Question

If $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are those mutually perpendicular vectors, then the projection of the vector $(l\frac{\overline{a}}{|\overline{a}|}+m\frac{\overline{b}}{|\overline{b}|}+n\frac{(\overline{a}\times \overline{b})}{|\overline{a}\times \overline{b}|})$ along bisector of vectors $\overrightarrow{a}$ and $\overrightarrow{a}$ may be given as ?

• A. $\frac{l^2+m^2}{\sqrt{l^2+m^2+n^2}}$
• B. $\sqrt{l^2+m^2+n^2}$
• C. $\frac{\sqrt{l^2+m^2}}{\sqrt{l^2+m^2+n^2}}$
• D. $\frac{l+m}{\sqrt{2}}$

Answer 1

The correct option is D $\frac{l+m}{\sqrt{2}}$

Question :- If $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c}$ are three mutually perpendicular

vectors, then projection of vector $(l\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}+m\frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}+n\frac{(\overrightarrow{a}\times \overrightarrow{b})}{|\overrightarrow{a}\times \overrightarrow{b}|})$

along bisector of sectors $\overrightarrow{a}$ and $\overrightarrow{b}$ may be given as?

Solution:-

We know for mutually perp. vectors.

$\overrightarrow{a}\times \overrightarrow{b}=\overrightarrow{c}$ $\overrightarrow{b}\times \overrightarrow{c}=\overrightarrow{a}$ $\overrightarrow{c}\times \overrightarrow{a}=\overrightarrow{b}$

$\overrightarrow{a}.\overrightarrow{b}=\overrightarrow{b}.\overrightarrow{c}=\overrightarrow{c}.\overrightarrow{a}=0$ and ${\left|\hat{a}\right|}^2=1$

Now, A vector parallel to bisector of angle between

vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ is given as $\overrightarrow{x}$

$\overrightarrow{x}=\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}+\frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}=\hat{a}+\hat{b}$ $(\hat{a}=\hat{b}$ = unit vectors)

say $\overrightarrow{y}=(l\frac{\overrightarrow{a}}{\left|\overrightarrow{a}\right|}+m\frac{\overrightarrow{b}}{\left|\overrightarrow{b}\right|}+n\frac{(\overrightarrow{a}\times \overrightarrow{b})}{|\overrightarrow{a}\times \overrightarrow{b}|})$

$\overrightarrow{y}=(l\hat{a}+m\hat{b}+n\frac{\overrightarrow{c}}{\left|\overrightarrow{c}\right|})$

$\overrightarrow{y}=(l\hat{a}+m\hat{b}+n\hat{c})$

Now projection of y along x is given as

$=\overrightarrow{y}.\frac{\overrightarrow{x}}{\left|\overrightarrow{x}\right|}$

$=\overrightarrow{y}.\hat{x}$

$=(l\hat{a}+m\hat{b}+n\hat{c}).(\frac{\hat{a}}{\sqrt{2}}+\frac{\hat{b}}{\sqrt{2}})$

$=\frac{l}{\sqrt{2}}+\frac{m}{\sqrt{2}}$

$=\frac{l+m}{\sqrt{2}}$

Ans (D) $\frac{l+m}{\sqrt{2}}$

Using $\overrightarrow{x}=\hat{a}+\hat{b}$

$\hat{x}=\frac{\widehat{a+\hat{b}}}{|\hat{a}+\hat{b}|}=\frac{1}{\sqrt{2}}(\hat{a}+\hat{b})$

${|\hat{a}+\hat{b}|}^2={\left|\hat{a}\right|}^2+{\left|\hat{b}\right|}^2+2\hat{a}.\hat{b}$

$=1+1+0=2$