Maths M.C.Q (1 Marks)

3. $\frac{-2}{3}$

Solution:

Since, two non zero vector $\vec{\text{a}}\ \&\ \vec{\text{b}}$ are i.e.,

$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$

$\vec{\text{b}}=2\hat{\text{i}}+\lambda\hat{\text{k}}$

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

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

$2+3\lambda=0$

$-2=3\lambda$

$\lambda=\frac{-2}{3}$

3. $\frac{-2}{3}$

Solution:

Since, two non zero vector $\vec{\text{a}}\ \&\ \vec{\text{b}}$ are i.e.,

$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$

$\vec{\text{b}}=2\hat{\text{i}}+\lambda\hat{\text{k}}$

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

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

$2+3\lambda=0$

$-2=3\lambda$

$\lambda=\frac{-2}{3}$

3. $\frac{-2}{3}$

Solution:

Since, two non zero vector $\vec{\text{a}}\ \&\ \vec{\text{b}}$ are i.e.,

$\vec{\text{a}}=\hat{\text{i}}-2\hat{\text{j}}+3\hat{\text{k}}$

$\vec{\text{b}}=2\hat{\text{i}}+\lambda\hat{\text{k}}$

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

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

$2+3\lambda=0$

$-2=3\lambda$

$\lambda=\frac{-2}{3}$

2. $-\frac{3}{2}$

3. 1

Solution:

$\hat{\text{i}}.\big(\hat{\text{j}}\times\hat{\text{k}}\big)+\hat{\text{j}}.\big(\hat{\text{i}}\times\hat{\text{k}}\big)+\hat{\text{k}}.\big(\hat{\text{i}}\times\hat{\text{j}}\big)$

$=\hat{\text{i}}.\hat{\text{i}}+\hat{\text{j}}.(-\hat{\text{j}})+\hat{\text{k}}.\hat{\text{k}}$

$=|\hat{\text{i}}|^2-|\hat{\text{j}}|^2+|\hat{\text{k}}|^2$

$=1-1+1$

$=1$

4. 16.

Solution:

Here, $|\vec{{\text{a}}}|=10,|\vec{{\text{b}}}|=2$ and $\vec{{\text{a}}}\cdot\vec{\text{b}}=12$ [given]

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

$12=10\times2\cos\theta$

$\Rightarrow\cos\theta=\frac{12}{20}=\frac{3}{5}$

$\Rightarrow\sin\theta=\sqrt{1-\cos\theta}$

$=\sqrt{1-\frac{9}{25}}$

$\sin\theta=\pm\frac{4}{5}$

$\therefore|\vec{{\text{a}}}\times\vec{{\text{b}}}|=|\vec{{\text{a}}}||\vec{{\text{b}}}||\sin\theta|$

$=10\times2\times\frac{4}{5}$

$=16$

2. $4\overrightarrow{\text{OG}}$

Solution:

Let us consider the point O as origin.

G is the mid-point of AC.

$\therefore\ \overrightarrow{\text{OG}}=\frac{\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}}2$

$2\overrightarrow{\text{OG}}=\overrightarrow{\text{OA}}+\overrightarrow{\text{OC}}\ \dots(1)$

Also, G is the mid-point BD

$\therefore\ \overrightarrow{\text{OG}}=\frac{\overrightarrow{\text{OB}}+\overrightarrow{\text{OD}}}2$

$2\overrightarrow{\text{OG}}=\overrightarrow{\text{OB}}+\overrightarrow{\text{OD}}\ \dots(2)$

On adding (1) and (2) we get,

$2\overrightarrow{\text{OG}}+2\overrightarrow{\text{OG}}=\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}$

$4\overrightarrow{\text{OG}}=\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}$

$\therefore\overrightarrow{\text{OA}}+\overrightarrow{\text{OB}}+\overrightarrow{\text{OC}}+\overrightarrow{\text{OD}}=4\overrightarrow{\text{OG}}$​​​​​​​

1. $\pm1$

Solution:

We have

$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$

$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}$

$=\big|\vec{\text{a}}\times\vec{\text{b}}\big|\big|\vec{\text{c}}\big|\cos0^\circ$ or $\big|\vec{\text{a}}\times{\vec{\text{b}}}\big|\big|\vec{\text{c}}\big|\cos180^\circ$ $\big(\therefore\vec{\text{a}},\vec{\text{b}},\vec{\text{c}}$ are perpendicular to each other$)$

$=\big|\vec{\text{a}}\times\vec{\text{b}}\big|$ or $-\big|\vec{\text{a}}\times\vec{\text{b}}\big|$  $\big(\therefore\big|\vec{\text{c}}\big|=1,\cos0^\circ=1\text{ and }\cos180^\circ=-1\big)$  $$

$=\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin90^\circ$ or $-\big|\text{a}\big|\big|\vec{\text{b}}\big|\sin90$  $\big(\therefore\vec{\text{a}} \text{ is perpendicular to }\vec{\text{b}})$ 

$=1 \text{ or }-1$ $\big(\therefore\big|\vec{\text{a}}\big|=1 \text{ and }\big|\vec{\text{b}}\big|=1\big)$ $$

$=\pm1$

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 $\pi.$

2. $10\hat{\text{i}}+3\hat{\text{j}}+11\hat{\text{k}}$

Solution:

$\vec{\text{a}}\times\vec{\text{b}}=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\2&-3&-1\\1&4&-2 \end{vmatrix}$

$=10\hat{\text{i}}+3\hat{\text{j}}+11\hat{\text{k}}$

3. $-\frac{3}{2}.$

Solution:

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

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

$\Rightarrow\vec{\text{a}}^2+\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{a}}\cdot\vec{\text{c}}+\vec{\text{b}}\cdot\vec{\text{a}}+\vec{\text{b}}^2+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}+\vec{\text{c}}\cdot\vec{\text{b}}+\vec{\text{c}}^2=0$

$\Rightarrow\vec{\text{a}}^2+\vec{\text{b}}^2+\vec{\text{c}}^2+2(\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}})=0$ $[\because\vec{\text{a}}\cdot\vec{\text{b}}=\vec{\text{b}}\cdot\vec{\text{a}},\vec{\text{b}}\cdot\vec{\text{c}}=\vec{\text{c}}\cdot\vec{\text{b}}\text{ and }\vec{\text{c}}\cdot\vec{\text{a}}=\vec{\text{a}}\cdot\vec{\text{c}}]$

$\Rightarrow1+1+1+2(\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}})=0$

$\Rightarrow\vec{\text{a}}\cdot\vec{\text{b}}+\vec{\text{b}}\cdot\vec{\text{c}}+\vec{\text{c}}\cdot\vec{\text{a}}=-\frac{3}{2}$

3. $\frac{1}3\big(\vec{\text{a}}+\vec{\text{b}}\big)$

Solution:

Taking A as origin. Then, position vector of A, B and C are $\vec0,\vec{\text{a}}$ and $\vec{\text{b}}$ respectively. Then,

Centroid G has position vector $\frac{\vec0+\vec{\text{a}}+\vec{\text{b}}}3=\frac{\vec{\text{a}}+\vec{\text{b}}}3$

Therefore, $\text{AG}=\frac{\vec{\text{a}}+\vec{\text{b}}}3-\vec0=\frac{\vec{\text{a}}+\vec{\text{b}}}3$

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)

3. [0,12]

Solution:

We have $|\vec{\text{a}}|=4$ and $-3\leq\lambda\leq2$

Now $|\lambda\vec{\text{a}}|=|\lambda||\vec{\text{a}}|=4|\lambda|$

Now $-3\leq\lambda\leq2$

$\Rightarrow0\leq|\lambda|\leq3$

$\Rightarrow0\leq4|\lambda|\leq12$

$\Rightarrow0\leq|\lambda\vec{\text{a}}|\leq12$

4. None of these

Solution:

We have

$\big(\vec{\text{a}}-\vec{\text{b}}\big).\big[\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\big(\vec{\text{c}}-\vec{\text{a}}\big)\big]$

$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big[\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\vec{\text{c}}-\big(\vec{\text{b}}-\vec{\text{c}}\big)\times\vec{\text{a}}\big]$

$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}-\vec{\text{c}}\times\vec{\text{c}}-\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}\big)$

$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times{\vec{\text{c}}}-0-\vec{\text{b}}\times\vec{\text{a}}+\vec{\text{c}}\times\vec{\text{a}}\big)$

$=\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\big(\vec{\text{a}}-\vec{\text{b}}\big).\big(\vec{\text{c}}\times\vec{\text{a}}\big)$

$=\vec{\text{a}}.\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{b}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)-\vec{\text{a}}\big(\vec{\text{b}}\times\vec{\text{a}}\big)\\+\vec{\text{b}}.\big(\vec{\text{b}}\times\vec{\text{a}}\big)+\vec{\text{a}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)-\vec{\text{b}}.\big(\vec{\text{c}}\times\vec{\text{a}}\big)$

$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-0-0+0+0-\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$ $\big(\therefore\big[\vec{\text{b}}\vec{\text{b}}\vec{\text{c}}\big]=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\big]=\big[\vec{\text{b}}\vec{\text{b}}\vec{\text{a}}\big]=0\big)$

$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]=\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]\big)$

$=0$

4. $4\overrightarrow{\text{AB}}$

Solution:

$\overrightarrow{\text{AD}}=2\overrightarrow{\text{BC}}$

$\overrightarrow{\text{EB}}=2\overrightarrow{\text{FA}}$

$\overrightarrow{\text{FC}}=2\overrightarrow{\text{AB}}$

$\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}=2\Big(\overrightarrow{\text{BC}}+\overrightarrow{\text{FA}}\Big)$

$=2\Big(\overrightarrow{\text{AO}}+\overrightarrow{\text{FA}}\Big)$ $\Big(\because\ \overrightarrow{\text{BC}}=\overrightarrow{\text{AO}}\Big)$

In triangle AOF,

$\overrightarrow{\text{FA}}+\overrightarrow{\text{AO}}+\overrightarrow{\text{FO}}=0$

$\therefore\ \overrightarrow{\text{FA}}+\overrightarrow{\text{AO}}=-\overrightarrow{\text{FO}}$

$\therefore\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}=-2\overrightarrow{\text{FO}}$

And $\overrightarrow{\text{AB}}=-\overrightarrow{\text{FO}}$

$\therefore\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}=2\overrightarrow{\text{AB}}$

$\therefore\overrightarrow{\text{AD}}+\overrightarrow{\text{EB}}+\overrightarrow{\text{FC}}=2\overrightarrow{\text{AB}}+2\overrightarrow{\text{AB}}$

$=4\overrightarrow{\text{AB}}$

3. 3AC = 5CB

Solution:

Draw ON, the perpendicular to the line AB

Let $\vec{\text{i}}$ be the unit vector along ON

The resultant force $\vec{\text{R}}=3\overrightarrow{\text{OA}}+5\overrightarrow{\text{OB}}\ \dots(1)$

The angles between $\vec{\text{i}}$ and the forces $\vec{\text{R}},\ 3\overrightarrow{\text{OA}},\ 5\overrightarrow{\text{OB}}$ are $\angle\text{CON},\ \angle\text{AON},\ \angle\text{BON}$ respectively.

$\vec{\text{R}}.\vec{\text{i}}=3\overrightarrow{\text{OA}}.\vec{\text{i}}+5\overrightarrow{\text{OB}}.\vec{\text{i}}$

$\Rightarrow\text{R}.1.\cos\angle\text{CON}\\=3\overrightarrow{\text{OA}}.1.\cos\angle\text{AON}+5\overrightarrow{\text{OB}}.1.\cos\angle{\text{BON}}$

$\text{R}.\frac{\text{ON}}{\text{OC}}=3\text{OA}\times\frac{\text{ON}}{\text{OA}}+5\text{OB}\frac{\text{ON}}{\text{OB}}$

$\frac{\text{R}}{\text{OC}}=(3+5)$

$\text{R}=8\overrightarrow{\text{OC}}$

We know that,

$\overrightarrow{\text{OA}}=\overrightarrow{\text{OC}}+\overrightarrow{\text{CA}}$

$\Rightarrow3\overrightarrow{\text{OA}}=3\overrightarrow{\text{OC}}+3\overrightarrow{\text{CA}}\ \dots(\text{i})$

$\overrightarrow{\text{OB}}=\overrightarrow{\text{OC}}+\overrightarrow{\text{CB}}$

$\Rightarrow5\overrightarrow{\text{OB}}=5\overrightarrow{\text{OC}}+5\overrightarrow{\text{CB}}\ \dots(\text{ii})$

On adding (i) and (ii) we get,

$3\overrightarrow{\text{OA}}+5\overrightarrow{\text{OB}}=8\overrightarrow{\text{OC}}+3\overrightarrow{\text{CA}}+5\overrightarrow{\text{CB}}$

$\vec{\text{R}}=8\overrightarrow{\text{OC}}+3\overrightarrow{\text{CA}}+5\overrightarrow{\text{CB}}$

$8\overrightarrow{\text{OC}}=8\overrightarrow{\text{OC}}+3\overrightarrow{\text{CA}}+5\overrightarrow{\text{CB}}$

$\Big|3\overrightarrow{\text{AC}}\Big|=\Big|5\overrightarrow{\text{CB}}\Big|$

$\Rightarrow3\overrightarrow{\text{AC}}=5\overrightarrow{\text{CB}}$

1. 300

Solution:

$\big(\vec{\text{a}}+3\vec{\text{b}}\big)\times\big(3\vec{\text{a}}-\vec{\text{b}}\big)$

$=3\big(\vec{\text{a}}\times\vec{\text{a}}\big)-\vec{\text{a}}\times\vec{\text{b}}+9\big(\vec{\text{b}}\times\vec{\text{a}}\big)-3\big(\vec{\text{b}}\times\vec{\text{b}}\big)$

$=3(0)-\vec{\text{a}}\times\vec{\text{b}}-9\big(\vec{\text{a}}\times\vec{\text{b}}\big)-3(0)$

$=-10\big(\vec{\text{a}}\times\vec{\text{b}}\big)$

Now,

$\big|\big(\vec{\text{a}}\times3\vec{\text{b}}\big)\times\big(3\vec{\text{a}}-\vec{\text{b}}\big)\big|^2$

$=\big|-10\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|^2$

$=100\big|\big(\vec{\text{a}}\times\vec{\text{b}}\big)\big|^2$

$=100|\vec{\text{a}}|^2\big|\vec{\text{b}}\big|^2\sin^2120$

$=100(1)^2(2)^2\Big(\frac{\sqrt{3}}{2}\Big)^2$

$=400\times\frac{3}{4}$

$=300$

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

Solution:

We have

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

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

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

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

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

$\Rightarrow2\cos\theta=-1$

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

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

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

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

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

Solution:

$ \vec{ \text{AB}}=\langle{2-1, 1-2,-1+1}\rangle=\langle{1, -1, 0}\rangle\therefore \vec{ \text{AB}}=\hat{\text{i}}-\hat{\text{j}}$

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

Solution:

We have

$\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\times\vec{\text{b}}\big]+\big(\vec{\text{a}}.\vec{\text{b}}\big)^2$

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

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

$=\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\sin\theta\big)^2+\big(\big|\vec{\text{a}}\big|\big|\vec{\text{b}}\big|\cos\theta\big)^2$

$=\big|\vec{\text{a}}\big|^2\big|{\text{b}}\big|^2\sin^2\theta+\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\cos^2\theta$

$=\big|\vec{\text{a}}\big|^2\big|\vec{\text{b}}\big|^2\big(\sin^2\theta+\cos^2\theta\big)$

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

4. $-\frac{5}{2}$

Solution:

Given two non-zero vectors $\vec{\text{a}}$ and $\vec{\text{b}}$ are orthogonal

$\therefore\ \vec{\text{a}}\cdot\vec{\text{b}}=0$

$\therefore\ (2\hat{\text{i}}+\lambda\hat{\text{j}}+\hat{\text{k}})\cdot(\hat{\text{i}}+2\hat{\text{j}}+\hat{3\text{k}})=0$

$\Rightarrow2+2\lambda+3=0$

$\Rightarrow\lambda=-\frac{5}{2}$

1. 0

Solution:

We have

$\big[\vec{\text{a}}-\vec{\text{b}},\vec{\text{b}}-\vec{\text{c}},\vec{\text{c}}-\vec{\text{a}}\big]$

$=\big(\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big).\big(\vec{\text{c}}.\vec{\text{a}}\big)$ (By definition of scalar triple product)

$=\big(\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\vec{\text{b}}-\big(\vec{\text{a}}-\vec{\text{b}}\big)\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$

$=\big(\vec{\text{a}}\times\vec{\text{b}}-\vec{\text{b}}\times{\vec{\text{b}}}-\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$

$=\big(\vec{\text{a}}\times\vec{\text{b}}-0-\vec{\text{a}}\times\vec{\text{c}}+\vec{\text{b}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$

$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)-\big(\vec{\text{a}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)+\big(\vec{\text{b}}\times\vec{\text{c}}\big).\big(\vec{\text{c}}-\vec{\text{a}}\big)$

$=\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{c}}-\big(\vec{\text{a}}\times\vec{\text{b}}\big).\vec{\text{a}}-\big(\vec{\text{a}}\times\vec{\text{c}}\big).\vec{\text{c}}+\big(\vec{\text{a}}\times\vec{\text{c}}\big).\vec{\text{a}}+\big(\vec{\text{b}}\times\vec{\text{c}}\big).\vec{\text{c}}\big(\vec{\text{b}}\times\vec{\text{c}}\big)\\.\vec{\text{a}}\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{a}}\big]-\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{c}}\big]+\big[\vec{\text{a}}\vec{\text{c}}\vec{\text{a}}\big]+\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{c}}\big]-\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]$

$=\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]-0+0+0-\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]$ $\big(\therefore\big[\vec{\text{a}}\vec{\text{b}}\vec{\text{c}}\big]=\big[\vec{\text{b}}\vec{\text{c}}\vec{\text{a}}\big]=\big[\vec{\text{c}}\vec{\text{a}}\vec{\text{b}}\big]\big)$

$=0$

4. $2\vec{\text{a}}^2$

Solution:

Let $\vec{\text{a}}^2=\text{x}\hat{\text{i}}+\text{y}\hat{\text{j}}+\text{z}\hat{\text{k}}$

$\therefore\vec{\text{a}}^2=\text{x}^2+\text{y}^2+\text{z}^2$

$\therefore\ \vec{\text{a}}\times\vec{\text{i}}\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}} \\\text{x}&\text{y}&\text{z}\\1&0&0 \end{vmatrix}$

$=\hat{\text{i}}[0]-\hat{\text{j}}[-\text{z}]+\hat{\text{k}}[-\text{y}]$

$=\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}}$

$\therefore\ (\vec{\text{a}}\times\hat{\text{i}})=(\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}})(\text{z}\hat{\text{j}}-\text{y}\hat{\text{k}})$

$=\text{y}^2+\text{z}^2$

Similarly, $(\vec{\text{a}}\times\hat{\text{j}})^2=\text{x}^2+\text{z}^2$

And $(\vec{\text{a}}\times\hat{\text{k}})^2=\text{x}^2+\text{y}^2$

$\therefore\ (\vec{\text{a}}\times\hat{\text{i}})^2+(\vec{\text{a}}\times\hat{\text{j}})^2+(\vec{\text{a}}\times\hat{\text{k}})^2$ $=\text{y}^2+\text{z}^2+\text{x}^2+\text{z}^2+\text{x}^2+\text{y}^2$

$2(\text{x}^2+\text{y}^2+\text{z}^2)=2\vec{\text{a}}^2$

2. Density

Solution:

Density is a scalar quantity as it has only magnitude but no direction. Speed, force, velocity has both magnitude and direction.

$\therefore$ They all are vectors.

1. $\sqrt{34}$

Solution:

Vector, V = -3i + 5j

Magnitude of the vector, V

$\mid\text{v}\mid=\sqrt{{(-3)^2}+5^2}=\sqrt{(9+25)}=\sqrt{34}$

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)]

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

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

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

Solution:

$\vec{\text{a}}+\vec{\text{b}}=0\hat{\text{i}}+3\hat{\text{j}}+\hat{\text{k}}$

$\vec{\text{b}}-\vec{\text{c}}=0\hat{\text{i}}-0\hat{\text{j}}+3\hat{\text{k}}$

$\big(\vec{\text{a}}\times\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)=\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\0&3&1\\0&0&3 \end{vmatrix}$

$=9\hat{\text{i}}$

$\big|\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|=9|\hat{\text{i}}|$

$=9(1)$

$=9$

Unit vector perpendicular to both $\vec{\text{a}}+\vec{\text{b}}$ and $\vec{\text{b}}-\vec{\text{c}}=\frac{\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)}{\big|\big(\vec{\text{a}}+\vec{\text{b}}\big)\times\big(\vec{\text{b}}-\vec{\text{c}}\big)\big|}$

$=\frac{9\hat{\text{i}}}{9}$

$=\hat{\text{i}}$

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

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

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)]

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

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

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

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

4. $\frac{5\vec{\text{a}}}{4}$

Solution:

Let the given points be $\text{A}(2\vec{\text{a}}-3\vec{\text{b}})$ and $\text{B}(\vec{\text{a}}+\vec{\text{b}})$

Let C divides AB in ratio 3 : 1 

Now the position vector of a point C dividing the line segment joining the points P and Q, whose position vectors are p and q in the ratio m : n internally, is given by $\frac{\text{m}\vec{\text{q}}+\text{n}\vec{\text{p}}}{\text{m}+\text{n}}$

$\therefore$ Position vector $\text{C}=\frac{3(\vec{\text{a}}+\vec{\text{b}})+1(\vec{2\text{a}}-3\vec{\text{b}})}{3+1}$

$\Rightarrow\text{C}=\frac{5\vec{\text{a}}}{4}$

1. $\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$

Solution:

Let:

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

$\vec{\beta}=\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}$

Now,

$\vec{\text{b}}=3\hat{\text{i}}+4\hat{\text{k}}=\vec{\alpha}+\vec{\beta}$ (Given)

$\Rightarrow3\hat{\text{i}}+0\hat{\text{j}}+4\hat{\text{k}}=(\text{a}_1+\text{b}_1)\hat{\text{i}}+(\text{a}_2+\text{b}_2)\hat{\text{j}}+(\text{a}_3+\text{b}_3)\hat{\text{k}}$

$\Rightarrow\text{a}_1+\text{b}_1=3;\text{a}_2+\text{b}_2=0;\text{a}_3+\text{b}_3=4$

$\Rightarrow\text{a}_1+\text{b}_1=3;\text{a}_2=-\text{b}_2;\text{a}_3+\text{b}_3=4\dots(1)$

$\vec{\text{a}}=\hat{\text{i}}+\hat{\text{j}}$ (Given)

Also, $\vec{\alpha}$ is parallrl to $\vec{\text{a}}.$

$\Rightarrow\vec{\alpha}\times\vec{\text{a}}=\vec{0}$

$\Rightarrow\begin{vmatrix}\hat{\text{i}}&\hat{\text{j}}&\hat{\text{k}}\\\text{a}_1&\text{a}_2&\text{a}_3\\1&1&0\end{vmatrix}=\vec{0}$

$\Rightarrow-\text{a}_3\hat{\text{i}}+\text{a}_3\hat{\text{j}}+(\text{a}_1-\text{a}_2)\hat{\text{k}}=0\hat{\text{i}}+0\hat{\text{j}}+0\hat{\text{k}}$

$\Rightarrow\text{a}_3=0;\text{a}_1-\text{a}_2=0$

$\Rightarrow\text{a}_3=0;\text{a}_1=\text{a}_2\dots(2)$

Since $\vec{\beta}$ is perpendicular to $\vec{\text{a}},$ we get

$\Rightarrow\vec{\beta}.\vec{\text{a}}=0$

$\Rightarrow\big(\text{b}_1\hat{\text{i}}+\text{b}_2\hat{\text{j}}+\text{b}_3\hat{\text{k}}\big).\big(\hat{\text{i}}.\hat{\text{j}}\big)=0$

$\Rightarrow\text{b}_1+\text{b}_2=0$

$\Rightarrow\text{b}_1=-\text{b}_2\dots(3)$

Solving (1), (2) and (3), we get

$\text{a}_1=\frac{3}{2};\text{a}_2=\frac{3}{2};\text{a}_3=0$

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

$=\frac{3}{2}\hat{\text{i}}+\frac{3}{2}\hat{\text{j}}+0\hat{\text{k}}$

$=\frac{3}{2}\big(\hat{\text{i}}+\hat{\text{j}}\big)$