Vector Algebra

To find the angle between the vector $\vec{Q}$ and the resultant of $(2 \vec{Q}+2 \vec{P})$ and $(2 \vec{Q}-2 \vec{P})$, we first find the resultant vector of $(2 \vec{Q}+2 \vec{P})$ and $(2 \vec{Q}-2 \vec{P})$.

The resultant vector of $(2 \vec{Q}+2 \vec{P})$ and $(2 \vec{Q}-2 \vec{P})$ can be simply found by adding these two vectors:

$ \text{Resultant} = (2 \vec{Q}+2 \vec{P}) + (2 \vec{Q}-2 \vec{P}) = 4 \vec{Q} $

Now, we need to find the angle between the vector $\vec{Q}$ and this resultant vector $4\vec{Q}$. Since the resultant vector is just a scaled version of $\vec{Q}$, they are in the same direction. The angle between any vector and another vector that is a scaled version of the first vector is always $0^\circ$, because they are parallel to each other.

Therefore, the angle between $\vec{Q}$ and the resultant of $(2 \vec{Q}+2 \vec{P})$ and $(2 \vec{Q}-2 \vec{P})$ is $0^\circ$.

So, the correct option is:

Option B: $0^\circ$

The magnitude of resultant vector

$R^{\prime}=\sqrt{a^2+b^2+2 a b \cos \theta}$

Here $a=b=R$

Then $R^{\prime}=\sqrt{R^2+R^2+2 R^2 \cos \theta}$

$\begin{aligned} & =R \sqrt{2} \sqrt{1+\cos \theta} \\ & =\sqrt{2} R \sqrt{2 \cos ^2 \frac{\theta}{2}} \\ & =2 R \cos \frac{\theta}{2} \end{aligned}$

$ \begin{aligned} & \mathrm{A}_{\mathrm{y}}=\mathrm{A} \cos 30^{\circ}=2 \sqrt{3} \\\\ & \Rightarrow \mathrm{A} \frac{\sqrt{3}}{2}=2 \sqrt{3} \\\\ & \Rightarrow \mathrm{A}=4 \end{aligned} $

Now $A_x=A \sin 30^{\circ}=4 \times \frac{1}{2}=2$

Given that when vector $\vec{A}=2 \hat{i}+3 \hat{j}+2 \hat{k}$ is subtracted from vector $\overrightarrow{\mathrm{B}}$, it gives a vector equal to $2 \hat{j}$.

We can write this as:

$\vec{B} - \vec{A} = 2 \hat{j}$

Now, let's express the vector $\overrightarrow{\mathrm{B}}$ in terms of its components:

$\vec{B} = B_x \hat{i} + B_y \hat{j} + B_z \hat{k}$

Subtract vector $\vec{A}$ from vector $\vec{B}$:

$(B_x \hat{i} + B_y \hat{j} + B_z \hat{k}) - (2 \hat{i}+3 \hat{j}+2 \hat{k}) = 2 \hat{j}$

Comparing the components, we get:

$ B_x - 2 = 0 \\ B_y - 3 = 2 \\ B_z - 2 = 0 $

Solving these equations, we find the components of vector $\vec{B}$:

$ B_x = 2 \\ B_y = 5 \\ B_z = 2 $

Now, we can find the magnitude of vector $\vec{B}$:

$ |\vec{B}| = \sqrt{B_x^2 + B_y^2 + B_z^2} = \sqrt{2^2 + 5^2 + 2^2} = \sqrt{4 + 25 + 4} = \sqrt{33} $

Therefore, the magnitude of vector $\overrightarrow{\mathrm{B}}$ is $\sqrt{33}$

The resultant of two perpendicular vectors is given by the Pythagorean theorem.

If we have two vectors of magnitudes $A$ and $\frac{A}{2}$, the resultant $R$ is:

$R = \sqrt{A^{2} + \left(\frac{A}{2}\right)^{2}} = \sqrt{A^{2} + \frac{A^{2}}{4}} = \sqrt{\frac{5A^{2}}{4}} = \frac{\sqrt{5} A}{2}$.

$\vec{P} \,\&\, \vec{Q}$ are perpendicular

$ \begin{aligned} & \overrightarrow{\mathrm{P}} \cdot \overrightarrow{\mathrm{Q}}=0 \\\\ & (\hat{\mathrm{i}}+2 \mathrm{~m} \hat{\mathrm{j}}+\mathrm{m} \hat{\mathrm{k}}) \cdot(4 \hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\mathrm{m} \hat{\mathrm{k}})=0 \\\\ & \Rightarrow 4-4 \mathrm{~m}+\mathrm{m}^2=0 \\\\ & \Rightarrow(\mathrm{m}-2)^2=0 \Rightarrow \mathrm{m}=2 \end{aligned} $

$\sqrt {{A^2} + {A^2} + 2{A^2}\cos \theta } = 2\sqrt {{A^2} + {A^2} + 2{A^2}( - \cos \theta )} $

$ \Rightarrow 2{A^2} + 2{A^2}\cos \theta = 8{A^2} + 8{A^2} - ( - \cos \theta )$

$ \Rightarrow 5\cos \theta = 3$

$ \Rightarrow \theta = {\cos ^{ - 1}}\left( {{3 \over 5}} \right)$

$\overrightarrow A \times \overrightarrow A = A \times A \times \sin 0^\circ $

$ = 0$

$\because$ $\left| {\widehat A - \widehat B} \right| = 2\sin \left( {{\theta \over 2}} \right)$

and, $\left| {\widehat A + \widehat B} \right| = 2\cos \left( {{\theta \over 2}} \right)$

$ \Rightarrow {{\left| {\widehat A - \widehat B} \right|} \over {\left| {\widehat A + \widehat B} \right|}} = \tan \left( {{\theta \over 2}} \right)$

$\overrightarrow A = \overrightarrow P + \overrightarrow Q $

$\overrightarrow B = \overrightarrow P - \overrightarrow Q $

$\overrightarrow P \bot \overrightarrow Q $

$\left| {\overrightarrow A } \right| = \left| {\overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})(1 + \cos \theta )} $

For $\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {3({P^2} + {Q^2})} $

${\theta _1} = 60^\circ $

For $\left| {\overrightarrow A + \overrightarrow B } \right| = \sqrt {2({P^2} + {Q^2})} $

${\theta _2} = 90^\circ $

$\overrightarrow {{F_x}} = \left( {10 \times {{\sqrt 3 } \over 2} + 20\left( {{1 \over 2}} \right) + 20\left( {{1 \over {\sqrt 2 }}} \right) - 15\left( {{1 \over {\sqrt 2 }}} \right) - 15\left( {{{\sqrt 3 } \over 2}} \right)} \right)\widehat i$

$ = 9.25\widehat i$

$\overrightarrow {{F_y}} = \left( {15\left( {{1 \over 2}} \right) + 20\left( {{{\sqrt 3 } \over 2}} \right) + 10\left( {{1 \over 2}} \right) - 15\left( {{1 \over {\sqrt 2 }}} \right) - 20\left( {{1 \over {\sqrt 2 }}} \right)} \right)\widehat j$

$ = 5\widehat j$

Angle between ${\overrightarrow A }$ and ${\overrightarrow B }$, $\theta$ = 60$^\circ$

Angle between ${\overrightarrow A }$ and -${\overrightarrow B }$, $\theta$ = 120$^\circ$

If angle between ${\overrightarrow A }$ and ${\overrightarrow A }$ $-$ ${\overrightarrow B }$ is $\alpha $

then $\tan \alpha = $${{\left| { - \overrightarrow B } \right|\sin \theta } \over {\overrightarrow A + \left| { - \overrightarrow B } \right|\cos \theta }}$

= ${{B\sin 120^\circ } \over {A + B\cos 120^\circ }}$

=${{B{{\sqrt 3 } \over 2}} \over {A - {B \over 2}}}$

$ \Rightarrow $ $\tan \alpha = {{\sqrt 3 B} \over {2A - B}}$

Let magnitude be equal to $\lambda$

$\overrightarrow {OA} = \lambda \left[ {\cos 30^\circ \widehat i + \sin 30\widehat j} \right] = \lambda \left[ {{{\sqrt 3 } \over 2}\widehat i + {1 \over 2}\widehat j} \right]$

$\overrightarrow {OB} = \lambda \left[ {\cos 60^\circ \widehat i - \sin 60\widehat j} \right] = \lambda \left[ {{1 \over 2}\widehat i - {{\sqrt 3 } \over 2}\widehat j} \right]$

$\overrightarrow {OC} = \lambda \left[ {\cos 45^\circ ( - \widehat i) + \sin 45\widehat j} \right] = \lambda \left[ { - {1 \over {\sqrt 2 }}\widehat i + {1 \over {\sqrt 2 }}\widehat j} \right]$

$\therefore$ $\overrightarrow {OA} + \overrightarrow {OB} - \overrightarrow {OC} $

$ = \lambda \left[ {\left( {{{\sqrt 3 + 1} \over 2} + {1 \over {\sqrt 2 }}} \right)\widehat i + \left( {{1 \over 2} - {{\sqrt 3 } \over 2} - {1 \over {\sqrt 2 }}} \right)\widehat j} \right]$

$\therefore$ Angle with x-axis

${\tan ^{ - 1}}\left[ {{{{1 \over 2} - {{\sqrt 3 } \over 2} - {1 \over {\sqrt 2 }}} \over {{{\sqrt 3 } \over 2} + {1 \over 2} + {1 \over {\sqrt 2 }}}}} \right] = {\tan ^{ - 1}}\left[ {{{\sqrt 2 - \sqrt 6 - 2} \over {\sqrt 6 + \sqrt 2 + 2}}} \right]$

$ = {\tan ^{ - 1}}\left[ {{{1 - \sqrt 3 - \sqrt 2 } \over {\sqrt 3 + 1 + \sqrt 2 }}} \right]$

Hence option (a).

Let the position vectors of $A,B,C,D,O$ be $\vec a,\vec b,\vec c,\vec d,\vec 0$ respectively.

Check Assertion (A)

$ \overrightarrow{AB}+\overrightarrow{AC}+\overrightarrow{AD} =(\vec b-\vec a)+(\vec c-\vec a)+(\vec d-\vec a) =\vec b+\vec c+\vec d-3\vec a $

Also,

$ 4\overrightarrow{AO}+\overrightarrow{OB}+\overrightarrow{OC} =4(\vec 0-\vec a)+(\vec b-\vec 0)+(\vec c-\vec 0) =-4\vec a+\vec b+\vec c $

Equating both sides gives

$ \vec b+\vec c+\vec d-3\vec a=-4\vec a+\vec b+\vec c \;\Rightarrow\; \vec d=-\vec a $

But $\vec d=-\vec a$ means $O$ is the midpoint of $AD$, i.e. $AD$ is a diameter. Since $A$ and $D$ are the endpoints of the given semicircle (as in the figure), this is true.

So Assertion A is correct (and it doesn’t actually need $|AB|=|BC|=|CD|$).

Check Reason (R)

By the polygon law along the path $A\to B\to C\to D$,

$ \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}=\overrightarrow{AD} $

Hence

$ \overrightarrow{AB}+\overrightarrow{BC}+\overrightarrow{CD}+\overrightarrow{AD} =2\overrightarrow{AD}\neq 2\overrightarrow{AO} $

So Reason R is incorrect.

Correct option

Option A: A is correct but R is not correct.

Given X = Y

$\sqrt {{X^2} + {Y^2} - 2 \times Y\cos \theta } $

$ = n\sqrt {{X^2} + {Y^2} + 2 \times Y\cos \theta } $

Square both sides

$2{X^2}(1 - \cos \theta ) = {n^2}.2{X^2}(1 + \cos \theta )$

$1 - \cos \theta = {n^2} + {n^2}\cos \theta $

$\cos \theta = {{1 - {n^2}} \over {1 + {n^2}}}$

$\theta = {\cos ^{ - 1}}\left[ {{{{n^2} - 1} \over { - {n^2} - 1}}} \right]$

(a) $\overrightarrow C = \overrightarrow A + \overrightarrow B $

Option (iv)

(b) $\overrightarrow A = \overrightarrow B + \overrightarrow C = \overrightarrow C + \overrightarrow B $

Option (iii)

(c) $\overrightarrow B = \overrightarrow A + \overrightarrow C $

Option (i)

(d) $\overrightarrow A + \overrightarrow B + \overrightarrow C = 0$

Option (ii)

Projection = ${{\overrightarrow A .\overrightarrow B } \over {\left| {\overrightarrow B } \right|}}(\widehat B)$

$ = {{(\widehat i + \widehat j + \widehat k).(\widehat i + \widehat j)} \over {\sqrt 2 }}{{(\widehat i + \widehat j)} \over {\sqrt 2 }}$

$ = {2 \over {\sqrt 2 }} $$ \times $$ {{(\widehat i + \widehat j)} \over {\sqrt 2 }}$

$ = (\widehat i + \widehat j)$

$\left| {\overrightarrow P } \right| = \left| {\overrightarrow Q } \right| = x$ ..... (i)

$\left| {\overrightarrow P + \overrightarrow Q } \right| = n\left| {\overrightarrow P - \overrightarrow Q } \right|$

${P^2} + {Q^2} + 2PQ\cos \theta = {n^2}({P^2} + {Q^2} - 2PQ\cos \theta )$

Using (i) in above equation

$\cos \theta = {{{n^2} - 1} \over {1 + {n^2}}}$

$\theta = {\cos ^{ - 1}}\left( {{{{n^2} - 1} \over {{n^2} + 1}}} \right)$

Given, $\overrightarrow A $ . $\overrightarrow B $ = $\left| {\overrightarrow A \times \overrightarrow B } \right|$ ..... (i)

Also, we know that

$\overrightarrow A $ . $\overrightarrow B $ = $\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|$ cos$\theta$ .... (ii)

and $\overrightarrow A \times \overrightarrow B $ = $\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|$ sin$\theta$ ..... (iii)

From Eqs. (i), (ii) and (iii), we get

$\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|$ cos$\theta$ = $\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|$ sin$\theta$

$ \Rightarrow \cos \theta = \sin \theta \Rightarrow {{\sin \theta } \over {\cos \theta }} = 1$

$ \Rightarrow \tan \theta = 1$

$ \Rightarrow \tan \theta = \tan 45^\circ \Rightarrow \theta = 45^\circ $

$\therefore$ $\left| {\overrightarrow A - \overrightarrow B } \right| = \sqrt {{A^2} + {B^2} - 2\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos \theta } $

$ = \sqrt {{A^2} + {B^2} - 2\left| {\overrightarrow A } \right|\left| {\overrightarrow B } \right|\cos (45^\circ )} $

$ = \sqrt {{A^2} + {B^2} - \sqrt 2 AB} $

We know,

$ \because $ $\overrightarrow {OA} + \overrightarrow {OB} + \overrightarrow {OC} + \overrightarrow {OD} + \overrightarrow {OE} + \overrightarrow {OF} + \overrightarrow {OG} + \overrightarrow {OH} = \overrightarrow 0 $

By triangle law of vector addition, we can write

$\overrightarrow {AB} = \overrightarrow {AO} + \overrightarrow {OB} \,;\,\overrightarrow {AC} = \overrightarrow {AO} + \overrightarrow {OC} $

$\overrightarrow {AD} = \overrightarrow {AO} + \overrightarrow {OD} \,;\,\overrightarrow {AE} = \overrightarrow {AO} + \overrightarrow {OE} $

$\overrightarrow {AF} = \overrightarrow {AO} + \overrightarrow {OF} \,;\,\overrightarrow {AG} = \overrightarrow {AO} + \overrightarrow {OG} $

$\overrightarrow {AH} = \overrightarrow {AO} + \overrightarrow {OH} $

Now

$\overrightarrow {AB} + \overrightarrow {AC} + \overrightarrow {AD} + \overrightarrow {AE} + \overrightarrow {AF} + \overrightarrow {AG} + \overrightarrow {AH} $

$ = (7\overrightarrow {AO} ) + \overrightarrow {OB} + \overrightarrow {OC} + \overrightarrow {OD} + \overrightarrow {OE} + \overrightarrow {OF} + \overrightarrow {OG} + \overrightarrow {OH} $

$ = (7\overrightarrow {AO} ) + \overrightarrow 0 - \overrightarrow {OA} $

$ = (7\overrightarrow {AO} ) + \overrightarrow {AO} $

$ = 8\overrightarrow {AO} = 8(2\widehat i + 3\widehat j - 4\widehat k)$

$ = 16\widehat i + 24\widehat j - 32\widehat k$

$\left| {\overrightarrow {{A_1}} } \right| = 3,\left| {\overrightarrow {{A_2}} } \right| = 5\,and\,\left| {\overrightarrow {{A_1}} + \overrightarrow {{A_2}} } \right| = 5$

$\,\left| {\overrightarrow {{A_1}} + \overrightarrow {{A_2}} } \right| = {\left| {\overrightarrow {{A_1}} } \right|^2} + {\left| {\overrightarrow {{A_2}} } \right|^2} + 2\left| {\overrightarrow {{A_1}} } \right|\left| {\overrightarrow {{A_2}} } \right|\cos \theta $

$\cos \theta = - {3 \over {10}}$

$\left( {2\overrightarrow {{A_1}} + 3\overrightarrow {{A_2}} } \right).\left( {3\overrightarrow {{A_1}} - 2\overrightarrow {{A_2}} } \right)$

= $6{\left| {\overrightarrow {{A_1}} } \right|^2} + 9\overrightarrow {{A_1}} .\overrightarrow {{A_2}} - 4\overrightarrow {{A_1}} .\overrightarrow {{A_2}} - 6{\left| {\overrightarrow {{A_2}} } \right|^2}$

= - 118.5

$\left| {\overrightarrow A + \overrightarrow B } \right| = 2a\cos \theta /2$      . . . (1)

$\left| {\overrightarrow A - \overrightarrow B } \right| = 2a\cos {{\left( {\pi - \theta } \right)} \over 2} = 2a\sin \theta /2$      . . . (2)

$ \Rightarrow \,\,\,n\left( {2a\cos {\theta \over 2}} \right) = 2a{{\sin \theta } \over 2}$

$ \Rightarrow \,\,\,\tan {\theta \over 2} = n$

$\overrightarrow {{r_g}} = {a \over 2}\widehat i + {a \over 2}\widehat k$

$\overrightarrow {{r_H}} = {a \over 2}\widehat i + {a \over 2}\widehat k$

$\overrightarrow {{r_H}} - \overrightarrow {{r_g}} = {a \over 2}\left( {\widehat j - \widehat i} \right)$

Let $\overrightarrow C $ = a$\widehat i$ + b$\widehat j$

Given, $\overrightarrow A .\overrightarrow C = \overrightarrow A .\overrightarrow B $

$ \Rightarrow $ $\,\,\,\,$ a + b = 2 $-$ 1

$ \Rightarrow $ $\,\,\,\,$ a + b = 1 . . . . .(1)

also given

$\overrightarrow B .\overrightarrow C = \overrightarrow A .\overrightarrow B $

$ \Rightarrow $ $\,\,\,\,$ 2a $-$ b = 1 . . . . (2)

Solving (1) and (2), we get,

a = ${1 \over 3}$ and b = ${2 \over 3}$

$\therefore\,\,\,\,$ $\overrightarrow C = {1 \over 3}\widehat i + {2 \over 3}\widehat j$

$\left| {\overrightarrow C } \right| = \sqrt {{{\left( {{1 \over 3}} \right)}^2} + {{\left( {{2 \over 3}} \right)}^2}} $

= $\sqrt {{5 \over 9}} $

$\overrightarrow A \times \overrightarrow B = \overrightarrow B \times \overrightarrow A $

$\overrightarrow A \times \overrightarrow B - \overrightarrow B \times \overrightarrow A = 0$

$ \Rightarrow \overrightarrow A \times \overrightarrow B + \overrightarrow A \times \overrightarrow B = 0$

$\therefore$ $\overrightarrow A \times \overrightarrow B = 0$

$ \Rightarrow AB\sin \theta = 0$

$\theta = $ $0,\pi ,\,\,$$2\pi $ ........

from the given options, $\theta = \pi $