Vector Algebra

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$

Since $\hat{v}$ unit vector along the incident ray and $\widehat{w}$ unit vector along the reflected ray

Hence, $\hat{a}$ is a unit vector along the external bisector of $\hat{v}$ and $\widehat{w}$

$\therefore \quad \hat{w}-\hat{v}=\lambda \hat{a}$

Squaring on both sides

${(\widehat w - \widehat v)^2} = {(\lambda \widehat a)^2}$

$\Rightarrow 1+1-2\cos2\theta=\lambda^2$

$\Rightarrow \quad \lambda^2=2(1-cos2\theta)$

$\Rightarrow \quad \lambda^{2}=2 \cdot 2 \sin ^{2} \theta$

$\Rightarrow \quad \lambda=2 \sin \theta$

Hence, $\quad \widehat{w}-\hat{v}=2 \sin \theta \cdot \hat{a}$

$\begin{aligned} & =2 \cos \left(90^{\circ}-\theta\right) \cdot \hat{a} \\ & =-(2 \hat{a} \cdot \hat{v}) \cdot \hat{a} \\ \widehat{w} & =\hat{v}-2(\hat{a} \cdot \hat{v}) \cdot \hat{a} \end{aligned}$

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.$

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