Vector Algebra

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

$ \Rightarrow \overrightarrow b .\overrightarrow c = \overrightarrow b .\left( {\overrightarrow b \times \overrightarrow a } \right) \Rightarrow \overrightarrow b .\overrightarrow c = 0$

$ \Rightarrow \left( {{b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k} \right).\left( {\widehat i - \widehat j - \widehat k} \right) = 0,$

where $\overrightarrow b = {b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k$

${b_1} - {b_2} - {b_3} = 0\,\,\,\,\,\,\,\,\,\,\,....\left( i \right)$

and $\,\,\,\,\overrightarrow a .\overrightarrow b = 3$

$ \Rightarrow \left( {\widehat j - \widehat k} \right).\left( {{b_1}\widehat i + {b_2}\widehat j + {b_3}\widehat k} \right) = 3$

$ \Rightarrow {b_2} - {b_3} = 3$

From equation $(i)$

${b_1} = {b_2} + {b_3} = \left( {3 + {b_3}} \right) + {b_3} = 3 + 2{b_3}$

$\overrightarrow b = \left( {3 + 2{b_3}} \right)\widehat i + \left( {3 + {b_3}} \right)\widehat j + {b_3}\widehat k$

From the option given, it is clear that ${b_3}$ equal to either $2$ or $-2.$

${b_3} = 2$

then $\overrightarrow b = 7\widehat i + 5\widehat j + 2\widehat k$ which is not possible

If ${b_3} = - 2,$ then $\overrightarrow b = - \widehat i + \widehat j - 2\widehat k$

$\left[ {3\overrightarrow u \,\,p\overrightarrow v \,\,p\overrightarrow \omega } \right] - \left[ {p\overrightarrow v \,\,\overrightarrow \omega \,\,q\overrightarrow u } \right] - \left[ {2\overrightarrow \omega \,\,q\overrightarrow v \,\,q\overrightarrow u } \right] = 0$

$ \Rightarrow \left( {3{p^2} - pq + 2{q^2}} \right)\left[ {\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow \omega } \right] = 0$

$ \Rightarrow 3{p^2} - pq + 2{q^2} = 0\,\,$ $\,\,\,\,\left( \, \right.$ As $\,\,\,\,\left[ {\overrightarrow u \,\,\overrightarrow v \,\,\overrightarrow \omega } \right] \ne 0$ $\left. {} \right)$

$ \Rightarrow 2{p^2} + {p^2} - pq + {{{q^2}} \over 4} + {{7{q^2}} \over 4} = 0$

$ \Rightarrow 2{p^2} + {\left( {p - {q \over 2}} \right)^2} + {7 \over 4}{q^2} = 0$

$ \Rightarrow p = 0,q = 0,p = {q \over 2}$ $ \Rightarrow p = 0,q = 0$

$\therefore$ Exactly one value of $\left( {p,q} \right)$

As $\overrightarrow a $ lies in the plane of $\overrightarrow b $ and $\overrightarrow c $

$\therefore$ $\overrightarrow a = \overrightarrow b + \lambda \overrightarrow c $

$ \Rightarrow \alpha \widehat i + 2\widehat j + \beta \widehat k = \widehat i + \widehat j + \lambda \left( {\widehat j + \widehat k} \right)$

$ \Rightarrow \alpha = 1,2 = 1 + \lambda ,\,\beta = \lambda \Rightarrow \alpha = 1,\beta = 1$

Clearly $\overrightarrow a = - {8 \over 7}\overrightarrow c $

$ \Rightarrow \overrightarrow a ||\overrightarrow c $ and are opposite in direction

$\therefore$ Angle between $\overrightarrow a $ and $\overrightarrow c $ is $\pi .$

Given $\left| {2\widehat u \times 3\widehat v} \right| = 1$

and $\theta $ is acute angle between $\widehat u$

and $\widehat v,\,\,\left| {\widehat u} \right| = 1,\,\,\left| {\widehat v} \right| = 1\,\,\,$

$ \Rightarrow \,\,\,6\left| {\widehat u} \right|\left| {\widehat v} \right|\left| {\sin \theta } \right| = 1$

$ \Rightarrow 6\left| {\sin \theta } \right| = 1 \Rightarrow \sin \theta = {1 \over 6}$

Hence, there is exactly one value of $\theta $

for which $2\widehat u \times 3\widehat v$ is a unit vector.

Given $\overrightarrow a = \widehat i + \widehat j + \widehat k,\overrightarrow b = \widehat i - \widehat j + 2\widehat k$

and $\overrightarrow c = x\widehat i + \left( {x - 2} \right)\widehat j - \widehat k$

If $\overrightarrow c $ lies in the plane of $\overrightarrow a $ and $\overrightarrow b ,$

then $\left[ {\overrightarrow a \,\overrightarrow {b\,} \overrightarrow c } \right] = 0$

i.e.$\left| {\matrix{ 1 & 1 & 1 \cr 1 & { - 1} & 2 \cr x & {\left( {x - 2} \right)} & { - 1} \cr } } \right| = 0$

$ \Rightarrow 1\left[ {1 - 2\left( {x - 2} \right)} \right] - 1\left[ { - 1 - 2x} \right]$

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, + 1\left[ {x - 2 + x} \right] = 0$

$ \Rightarrow 1 - 2x + 4 + 1 + 2x + 2x - 2 = 0$

$ \Rightarrow 2x = - 4\,\,\,\,\,\,\,\,\, \Rightarrow x = - 2$

$\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c $

$\,\,\,\,\,\,\,\,\,\,\,\,\, = \overrightarrow a \times \left( {\overrightarrow b \times \overrightarrow c } \right),\overrightarrow a .\overrightarrow b \ne 0,\,\,\overrightarrow b .\overrightarrow c \ne 0$

$ \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right).\overrightarrow b - \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow a $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {\overrightarrow a .\overrightarrow c } \right).\overrightarrow b - \left( {\overrightarrow a .\overrightarrow b } \right).\overrightarrow c $

$ \Rightarrow \left( {\overrightarrow a .\overrightarrow b } \right).\overrightarrow c = \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow a $

$ \Rightarrow \overrightarrow a ||\overrightarrow c .$

$\overrightarrow {CA} = \left( {2 - a} \right)\widehat i + 2\widehat j;$

$\overrightarrow {CB} = \left( {1 - a} \right)\widehat i - 6\widehat k$

$\overrightarrow {CA} .\overrightarrow {CB} = 0$

$\,\,\,\,\,\,\,\, \Rightarrow \left( {2 - a} \right)\left( {1 - a} \right) = 0$

$ \Rightarrow a = 2,1$

Vector $a\overrightarrow i + a\overrightarrow j + c\overrightarrow k ,\,\,\overrightarrow i + \overrightarrow k $

and $c\overrightarrow i + c\overrightarrow j + b\overrightarrow k $ are coplanar

$\left| {\matrix{ a & a & c \cr 1 & 0 & 1 \cr c & c & b \cr } } \right| = 0 \Rightarrow {c^2} = ab$

$ \Rightarrow c = \sqrt {ab} $

$\therefore$ $c$ is $G.M.$ of $a$ and $b.$

$\overrightarrow a = \widehat j - \widehat k,\overrightarrow b = x\widehat i + \overrightarrow j + \left( {1 - x} \right)\widehat k$

and $\overrightarrow c = y\widehat i + x\widehat j + \left( {1 + x - y} \right)\widehat k$

$\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] = \overrightarrow a .\overrightarrow b \times \overrightarrow c = \left| {\matrix{ 1 & 0 & { - 1} \cr x & 1 & {1 - x} \cr y & x & {1 + x - y} \cr } } \right|$

$ = 1\left[ {1 + x - y - x + {x^2}} \right] - \left[ { - {x^2} - y} \right]$

$ = 1 - y + {x^2} - {x^2} + y = 1$

Hence $\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$ is independent of $x$ and $y$ both.

$\left[ {\lambda \left( {\overrightarrow a + \overrightarrow b } \right){\lambda ^2}\overrightarrow b \,\,\,\lambda \overrightarrow c } \right] = \left[ {\overrightarrow a \,\,\overrightarrow b + \overrightarrow c \,\,\overrightarrow b } \right]$

$ \Rightarrow {\lambda ^4}\left[ {\overrightarrow a + \overrightarrow b \,\,\overrightarrow b \overrightarrow c } \right] = \left[ {\overrightarrow a \,\,\overrightarrow b + \overrightarrow c \,\,\overrightarrow b } \right]$

$ \Rightarrow {\lambda ^4}\left\{ {\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] + \left[ {\overrightarrow b \,\overrightarrow b \,\overrightarrow c } \right]} \right\} = \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow b } \right] + \left[ {\overrightarrow a \,\overrightarrow c \,\overrightarrow b } \right]$

$ \Rightarrow {\lambda ^4}\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right] = - \left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]$

$ \Rightarrow {\lambda ^4} = - 1$

$ \Rightarrow \lambda \,\,$ has no real values.

$\overrightarrow {PA} + \overrightarrow {AP = 0} $ and $\overrightarrow {PC} + \overrightarrow {CP} = 0$

$ \Rightarrow \overrightarrow {PA} + \overrightarrow {AC} + \overrightarrow {CP} = 0$

and

$\overrightarrow {PB} + \overrightarrow {BC} + \overrightarrow {CP} = 0$

Adding, we get

$\overrightarrow {PA} + \overrightarrow {PB} + \overrightarrow {AC} + \overrightarrow {BC} + 2\overrightarrow {CP} = 0.$

Since

$\overrightarrow {AC} = - \overrightarrow {BC} $ $\,\,\,\,\,\,$ & $\,\,\,\,\,\,$ $\overrightarrow {CP} = - \overrightarrow {PC} $

$ \Rightarrow \overrightarrow {PA} + \overrightarrow {PB} - 2\overrightarrow {PC} = 0.$

Let $\overrightarrow a = x\overrightarrow i + y\overrightarrow j + z\overrightarrow k $

$\overrightarrow a \times \overrightarrow i = z\overrightarrow j - y\overrightarrow k $

$ \Rightarrow {\left( {\overrightarrow a \times \overrightarrow i } \right)^2} = {y^2} + {z^2}$

Similarly, ${\left( {\overrightarrow a \times \overrightarrow j } \right)^2} = {x^2} + {z^2}\,\,$

and ${\left( {\overrightarrow a \times \overrightarrow k } \right)^2} = {x^2} + {y^2}$

$ \Rightarrow {\left( {\overrightarrow a \times \overrightarrow i } \right)^2} + {\left( {\overrightarrow a \times \overrightarrow j } \right)^2} + {\left( {\overrightarrow a \times \overrightarrow k } \right)^2}$

$ = 2\left( {{x^2} + {y^2} + {z^2}} \right) = 2\overrightarrow a 2$

Given $\left( {\overrightarrow a \times \overrightarrow b } \right) \times \overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $

Clearly $\overrightarrow a $ and $\overrightarrow b $ are noncollinear

$ \Rightarrow \left( {\overrightarrow a .\overrightarrow c } \right)\overrightarrow b - \left( {\overrightarrow b .\overrightarrow c } \right)\overrightarrow a = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $

$\therefore$ $\overrightarrow a .\overrightarrow c = 0$

and $ - \overrightarrow b .\overrightarrow c = {1 \over 3}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right| \Rightarrow \cos \theta = {{ - 1} \over 3}$

$\therefore$ $\sin \theta = \sqrt {1 - {1 \over 9}} = {{2\sqrt 2 } \over 3}$

$\,\,\,\,\,\,\,\left[ {} \right.$ $\theta $ is acute angle between $\overrightarrow b $ and $\overrightarrow c $ $\left. {} \right]$

If $\overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}$ is collinear with $\overrightarrow{\mathbf{c}}$, then

$ \overrightarrow{\mathbf{a}}+2 \overrightarrow{\mathbf{b}}=t \overrightarrow{\mathbf{c}} $

Also, if $\overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}$ is collinear with $\overrightarrow{\mathbf{a}}$, then

$ \begin{aligned} & \overrightarrow{\mathbf{b}}+3 \overrightarrow{\mathbf{c}}=\lambda \overrightarrow{\mathbf{a}} \\\\ & \Rightarrow \overrightarrow{\mathbf{b}}=\lambda \overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{c}} \end{aligned} $

On putting this value in Eq. (i), we get

$ \overrightarrow{\mathbf{a}}+2(\lambda \overrightarrow{\mathbf{a}}-3 \overrightarrow{\mathbf{c}})=t \overrightarrow{\mathbf{c}} $

$ \Rightarrow \overrightarrow{\mathbf{a}}+2 \lambda \overrightarrow{\mathbf{a}}-6 \overrightarrow{\mathbf{c}}=t \overrightarrow{\mathbf{c}} $

$ \Rightarrow (\overrightarrow{\mathbf{a}}-6 \overrightarrow{\mathbf{c}})=t \overrightarrow{\mathbf{c}}-2 \lambda \overrightarrow{\mathbf{a}} $

On comparing, we get

and $-6=t $

$\Rightarrow t=-6$

From Eq. (i),

$\vec{a}+2 \vec{b}=-6 \vec{c} $

$\Rightarrow \vec{a}+2 \vec{b}+6 \vec{c}=\overrightarrow{0}$

The work done by a force on a particle is given by the dot product of the force and the displacement vector of the particle. The displacement vector can be found by subtracting the initial position from the final position:

$\mathbf{displacement} = \mathbf{final\ position} - \mathbf{initial\ position} = (5\widehat i + 4\widehat j + \widehat k) - (\widehat i + 2\widehat j + 3\widehat k) = 4\widehat i + 2\widehat j - 2\widehat k$

The total work done by the two forces is equal to the sum of the work done by each force. The work done by each force can be calculated as the dot product of the force and the displacement:

$\mathbf{work\ done\ by\ force\ 1} = (4\widehat i + \widehat j - 3\widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k) = 4 \cdot 4 + 1 \cdot 2 - 3 \cdot -2 = 16 + 2 + 6 = 24$

$\mathbf{work\ done\ by\ force\ 2} = (3\widehat i + \widehat j - \widehat k) \cdot (4\widehat i + 2\widehat j - 2\widehat k) = 3 \cdot 4 + 1 \cdot 2 - 1 \cdot -2 = 12 + 2 + 2 = 16$

The total work done by the forces is the sum of the work done by each force:

$\mathbf{total\ work\ done} = 24 + 16 = 40$

Therefore, the total work done by the forces is 40 J (joules) or 40 units.

Projection of $\overrightarrow v $ along $\overrightarrow u = {{\overrightarrow v .\overrightarrow u } \over {\left| {\overrightarrow u } \right|}} = {{\overrightarrow v .\overrightarrow u } \over 2}$

projection of $\overrightarrow w $ along $\overrightarrow u = {{\overrightarrow w .\overrightarrow u } \over {\left| {\overrightarrow u } \right|}} = {{\overrightarrow w .\overrightarrow u } \over 2}$

Given ${{\overrightarrow v .\overrightarrow u } \over 2} = {{\overrightarrow w .\overrightarrow u } \over 2}\,\,\,\,\,\,\,\,\,\,\,\,...\left( 1 \right)$

Also, $\overrightarrow v .\overrightarrow w = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...\left( 2 \right)$

Now ${\left| {\overrightarrow u - \overrightarrow v + \overrightarrow w } \right|^2}$

$ = {\left| {\overrightarrow u } \right|^2} + {\left| {\overrightarrow v } \right|^2} + {\left| {\overrightarrow w } \right|^2} - $

$\,\,\,\,\,\,\,\,\,\,\,\,\,\,2\overrightarrow u .\overrightarrow v - 2\overrightarrow v .\overrightarrow w + 2\overrightarrow u .\overrightarrow w $

$ = 1 + 4 + 9 + 0$ [ From $(1)$ and $(2)$ ] $=14$

$\therefore$ $\left| {\overrightarrow u - \overrightarrow v + \overrightarrow w } \right| = \sqrt {14} $

Vectors $\overrightarrow a + 2\overrightarrow b + 3\overrightarrow c ,\lambda \overrightarrow b + 4\overrightarrow c ,\,\,\,$

and $\left( {2\lambda - 1} \right)\overrightarrow c $ are

coplanar if $\left| {\matrix{ 1 & 2 & 3 \cr 0 & \lambda & 4 \cr 0 & 0 & {2\lambda - 1} \cr } } \right| = 0$

$ \Rightarrow \lambda \left( {2\lambda - 1} \right) = 0 \Rightarrow \lambda = 0$ or ${1 \over 2}$

$\therefore$ Forces are noncoplanar for all $\lambda ,$

except $\lambda = 0,{1 \over 2}$

$\overrightarrow a + \overrightarrow b + \overrightarrow c = 0$

$ \Rightarrow \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right).\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) = 0$

${\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2} + 2\left( {\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a } \right) = 0$

$\overrightarrow a .\overrightarrow b + \overrightarrow b .\overrightarrow c + \overrightarrow c .\overrightarrow a = {{ - 1 - 4 - 9} \over 2}$

$ = - 7$

Vector perpendicular to the face $OAB$

$ = \overrightarrow {OA} \times \overrightarrow {OB} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & 2 & 1 \cr 2 & 1 & 3 \cr } } \right| = 5\widehat i - \widehat j - 3\widehat k$

Vector perpendicular to the face $ABC$

$ = \overrightarrow {AB} \times \overrightarrow {AC} = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 1 & { - 1} & 2 \cr { - 2} & { - 1} & 1 \cr } } \right| = \widehat i - 5\widehat j - 3\widehat k$

Angle between the faces $=$ angle between their normals

$\cos \theta = \left| {{{5 + 5 + 9} \over {\sqrt {35} \sqrt {35} }}} \right| = {{19} \over {35}}$

or $\theta = {\cos ^{ - 1}}\left( {{{19} \over {35}}} \right)$

$\left( {\overrightarrow u + \overrightarrow v - \overrightarrow w } \right).\left( {\overrightarrow u \times \overrightarrow v - \overrightarrow u \times \overrightarrow w - \overrightarrow v \times \overrightarrow v + \overrightarrow v \times \overrightarrow w } \right)$

$ = \left( {\overrightarrow u + \overrightarrow v - \overrightarrow w } \right).\left( {\overrightarrow u \times \overrightarrow v - \overrightarrow u \times \overrightarrow w + \overrightarrow v \times \overrightarrow w } \right) = \overrightarrow u .\left( {\overrightarrow u \times \overrightarrow v } \right)$

$ - \overrightarrow u .\left( {\overrightarrow u \times \overrightarrow w } \right) + \overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right) + \overrightarrow v .\left( {\overrightarrow u \times \overrightarrow v } \right) - \overrightarrow v .\left( {\overrightarrow u \times \overrightarrow w } \right)$

$ + \overrightarrow v .\left( {\overrightarrow v \times \overrightarrow w } \right) - \overrightarrow w .\left( {\overrightarrow u \times \overrightarrow v } \right) + \overrightarrow w .\left( {\overrightarrow u \times \overrightarrow w } \right) - \overrightarrow w .\left( {\overrightarrow u \times \overrightarrow w } \right)$

$ = \overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right) - \overrightarrow v .\left( {\overrightarrow u \times \overrightarrow w } \right) - \overrightarrow w .\left( {\overrightarrow u \times \overrightarrow v } \right)$

$ = \left[ {\left. {\overrightarrow u \overrightarrow v \overrightarrow w } \right)} \right] + \left[ {\left. {\overrightarrow v \overrightarrow w \overrightarrow u } \right)} \right] - \left[ {\overrightarrow w \overrightarrow u \overrightarrow v } \right]$

$ = \overrightarrow u .\left( {\overrightarrow v \times \overrightarrow w } \right)$

$\left| {\matrix{ a & {{a^2}} & {1 + {a^3}} \cr b & {{b^2}} & {1 + {b^3}} \cr c & {{c^2}} & {1 + {c^3}} \cr } } \right| = 0$

$ \Rightarrow \left| {\matrix{ a & {{a^2}} & 1 \cr b & {{b^2}} & 1 \cr c & {{c^2}} & 1 \cr } } \right| + \left| {\matrix{ a & {{a^2}} & {{a^3}} \cr b & {{b^2}} & {{b^3}} \cr c & {{c^2}} & {{c^3}} \cr } } \right| = 0$

$ \Rightarrow \left( {1 + abc} \right)\left| {\matrix{ 1 & a & {{a^2}} \cr 1 & b & {{b^2}} \cr 1 & c & {{c^2}} \cr } } \right| = 0$

As $\,\,\,\left| {\matrix{ 1 & a & {{a^2}} \cr 1 & b & {{b^2}} \cr 1 & c & {{c^2}} \cr } } \right| \ne 0$ (given condition)

$\therefore$ $abc=-1$

$A = \left( {7, - 4,7} \right),B = \left( {1, - 6,10} \right),$

$C = \left( { - 1, - 3,4} \right)$ and $D = \left( {5, - 1,5} \right)$

$AB = \sqrt {{{\left( {7 - 1} \right)}^2} + {{\left( { - 4 + 6} \right)}^2} + {{\left( {7 - 10} \right)}^2}} $

$ = \sqrt {36 + 4 + 9} = 7$

Similarly $BC = 7,\,\,CD = \sqrt {41} ,\,\,DA = \sqrt {17} $

$\therefore$ None of the options is satisfied

Since, $\hat{\mathbf{n}} \perp \overrightarrow{\mathbf{u}}$ and $\hat{\mathbf{n}} \perp \overrightarrow{\mathbf{v}}$

$ \begin{aligned} \Rightarrow \hat{\mathbf{n}} & =\frac{\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|} \\\\ \therefore|\overrightarrow{\mathbf{w}} \cdot \hat{\mathbf{n}}| & =\left|\frac{\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|}\right| \\\\ & =\frac{|\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|}{|\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|} \end{aligned} $

$ \begin{aligned} & \text { Now, } \overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})=\left|\begin{array}{rrr} 1 & 2 & 3 \\ 1 & 1 & 0 \\ 1 & -1 & 0 \end{array}\right|=-6 \\\\ & \begin{array}{ll} \Rightarrow |\overrightarrow{\mathbf{w}} \cdot(\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}})|=6 \\\\ \text { and } \overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}=(\hat{\mathbf{i}}+\hat{\mathbf{j}}) \times(\hat{\mathbf{i}}-\hat{\mathbf{j}})-2 \hat{\mathbf{k}} \\\\ \Rightarrow |\overrightarrow{\mathbf{u}} \times \overrightarrow{\mathbf{v}}|=2 \\\\ \therefore |\overrightarrow{\mathbf{w}} \cdot \hat{\mathbf{n}}|=\frac{6}{2}=3 \end{array} \end{aligned} $

$PV\,\,$ of $\,\,\overrightarrow {AD} $ $ = {{\left( {3 + 5} \right)i + \left( {0 - 2} \right)j + \left( {4 + 4} \right)k} \over 2}$

$ = 4i - j + 4k\,\,$

or $\,\,\,\left| {\overrightarrow {AD} } \right| = \sqrt {16 + 16 + 1} = \sqrt {33} $

Let $\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow r .$ Then

$\overrightarrow a \times \left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right) = \overrightarrow a \times \overrightarrow r $

$ \Rightarrow 0 + \overrightarrow a \times \overrightarrow b + \overrightarrow a \times \overrightarrow c = \overrightarrow a \times \overrightarrow r $

$ \Rightarrow \overrightarrow a \times \overrightarrow b - \overrightarrow c \times \overrightarrow a = \overrightarrow a \times \overrightarrow r $

$ \Rightarrow \overrightarrow a \times \overrightarrow r = \overrightarrow 0 $

Similarly $\overrightarrow b \times \overrightarrow r = \overrightarrow 0 \,\,\,\& \,\,\,\overrightarrow c \times \overrightarrow r = \overrightarrow 0 $

Above three conditions will be satisfied for non-zero vectors if and only if $\overrightarrow r = \overrightarrow 0 $

We have, $\left[ {\overrightarrow a \times \overrightarrow b \,\,\overrightarrow b \times \overrightarrow c \,\,\overrightarrow c \times \overrightarrow a } \right]$

$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\,\,\left\{ {\left( {\overrightarrow b \times \overrightarrow c } \right) \times \left( {\overrightarrow c \times \overrightarrow a } \right)} \right\}$

$ = \left( {\overrightarrow a \times \overrightarrow b } \right).\,\,\left\{ {\left( {\overrightarrow m \,.\,\overrightarrow a } \right)\overrightarrow c - \left( {\overrightarrow m \,.\,\overrightarrow c } \right)\overrightarrow a } \right\}$

( where $\overrightarrow m = \overrightarrow b \times \overrightarrow c $ )

$ = \left\{ {\left( {\overrightarrow a \times \overrightarrow b } \right).\overrightarrow c } \right\}.\,\left\{ {\overrightarrow a \,.\,\left( {\overrightarrow b \times \overrightarrow c } \right)} \right\}$

$ = {\left[ {\overrightarrow a \,\overrightarrow b \,\overrightarrow c } \right]^2} = {4^2} = 16.$

Since $\overrightarrow a ,\overrightarrow c ,\overrightarrow b $ form a right handed system,

$\therefore$ $\overrightarrow c = \overrightarrow b \times \overrightarrow a = \left| {\matrix{ {\widehat i} & {\widehat j} & {\widehat k} \cr 0 & 1 & 0 \cr x & y & z \cr } } \right|$

$ = z\widehat i - x\widehat k$

If $\overrightarrow{\mathbf{a}}, \overrightarrow{\mathbf{b}}$ and $\overrightarrow{\mathbf{c}}$ are the sides of $\mathbf{a}$ triangle, then

$\overrightarrow{\mathbf{a}}+\overrightarrow{\mathbf{b}}+\overrightarrow{\mathbf{c}}=\overrightarrow{\mathbf{0}}$

Since,

$ \begin{aligned} \vec{a}+\vec{b}+\vec{c} & =\overrightarrow{0} \\\\ \vec{a}+\vec{b} & =-\vec{c} \end{aligned} $

$ \begin{array}{ll} \Rightarrow & (\vec{a}+\vec{b}) \times \vec{c}=-\vec{c} \times \vec{c} \\\\ \Rightarrow & \vec{a} \times \vec{c}+\vec{b} \times \vec{c}=\overrightarrow{0} \\\\ \Rightarrow & \vec{b} \times \vec{c}=\vec{c} \times \vec{a} \\\\ \text { Similarly, } & \vec{a} \times \vec{b}=\vec{b} \times \vec{c} \\\\ \text { Hence, } & \vec{a} \times \vec{b}=\vec{b} \times \vec{c}=\vec{c} \times \vec{a} \end{array} $

${\left( {\overrightarrow a \times \overrightarrow b } \right)^2} = {\left| {\overrightarrow a } \right|^2}{\left| {\overrightarrow b } \right|^2}\,\,{\sin ^2}{\pi \over 6}$

$ = 16 \times 4 \times {1 \over 4} = 16$

To solve this problem, let's take it step by step, beginning with calculating each of the vector magnitudes (or norms) and then finding the magnitude of the cross product vector $\overrightarrow{c}$.

Given vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ are:

$\overrightarrow{a} = 3\widehat{i} - 5\widehat{j}$

$\overrightarrow{b} = 6\widehat{i} + 3\widehat{j}$

1. Magnitude of $\overrightarrow{a}$

The magnitude of vector $\overrightarrow{a}$ is calculated using the formula:

$\left| \overrightarrow{a} \right| = \sqrt{(3)^2 + (-5)^2} = \sqrt{9 + 25} = \sqrt{34}.$

2. Magnitude of $\overrightarrow{b}$

Similarly, the magnitude of vector $\overrightarrow{b}$ is calculated as:

$\left| \overrightarrow{b} \right| = \sqrt{(6)^2 + (3)^2} = \sqrt{36 + 9} = \sqrt{45}.$

3. Calculating $\overrightarrow{c} = \overrightarrow{a} \times \overrightarrow{b}$

The cross product of two vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ in 3-dimensional space is given by:

$\overrightarrow{c} = \overrightarrow{a} \times \overrightarrow{b} = \left| \begin{array}{ccc} \widehat{i} & \widehat{j} & \widehat{k} \\ 3 & -5 & 0 \\ 6 & 3 & 0 \end{array} \right|$

For our provided vectors, the $k$ component of both vectors is $0$, thus their cross product will be purely in the $k$ direction. The determinant simplifies to:

$\overrightarrow{c} = (3 \times 3 - (-5) \times 6)\widehat{k} = (9 + 30)\widehat{k} = 39\widehat{k}.$

4. Magnitude of $\overrightarrow{c}$

The magnitude of vector $\overrightarrow{c}$ is:

$\left| \overrightarrow{c} \right| = \sqrt{39^2} = 39.$

Finally, the ratio of the magnitudes of vectors $\overrightarrow{a}$, $\overrightarrow{b}$, and $\overrightarrow{c}$ is thus:

$\sqrt{34}:\sqrt{45}:39.$

Therefore, the correct answer is:

Option B) $\sqrt{34}:\sqrt{45}:39.$

We have, $\overrightarrow a + \overrightarrow b + \overrightarrow c = \overrightarrow 0 $

$ \Rightarrow {\left( {\overrightarrow a + \overrightarrow b + \overrightarrow c } \right)^2} = 0$

$ \Rightarrow {\left| {\overrightarrow a } \right|^2} + {\left| {\overrightarrow b } \right|^2} + {\left| {\overrightarrow c } \right|^2}$

$\,\,\,\,\,\,\,\,\,\,\,\,\, + 2\left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = 0$

$ \Rightarrow 25 + 16 + 9 + 2\left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = 0$

$ \Rightarrow \left( {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right) = - 25.$

$\therefore$ $\left| {\overrightarrow a \,.\,\overrightarrow b + \overrightarrow b \,.\,\overrightarrow c + \overrightarrow c \,.\,\overrightarrow a } \right| = 25.$