Maths M.C.Q (1 Marks)

3. 4

Solution:

We have

$\sqrt{3}\big|\vec{\text{a}}+\vec{\text{b}}\big|+\big|\vec{\text{a}}-\vec{\text{b}}\big|$

$=\sqrt{3}\times\sqrt{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2|\vec{\text{a}}|​​\big|\vec{\text{b}}\big|\cos\theta}+\sqrt{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta}$

$=\sqrt{3}\times\sqrt{1^2+1^2+2\times1\times1\cos\theta}+\sqrt{1^2+1^2-2\times1\times1\cos\theta}$ (As $\vec{\text{a}}$ and $\vec{\text{b}}$ unit vectors)

$=\sqrt{3}\times\sqrt{2+2\cos\theta}+\sqrt{2-2\cos\theta}$

$=\sqrt{3}\times\sqrt{2(1+\cos\theta)}+\sqrt{2(1-\cos\theta)}$

$=\sqrt{3}\times\sqrt{2\times2\cos^2\frac{\theta}{2}}+\sqrt{2\times2\sin^2\frac{\theta}{2}}$

$=2\sqrt{3}\cos\frac{\theta}{2}+2\sin\frac{\theta}{2}$

$=2\big(\sqrt{3}\cos\frac{\theta}{2}+\sin\frac{\theta}{2}\big)$

$=2\times2\big(\frac{\sqrt{3}}{2}\cos\frac{\theta}{2}+\frac{1}{2}\sin\frac{\theta}{2}\big)$

$=2\times2\big(\sin\frac{\pi}{3}\cos\frac{\theta}{2}+\cos\frac{\pi}{3}\sin\frac{\theta}{2}\big)$

$=4\sin\big(\frac{\pi}{3}+\frac{\theta}{2}\big)$

Now, maximum value of $\sin\text{a}=1$

⇒ Maximum value of $\sin\big(\frac{\pi}{3}+\frac{\theta}{2}\big)=1$

⇒ Maximum value of $4\sin\big(\frac{\pi}{3}+\frac{\theta}{2}\big)=4$

$\therefore$ Maximum velue of $\sqrt{3}\big|\vec{\text{a}}+\vec{\text{b}}\big|+\big|\vec{\text{a}}-\vec{\text{b}}\big|=4$

2. an ellipse

Solution:

Let, $\vec{\text{a}}=\hat{\text{i}}-2\text{x}\hat{\text{j}}+3\text{y}\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{i}}+2\text{x}\hat{\text{j}}-3\text{y}\hat{\text{k}}$

It is given that the vectors are perpendicular. so, their dot product is zero.

$\vec{\text{a}}.\vec{\text{b}}=0$

$\Rightarrow\big(\hat{\text{i}}-2\text{x}\hat{\text{j}}+3\text{y}\hat{\text{k}}\big).\big(\hat{\text{i}}+2\text{x}\hat{\text{j}}-3\text{y}\hat{\text{k}}\big)=0$

$\Rightarrow1-4\text{x}^2-9\text{y}^2=0$

$\Rightarrow4\text{x}^2+9\text{y}^2=1$

Dividing both sides by 36, we get

$\frac{\text{x}^2}{9}+\frac{\text{y}^2}{4}=1$

This is an ellipse.

3. $\sqrt{593}$

Solution:

Let ABCD be a parallelogram in which

side $\overrightarrow{\text{AB}}=\overrightarrow{\text{DC}}=5\vec{\text{a}}+2\vec{\text{b}}$

and $\overrightarrow{\text{AD}}=\overrightarrow{\text{BC}}=\vec{\text{a}}-3\vec{\text{b}}$

and diagonals are AC and BD.

Now, $\overrightarrow{\text{AC}}=\overrightarrow{\text{AB}}+\overrightarrow{\text{BC}}$

$=\big(5\vec{\text{a}}+2\vec{\text{b}\big)}+\big(\vec{\text{a}}-3\vec{\text{b}}\big)$

$=6\vec{\text{a}}-\vec{\text{b}}$

$\therefore\big|\overrightarrow{\text{AC}}\big|=\big|6\vec{\text{a}}-\vec{\text{b}\big|}$

$=\sqrt{|6\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\times|6\vec{\text{a}}|\times\big|\vec{\text{b}}\big|\cos\theta}$

$=\sqrt{36|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-12\times|\vec{\text{a}}|\times\big|\vec{\text{b}}\big|\cos\frac{\pi}{4}}$

$=\sqrt{36|2\sqrt{2}|^2+|3|^2-12\times|2\sqrt{2}|\times|3|\times\frac{1}{\sqrt{2}}}$

$=\sqrt{288+9-72}$

$=\sqrt{225}=15\text{ units}$

$\overrightarrow{\text{BD}}=\overrightarrow{\text{BA}}+\overrightarrow{\text{BD}}$

$=-\overrightarrow{\text{AB}}+\overrightarrow{\text{BD}}$

$=-\big(5\vec{\text{a}}+2\vec{\text{b}}\big)+\big(\vec{\text{a}}-3\vec{\text{b}}\big)$

$=-4\vec{\text{a}}-5\vec{\text{b}}$

$\therefore|\overrightarrow{\text{BD}}|=\big|-4\vec{\text{a}}-5\vec{\text{b}}\big|$

$=\big|4\vec{\text{a}}+5\vec{\text{b}}\big|$

$=\sqrt{|4\vec{\text{a}}|^2+|5\vec{\text{b}}|^2+2|4\vec{\text{a}}|\times|5\vec{\text{b}|}\cos\theta}$

$=\sqrt{16|\vec{\text{a}}|^2+25\big|\vec{\text{b}}\big|^2+40\times|\vec{\text{a}}|\times\big|\vec{\text{b}}\big|\cos\frac{\pi}{4}}$

$=\sqrt{16|2\sqrt{2}|^2+25|3|^2+40\times|2\sqrt{2}|\times|3|\times\frac{1}{\sqrt{2}}}$

$=\sqrt{128+25+240}$

$=\sqrt{593}\text{ units}$ 

Therefore, the larger diagonal $=\sqrt{593}$

3. $\sqrt{3}\text{a}$

Solution:

Given that

So, $|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=|\vec{\text{c}}|=\text{a}\dots(1)$

Since they are mutually perpendicular,

$\vec{\text{a}}.\vec{\text{b}}=\vec{\text{b}}.\vec{\text{c}}=\vec{\text{c}}.\vec{\text{a}}=0\dots(2)$

Now,

$\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+|\vec{\text{c}}|^2+2\vec{\text{a}}.\vec{\text{b}}+2\vec{\text{b}}.\vec{\text{c}}+2\vec{\text{c}}.\vec{\text{a}}$

$=\text{a}^2+\text{a}^2+\text{a}^2+0+0+0$ [using (1) and (2)]

$=3\text{a}^2$

$\therefore\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=\sqrt{3}\text{a}$

1. 1

Solution:

Let $\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}$ and $\vec{\text{b}}=\hat{\text{j}}$

The projection of $\vec{\text{a}}$ on $\vec{\text{b}}$ is

$\frac{\vec{\text{a}}.\vec{\text{b}}}{\big|\vec{\text{b}}\big|}$

$=\frac{\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big).\hat{\text{j}}}{|\hat{\text{j}}|}$

$=\frac{0+1+0}{1}$

$=1$

2. a = 4, b = 4, c = 5

Solution:

It is given that vectors $2\hat{\text{i}}+3\hat{\text{j}}-4\hat{\text{k}}$ and $\text{a}\hat{\text{i}}+\text{b}\hat{\text{j}}+\text{c}\hat{\text{k}}$ are perpendicular.

So, their dot product is zero.

$\Rightarrow2\text{a}+3\text{b}-4\text{c}=0$

$(\text{b})\text{a}=4;\text{b}=4;\text{c}=5$

$\Rightarrow2(4)+3(4)-4(5)=0$

$8+12-20=0$

$0=0,$ which is true.

4. $-1$

Solution:

Given that $\vec{\text{a}},\vec{\text{b}}$ and $\vec{\text{c}}$ are unit vectors.

So, $|\vec{\text{a}}|=1,\big|\vec{\text{b}}\big|=1$ and $\vec{\text{c}}=1.$

Since $\vec{\text{a}}$ and $\vec{\text{b}}$ are mutually perpendicular,

$\vec{\text{a}}.\vec{\text{b}}=0$

Now,

$\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|=1$

$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}\big|^2=1$

$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+|\vec{\text{c}}|^2+2\vec{\text{a}}.\vec{\text{b}}+2\vec{\text{b}}.\vec{\text{c}}+2\vec{\text{c}}.\vec{\text{a}}=1$

$\Rightarrow1+1+1+2(0)+2|\vec{\text{a}}|\big|\vec{\text{b}\big|}\cos\beta+2|\vec{\text{c}}||\vec{\text{a}}|\cos\alpha=1$

$\Rightarrow3+2(\cos\alpha+\cos\beta)=1$

$\Rightarrow2(\cos\alpha+\cos\beta)=-2$

$\Rightarrow\cos\alpha+\cos\beta=-1$

4. $\text{a}=\frac{1}{|\lambda|}$

Solution:

Given that

$|\vec{\text{a}}|=\text{a};$

Now,

$|\lambda\vec{\text{a}}|=1$

$\Rightarrow|\lambda||\vec{\text{a}}|=1$

$\Rightarrow|\lambda|\text{a}=1$

$\Rightarrow\text{a}=\frac{1}{|\lambda|}$

2. $0\leq\theta\leq\frac{\pi}{2}$

Solution:

$\vec{\text{a}}.\vec{\text{b}}\geq0$

$\Rightarrow|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta\geq0$

$\Rightarrow\cos\theta\geq0$

$\Rightarrow0\leq\theta\leq\frac{\pi}{2}$

1. $\sqrt{3}$

Solution:

It is given that $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors.

$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1$

Now,

$\vec{\text{a}}.\vec{\text{b}}$

 $=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$

$=(1)(2)\cos\theta$

$=\cos\theta$

The range of $\cos\theta$ is [-1,1].

$\therefore\sqrt{3}$ is not a possible value of $\cos\theta$ as it is greater than 1.

4. $\frac{\pi}{3}$

Solution:

The given vectors are perpendicular. so, their dot product is zero.

$\big[(\sin\theta)\hat{\text{i}}+(\cos\theta)\hat{\text{j}}\big].\big(\hat{\text{j}}-\sqrt{3}\hat{\text{j}}\big)=0$

$\Rightarrow\sin\theta-\sqrt{3}\cos\theta=0$

$\Rightarrow\sin\theta=\sqrt{3}\cos\theta$

$\Rightarrow\tan\theta=\sqrt{3}$

$\Rightarrow\theta=\frac{\pi}{3}$ (because $\theta$ is acute)

2. $\frac{\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{a}}\big)}{|\vec{\text{a}}|^2}$

Solution:

The vector component of $\vec{\text{b}}$ perpendicular to $\vec{\text{a}}$ is

$\frac{\vec{\text{a}}\times\big(\vec{\text{b}}\times\vec{\text{a}}\big)}{|\vec{\text{a}}|^2}$

3. $14$

Solution:

It is given that vectors $3\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}$ and $2\hat{\text{i}}-\hat{\text{j}}+8\hat{\text{k}}$ are perpendicular.

So, their dot product is zero.

$\big(3\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}}\big).\big(2\hat{\text{i}}-\hat{\text{j}}+8\hat{\text{k}}\big)=0$

$\Rightarrow6-\lambda+8=0$

$\Rightarrow14-\lambda=0$

$\therefore\lambda=14$

3. $\text{a}=\frac{2\pi}{3}$

Solution:

$\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors.

$\Rightarrow|\vec{\text{a}}|=\big|\vec{\text{b}}\big|=1\dots(1)$

Now,

$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\text{a}$

$\Rightarrow\vec{\text{a}}.\vec{\text{b}}=\cos\text{a}\dots(2)$

[using (1)]

Given that

$\Big|\vec{\text{a}}+\vec{\text{b}}\big|=1$

Squaring both sides, we get

$\big|\vec{\text{a}}+\vec{\text{b}}\big|^2=1$

$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=1$

$\Rightarrow1+1+2\cos\text{a}=1$ [using (1) and (2)]

$\Rightarrow2+2\cos\text{a}=1$

$\Rightarrow2\cos\text{a}=-1$

$\Rightarrow\cos\text{a}=\frac{-1}{2}$

$\Rightarrow\text{a}=\frac{2\pi}{3}$

1. $2\sin\frac{\theta}{2}$

Solution:

$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$

$=1\times1\cos\theta$ (Because $\vec{\text{a}}$ and $\vec{\text{b}}$ are unit vectors)

$=\cos\theta\dots(1)$

$\big|\vec{\text{a}}-\vec{\text{b}}\big|^2=|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2-2\vec{\text{a}}.\vec{\text{b}}$

 $=1+1-2\cos\theta$ [using (1)]

$=2-2\cos\theta$

$=2(1-\cos)$

$=2\big(2\sin^2\frac{\theta}{2}\big)$

$=4\sin^2\frac{\theta}{2}$

$\therefore\big|\vec{\text{a}}-\vec{\text{b}}\big|=2\sin\frac{\theta}{2}$

4. $\frac{\pi}{3}$

Solution:

Given, $|\vec{\text{a}}|=3,\big|\vec{\text{b}}\big|=5$ and $|\vec{\text{c}}|=7\dots(1)$

Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$

Given that

$\vec{\text{a}}+\vec{\text{b}}+\vec{\text{c}}=0$

$\Rightarrow\vec{\text{a}}+\vec{\text{b}}=-\vec{\text{c}}$

$\Rightarrow\big|\vec{\text{a}}+\vec{\text{b}}\big|=|-\vec{\text{c}}|^2$

$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{c}}|^2$

$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{c}}|^2-|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$

$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=7^2-3^3-5^2$ [using (1)]

$\Rightarrow2\vec{\text{a}}.\vec{\text{b}}=15$

$\Rightarrow2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=15$

$\Rightarrow2(3)(5)\cos\theta=15$ [using (1)]

$\Rightarrow\cos\theta=\frac{1}{2}$

$\therefore\theta=\frac{\pi}{3}$

2. $0<\text{x}<\frac{1}{2}$

Solution:

$\vec{\text{a}}=2\text{x}^2\hat{\text{i}}+4\text{x}\hat{\text{j}}+\hat{\text{k}}, \vec{\text{b}}=7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}$

Let the angle between vector a and vector b be A.

$\therefore\cos\text{A}=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}=\frac{\big(2\text{x}^2\hat{\text{i}}+4\text{x}\hat{\text{j}}+\hat{\text{k}}\big).\big(7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}\big)}{\big|2\text{x}^2\hat{\text{i}}+4\text{x}\hat{\text{j}}+\hat{\text{k}}\big|\big|7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}\big|}$

$=\frac{14\text{x}^2-8\text{x}+\text{x}}{\sqrt{4\text{x}^4+16\text{x}^2+1}\sqrt{49+4+\text{x}^2}}$

$=\frac{14\text{x}^2-8\text{x}+\text{x}}{\sqrt{4\text{x}^4+16\text{x}^2+1}\sqrt{53+\text{x}^2}}$

Now, $\angle\text{A}$ is an obtuse angle.

$\therefore\cos\text{A}<0$

$\Rightarrow\frac{14\text{x}^2-7\text{x}}{\sqrt{4\text{x}^4+16\text{x}^2+1}\sqrt{53+\text{x}^2}}<0$

$\Rightarrow14\text{x}^2-7\text{x}<0$

$\Rightarrow2\text{x}^2-\text{x}<0$

$\Rightarrow\text{x}(2\text{x}-1)<0$

$\Rightarrow\text{x}<0\ \&\ 2\text{x}-1>0$ or $\text{x}>0\ \&\ 2\text{x}-1<0$

$\Rightarrow\text{x}<0\ \&\ \text{x}>\frac{1}{2}$ or $\text{x}>0\ \&\ \text{x}<\frac{1}{2}$

$\Rightarrow\text{x}>0\ \&\ \text{x}<\frac{1}{2}$ (As there cannot be any number less than zero and greater than $\frac{1}{2}$)

$\Rightarrow\text{x}\in(0,\frac{1}{2})\dots(1)$

Let the equation of the z-axis be $\text{z}\hat{\text{k}}.$

And let the angle between $\vec{\text{b}}$ and z-axis be B.

$\therefore\cos\text{B}=\frac{\big(7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}\big).\big(\text{z}\hat{\text{k}}\big)}{\big|7\hat{\text{i}}-2\hat{\text{j}}+\text{x}\hat{\text{k}}\big|\big|\text{z}\hat{\text{k}}\big|}$

$=\frac{\text{xz}}{\text{z}\sqrt{49+4+\text{x}^2}}$

$=\frac{\text{x}}{\sqrt{53+\text{x}^2}}$

Now, angle B is acute and less than $\frac{\pi}{6}.$

$\therefore0<\frac{\text{x}}{\sqrt{53+\text{x}^2}}<\cos\frac{\pi}{6}$

$\Rightarrow0<\text{x}<\frac{\sqrt{3}}{2}\sqrt{53+\text{x}^2}\dots(2)$

From (1) and (2) we get

$0<\text{x}<\frac{1}{2}$

2. Unit vector

Solution:

Let $\vec{\text{a}}=(\cos\text{a}\cos\beta)\hat{\text{i}}+(\cos\text{a}\sin\beta)\hat{\text{j}}+(\sin\text{a})\hat{\text{k}}$

$|\vec{\text{a}}|=\sqrt{\cos^2\text{a}\cos^2\beta+\cos^2\text{a}\sin^2\beta+\sin^2\text{a}}$

$=\sqrt{\cos^2\text{a}(\cos^2\beta+\sin^2\beta)+\sin^2\text{a}}$

$=\sqrt{\cos^2\text{a}(1)+\sin^2\text{a}}$

$=\sqrt{\cos^2\text{a}+\sin^2\text{a}}$

$=\sqrt{1}$

$=1$

So, $\vec{\text{a}}$ is a unit vector.

3. 0

Solution:

Given that

$|\vec{\text{a}}|=|\vec{\text{a}}|$

$\Rightarrow\big(\vec{\text{a}}+\vec{\text{b}}\big).\big(\vec{\text{a}}-\vec{\text{b}}\big)=|\vec{\text{a}}|^2-\big|\vec{\text{b}}\big|^2$

$|\vec{\text{a}}^2-|\vec{\text{a}}|^2$

$=0$

2. $\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{b}}}{\big|\vec{\text{b}}\big|^2}$

Solution:

The orthogonal projection of $​\vec{\text{a}}$ on $​\vec{\text{b}}$ is

$\frac{\big(\vec{\text{a}}.\vec{\text{b}}\big)\vec{\text{b}}}{\big|\vec{\text{b}}\big|^2}$

2. $\hat{\text{i}}$

Solution:

Let $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}$

$\vec{\text{a}}.\hat{\text{i}}=\text{a}_1$

and $\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\text{a}_1+\text{a}_2$

and $\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=\text{a}_1+\text{a}_2+\text{a}_3$

Given,

$\vec{\text{a}}.\hat{\text{i}}=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}\big)=\vec{\text{a}}.\big(\hat{\text{i}}+\hat{\text{j}}+\hat{\text{k}}\big)=1$

$\Rightarrow\text{a}_1=\text{a}_1+\text{a}_2=\text{a}_1+\text{a}_2+\text{a}_3=1$

$\Rightarrow\text{a}_1=1,\text{a}_2=0,\text{a}_3=0$

So, $\vec{\text{a}}=\text{a}_1\hat{\text{i}}+\text{a}_2\hat{\text{j}}+\text{a}_3\hat{\text{k}}=1\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}=\hat{\text{i}}$

3. $\cos \theta = -\frac{4}{5}$

Solution:

Given that

$2\vec{\text{a}}+\vec{\text{b}}=\vec{\text{p}}\dots(1)$

$\vec{\text{a}}+2\vec{\text{b}}=\vec{\text{q}}\dots(2)$

Solving these two we get

$\vec{\text{a}}=\frac{2\vec{\text{p}}-\vec{\text{q}}}{3},\vec{\text{b}}=\frac{2\vec{\text{q}}-\vec{\text{p}}}{3}$

And we have

$\vec{\text{p}}=\hat{\text{i}}+\hat{\text{j}}$ and $\vec{\text{q}}=\hat{\text{i}}-\hat{\text{j}}$

Substituting the values of $\vec{\text{p}}$ and $\vec{\text{q}},$ we get

$\vec{\text{a}}=\frac{2\vec{\text{p}}-\vec{\text{q}}}{3}=\frac{2\big(\hat{\text{i}}+\hat{\text{j}}\big)-\big(\hat{\text{i}}-\hat{\text{j}}\big)}{3}=\frac{\hat{\text{i}}+3\hat{\text{j}}}{3}$

$\Rightarrow|\vec{\text{a}}|=\frac{1}{3}\sqrt{1+9}=\frac{\sqrt{10}}{3}$

$\vec{\text{b}}=\frac{2\vec{\text{q}}-\vec{\text{p}}}{3}=\frac{2\big(\hat{\text{i}}-\hat{\text{j}}\big)-\big(\hat{\text{i}}+\hat{\text{j}}\big)}{3}=\frac{\hat{\text{i}}-3\hat{\text{j}}}{3}$

$\Rightarrow\big|\vec{\text{b}}\big|=\frac{1}{3}\sqrt{1+9}=\frac{\sqrt{10}}{3}$

$\vec{\text{a}}.\vec{\text{b}}=\frac{1}{9}(1-9)=\frac{-8}{9}$

We know that

$\vec{\text{a}}.\vec{\text{b}}=|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta$

$\Rightarrow\frac{-8}{9}=\frac{\sqrt{10}}{3}\times\frac{\sqrt{10}}{3}\cos\theta$

$\Rightarrow\frac{-8}{9}=\frac{10}{9}\cos\theta$

$\Rightarrow\cos\theta=\frac{-8}{9}\times\frac{9}{10}=\frac{-4}{5}$

4. $\frac{2\pi}{3}$

Solution:

We have

$|\vec{\text{a}}|=1$ and $\big|\vec{\text{b}\big|}=1$

Now, $\big|\vec{\text{a}}+\vec{\text{b}}\big|=1$

$\Rightarrow|\vec{\text{a}}|^2+\big|\vec{\text{b}\big|}^2+2\vec{\text{a}}.\vec{\text{b}}=1$

$\Rightarrow1+1+2|\vec{\text{a}}|\big|\vec{\text{b}}\big|\cos\theta=1$

$\Rightarrow2+2\cos\theta=1$

$\Rightarrow2\cos\theta=-1$

$\Rightarrow\cos\theta=\frac{-1}{2}$

$\Rightarrow\theta=\frac{2\pi}{3}$

3. R - [-4, 7]

Solution:

Let $\theta$ be the angle between $\vec{\text{a}}$ and $\vec{\text{b}}.$

$\cos\theta=\frac{\vec{\text{a}}.\vec{\text{b}}}{|\vec{\text{a}}|\big|\vec{\text{b}}\big|}=\frac{\text{x}^2-3\text{x}-28}{\sqrt{\text{x}^2+3^2+49}\sqrt{\text{x}^2+\text{x}^2+4^2}}$

For $\theta$ to be acute,

$\cos\theta>0$

$\Rightarrow\text{x}^2-3\text{x}-28>0$

$\Rightarrow(\text{x}-7)(\text{x}+4)>0$

$\Rightarrow\text{x}\in(-\infty,-4)\cup(7,\infty)$

$\Rightarrow\text{x}\in\text{R}-[-4,7]$

4. $\frac{2\pi}{3}<\theta<\pi$

Solution:

We have

$\big|\vec{\text{a}}+\vec{\text{b}}\big|<1$

$\Rightarrow\sqrt{|\vec{\text{a}}|^2+\big|\vec{\text{b}}\big|^2+2|\vec{\text{a}}|\times\big|\vec{\text{b}}\big|\cos\theta<1}$

$\Rightarrow\sqrt{1^2+1^2+2\times1\times1\times\cos\theta<1}$

$\Rightarrow\sqrt{2+2\cos\theta}<1$

$\Rightarrow\sqrt{2(1+\cos\theta)}<1$

$\Rightarrow\sqrt{2\times2\cos^2\frac{\theta}{2}}<1$

$\Rightarrow2\big|\cos\frac{\theta}{2}\big|<1$

$\Rightarrow\big|\cos\frac{\theta}{2}\big|<\frac{1}{2}$

$\Rightarrow\frac{\pi}{3}<\frac{\theta}{2}<\frac{2\pi}{3}$

$\Rightarrow\frac{2\pi}{3}<\theta<\frac{4\pi}{3}$

But here $\theta$ cannot be more than $\pi.$

3. $-\frac{5}{\sqrt{162}}$

Solution:

$\overrightarrow{\text{PQ}}=\overrightarrow{\text{OQ}}-\overrightarrow{\text{OP}}=5\hat{\text{i}}-2\hat{\text{j}}+4\hat{\text{k}}-\big(\hat{\text{i}}+3\hat{\text{j}}-7\hat{\text{k}}\big)=4\hat{\text{i}}-5\hat{\text{j}}+11\hat{\text{k}}$

The unit vector along y-axis is $\hat{\text{j}}.$

Let $\theta$ be the required angle.

$\cos\theta=\frac{\overrightarrow{\text{PQ}}.\hat{\text{j}}}{\big|\overrightarrow{\text{PQ}}\big|\big|\hat{\text{j}}\big|}=\frac{-5}{\sqrt{16+25+121}\sqrt{1}}=\frac{-5}{\sqrt{162}}$