Three forces of magnitudes 6N, 6N and $\sqrt {72} $ N at a corner of a cube along three sides as shown in figure. Resultant of these forces is

The resultant of 5 N along OC and 5 N along OA is , $R = \sqrt {{6^2} + {6^2}} = \sqrt {72} \;N\,along\,OB$ , The resultant of $\sqrt {72} N$ along OB and $\sqrt {72} $N along OG is $R' = \sqrt {72 + 72} = 12N$ along OE.

The correct Option is "12 N along OE"

If the resultant of two forces $\left( {A + B} \right)\,and\,\left( {A - D} \right)\,is\,\sqrt {{A^2} + {B^2}} $ , then the angle between these forces is

Here, $P = \left( {A + B} \right),\;\;Q = \left( {A - B} \right),\,R = \sqrt {{A^2} + {B^2}} ;$ , $\cos \;{\rm{\theta }} = \frac{{{R^2} - {P^2} - {Q^2}}}{{2PQ}}$ , $ = \frac{{\left( {{A^2} + {B^2}} \right) - {{\left( {A + B} \right)}^2} - {{\left( {A - B} \right)}^2}}}{{2\left( {A + B} \right)\left( {A - B} \right)}} = - \left[ {\frac{{{A^2} + {B^2}}}{{\begin{array}{*{20}{c}} {2\left( {{A^2} - {B^2}} \right)}\\ \; \end{array}}}} \right]$ , $\therefore \;\;\;{\rm{\theta }} = {\cos ^{ - 1}}\,\left( { - \frac{{{A^2} + {B^2}}}{{\begin{array}{*{20}{c}} {2\left( {{A^2} - {B^2}} \right)}\\ \; \end{array}}}} \right)$

${\cos ^{ - 1}}\left[ { - \frac{{\left( {{A^2} - {B^2}} \right)}}{{{A^2} + {B^2}}}} \right]$

${\cos ^{ - 1}}\left[ { - \frac{{\left( {{A^2} + {B^2}} \right)}}{{({A^2} - {B^2})}}} \right]$

The correct Option is "${\cos ^{ - 1}}\left[ { - \frac{{{A^2} + {B^2}}}{{2\left( {{A^2} - {B^2}} \right)}}} \right]$"

${\cos ^{ - 1}}\left[ { - \frac{{2\left( {{A^2} + {B^2}} \right)}}{{{A^2} - {B^2}}}} \right]$

Given $\theta $ is the angle between ${\rm{\vec A}}\,and\,{\rm{\vec B}}$. Then $\left| {\hat A \times \hat B\;} \right|$ is equal to

$\left| {\hat A \times \hat B} \right| = \left( 1 \right)\left( 1 \right)\,\sin \theta = \sin \theta $

The correct Option is "$\sin \theta $"

${\mathop{\rm Tan}\nolimits} \theta $

If ${{\rm{\vec A}}_1}\,and\,{{\rm{\vec A}}_2}$ are two non-collinear unit vectors and if $\left| {{{{\rm{\vec A}}}_1} + {{{\rm{\vec A}}}_2}} \right| = \sqrt 3 $, then the value of $\left( {{{{\rm{\vec A}}}_1} - {{{\rm{\vec A}}}_2}} \right).\left( {2{{{\rm{\vec A}}}_1} + {{{\rm{\vec A}}}_2}} \right)$

Here, ${A_1} = {A_2} = 1$ and $A_1^2 + A_2^2 + 2{A_1}{A_2}{\rm{cos\theta }} = {\left( {\sqrt 3 } \right)^2} = 3$ or $1 + 1 + 2 \times 1 \times 1 \times \cos \theta = 3\,or\,\cos \theta = \frac{1}{2}$ Now, $\left( {{{{\rm{\vec A}}}_1} - {{{\rm{\vec A}}}_2}} \right).\left( {2{{{\rm{\vec A}}}_1} + {{{\rm{\vec A}}}_2}} \right)$ , $ = 2A_1^2 - A_2^2 - {A_1}{A_2}\;{\rm{cos\theta }}$ , $ = 2 \times {1^2} - {1^2} - 1 \times 1 \times \frac{1}{2} = \frac{1}{2}$

The correct Option is "$\frac{1}{2}$"

The angle between ${\rm{\vec A}}\;\,and\,{\rm{\vec B}}\,is\,\theta $, the value of the triple product ${\rm{\vec A}}{\rm{.\vec B}} \times {\rm{\vec A}}$ is

If any two vectors are parallel or equal, then the scalar triple product is zero.

A body starts from rest from the origin with an acceleration of $6\;m/{s^2}$ along the x-axis and $8\;m/{s^2}$ along the y-axis. Its distance from the origin after 4 seconds will be

${S_x} = {u_x}t + \frac{1}{2}{a_x}{t^2} \Rightarrow {S_x} = \frac{1}{2} \times 6 \times 16 = 48\;m$ , ${S_y} = {u_y}t + \frac{1}{2}{a_y}{t^2} \Rightarrow {S_y} = \frac{1}{2} \times 8 \times 16 = 64\;m$ , $S = \sqrt {S_x^2 + S_y^2} = 80\;m$

The angle between the z-axis and the vector $\hat i + \hat j + \sqrt 2 \hat k$ is

${\rm{cos\theta }} = \frac{{\left( {\hat k} \right).\left( {\hat i + \hat j + \sqrt 2 \hat k} \right)}}{{1\sqrt {{1^2} + {1^2}{{\left( {\sqrt 2 } \right)}^2}} }}$ or ${\rm{cos\theta }} = \frac{{\sqrt 2 }}{2} = \frac{1}{{\sqrt 2 }}\,or\,\theta = {45^0}$

The correct Option is "${45^0}$"

If the magnitude of the sum of the two vectors is equal to the difference of their magnitudes, then the angle between vectors is

$\sqrt {{P^2} + {Q^2} + 2PQ{\rm{cos\theta }}} = \left( {P - Q} \right)$ , $ \Rightarrow \;\;{P^2} + {Q^2} + 2PQ{\rm{cos\theta }} = {P^2} + {Q^2} - 2PQ$ , $ \Rightarrow \;2PQ\left( {1 + \cos \theta } \right) = 0$ but $2PQ \ne 0$ ,$\therefore \;1 + {\rm{cos\theta }} = 0\,or\,\cos \theta = - 1\,or\,\theta = {180^0}$

The correct Option is "${180^0}$"

Two forces $\overrightarrow {{{\rm{F}}_1}} \;\,and\,\overrightarrow {{{\rm{F}}_2}} $ are acting at right angles to each other. Then their resultant is

${F^2} = F_1^2 + F_2^2 + 2{F_1}{F_2}{\rm{cos}}90^\circ $ or ${F^2} = F_1^2 + F_2^2 \Rightarrow F = \sqrt {F_1^2 + F_2^2} $

The correct Option is "$\sqrt {F_1^2 + F_2^2} $"

The vector which can give unit vector along x-axis with ${\rm{\vec A}} = 2\hat i - 4\hat j + 7\hat k,\,{\rm{\;\vec B}} = 7\hat i + 2\hat j - 5\hat k$ and ${\rm{\vec C}} = - 4\hat i + 7\hat j + 3\hat k$ is

The vector is $\hat i - \left[ {{\rm{\vec A}} + {\rm{\vec B}} + {\rm{\vec C}}} \right] = \hat i - [(2\hat i - 4\hat j + 7\hat k) + {\rm{\;}}\left( {7\hat i + 2\hat j - 5\hat k} \right) + \left( { - 4\hat i + 7\hat j + 3\hat k} \right)$] , $ = - 4\hat i - 5\hat j - 5\hat k$