vec x}}$.
${{\vec x = }}\dfrac{{{{\vec a}}}}{{{{|\vec a|}}}}{{ + }}\dfrac{{{{\vec b}}}}{{{{|\vec b|}}}}$
By adding the terms, we get
${{ = \hat a + \hat b}}$ ${{(}}\because {{\hat a = \hat b = }}$ unit vector)
Consider the given expression as,
Say ${{y = }}\left( {{{l }}\dfrac{{{{\bar a}}}}{{{{|a|}}}}{{ + m }}\dfrac{{{{\bar b}}}}{{{{|b|}}}}{{ + n }}\dfrac{{\left( {{{\bar a \times \bar b}}} \right)}}{{{{|a \times b|}}}}} \right)$
By substituting the known values by mutually perpendicular vector we get,
${{\vec y = }}\left( {{{l \hat a + m \hat b + n }}\dfrac{{{{\vec c}}}}{{{{|\vec c|}}}}} \right)$
Substituting the terms, we have
${{\vec y = }}\left( {{{l \hat a + m \hat b + n \hat c}}} \right)$
Now projection of ${\text{y along x}}$ is given as
$\Rightarrow {{ \vec y }}{{. }}\dfrac{{{{\vec x}}}}{{{{|\vec x|}}}}$
Here substituting the terms,
$\Rightarrow {{\vec y }}{{. \hat x}}$
Using ${{\vec x = \hat a + \hat b}}$
${{\hat x = }}\dfrac{{{{\hat a + \hat b}}}}{{{{|\hat a + \hat b|}}}}$
$(\because {{| \hat a + \hat b }}{{{|}}^{{2}}}{{ = | \hat a }}{{{|}}^{{2}}}{{ + | \hat b }}{{{|}}^{{2}}} + {{2 \hat a }}{{. \hat b = 1 + 1 + 0 = 2 )}}$
${{\hat x = }}\dfrac{{{1}}}{{\sqrt {{a}} }}\left( {{{\hat a + \hat b}}} \right)$
By Substituting the value of ${{\vec x}}$ we get,
$\Rightarrow \left( {{{l \hat a + m \hat b + n \hat c}}} \right){{.}}\left( {\dfrac{{{{\hat a}}}}{{\sqrt {{2}} }}{{ + }}\dfrac{{{{\hat b}}}}{{\sqrt {{2}} }}} \right)$
Multiplying the terms, we have
$\Rightarrow \dfrac{{{l}}}{{\sqrt {{2}} }}{{ + }}\dfrac{{{m}}}{{\sqrt {{2}} }}$
Here we are adding the terms,
$= \dfrac{{{{l + m}}}}{{\sqrt {{2}} }}$
$\left( {{{l }}\dfrac{{{{\bar a}}}}{{{{|a|}}}}{{ + m }}\dfrac{{{{\bar b}}}}{{{{|b|}}}}{{ + n }}\dfrac{{\left( {{{\bar a \times \bar b}}} \right)}}{{{{|a \times b|}}}}} \right)$ is $\dfrac{{{{l + m}}}}{{\sqrt {{2}} }}$
Hence the Projection of vector $\left( {{{l }}\dfrac{{{{\bar a}}}}{{{{|a|}}}}{{ + m }}\dfrac{{{{\bar b}}}}{{{{|b|}}}}{{ + n }}\dfrac{{\left( {{{\bar a \times \bar b}}} \right)}}{{{{|a \times b|}}}}} \right)$ is $\dfrac{{{{l + m}}}}{{\sqrt {{2}} }}$

Option D is the correct answer.

Note: A vector space is a collection of objects called vectors, which may be added together and multiplies by numbers, called scalars. Scalars are often taken to be real numbers, but there are also vector spaces with scalar multiplication by complex numbers, rational numbers, or generally any field. To specify that the scalars are real or complex numbers, the terms real vector space and complex vector space are often used.