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#### Let the volume of a parallelopiped whose coterminous edges are given by and be 1 cu. unit. If be the angle between the edges and then can be : Option: 1 Option: 2 Option: 3 Option: 4

Dot (Scalar) Product in Terms of Components -

Angle between two vectors

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Geometrical Interpretation of Scalar Triple Product -

Let vectors and represented the sides of a parallelepiped OA, OB and OC respectively. Then, is a vector perpendicular to the plane of  and . Let ? be the angle between vectors and and α be the angle between and .

If is a unit vector along , then α is the angle between and .

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Correct Option (3)

#### Let  and  be two vectors. If  is a vector such that  and  then  is equal to : Option: 1 Option: 2 Option: 3  Option: 4

Vector Triple Product -

For three  vectors and vector triple product is defined as .

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Correct Option (3)

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#### Let  and  be three unit vectors such that .If  and  then the ordered pair,   is equal to : Option: 1              Option:2  Option: 3 Option: 4

Addition and subtraction of Vectors -

The sum of two vectors is always a vector.

Properties of vector Subtraction

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Correct Option (3)

#### Let  be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle , with the vector . Then  is equal to ________.

Given, $|\vec{a}|=|\vec{b}|=|\vec{c}|$

and these are mutually perpendicular,

So, $\vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0$

Angle of $\vec{a}$ with $(\vec{a}+\vec{b}+\vec{c})=\theta$

\begin{aligned} &\therefore \quad \vec{a} \cdot(\vec{a}+\vec{b}+\vec{c})=|\vec{a}||\vec{a}+\vec{b}+\vec{c}| \cdot \cos \theta \\ &\Rightarrow \quad \vec{a} \cdot \vec{a}+\vec{a} \cdot \vec{b}+\vec{a} \cdot \vec{c}=|\vec{a}||\vec{a}+\vec{b}+\vec{c}| \cos \theta \\ &\Rightarrow \quad|\vec{a}|^{2}+0+0=|\vec{a}||\vec{a}+\vec{b}+\vec{c}| \cos \theta \\ &\Rightarrow \quad \frac{|\vec{a}|}{|\vec{a}+\vec{b}+\vec{c}|}=\cos \theta\; \; \; \; \; \; \; \; \; \; \; \; \; ...........(i) \end{aligned}

Now,

\begin{aligned} &|\vec{a}+\vec{b}+\vec{c}|^{2}=(\vec{a}+\vec{b}+\vec{c}) \cdot(\vec{a}+\vec{b}+\vec{c}) \\ &|\vec{a}+\vec{b}+\vec{c}|^{2}=|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2}+2(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a}) \\ \Rightarrow &|\vec{a}+\vec{b}+\vec{c}|^{2}=3|\vec{a}|^{2}+2(0) \\ \Rightarrow &|\vec{a}+\vec{b}+\vec{c}|=\sqrt{3}|\vec{a}| \end{aligned}

Using this in (i)

\begin{aligned} &\Rightarrow \quad \cos \theta=\frac{1}{\sqrt{3}} \\ &\Rightarrow \quad \cos 2 \theta=2 \cos ^{2} \theta-1=\frac{2}{3}-1=-\frac{1}{3} . \\ &\Rightarrow \quad 36 \cos ^{2} 2 \theta=\quad 36 \cdot\left(\frac{1}{9}\right)=4 . \end{aligned}

Hence, the correct answer is 4

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#### Let . Let a vector be in the plane containing . If is perpendicular to the vector and its projection on is 19 units, then is equal to___________

$Let\, \vec{\vartheta }= x\vec{a}+y\vec{b}$
$\vec{\vartheta }\cdot \left ( 3\hat{i}+2\hat{j}-\hat{k} \right )= 0$
$And\: \vec{\vartheta }\cdot \hat{a}= 19$
$Let\,\: 3\hat{i}+2\hat{j}-\hat{k}= \vec{c}$

$Let\,\: \vec{\vartheta } = \lambda \vec{c}\times \left ( \vec{a}\times \vec{b} \right )$
(a vector in the plane of $\vec{a}$ & $\vec{b}$ and perpendicular  to $\vec{c}$)
$\vec{c}\times \left ( \vec{a}\times \vec{b} \right )= 14\hat{i}-12\hat{j}+18\hat{k}$
$\vec{\vartheta }= \lambda\left ( 14\hat{i}-12\hat{j}+18\hat{k} \right )$
$Also\,\; \vec{\vartheta }\cdot \hat{a}= 19$
$\Rightarrow \lambda \left ( 14\hat{i}-12\hat{j}+18\hat{k} \right )\cdot \frac{\left ( 2\hat{i} -\hat{j}+2\hat{k}\right )}{\sqrt{2^{2}+1^{1}+2^{2}}}= 19$
$\Rightarrow \lambda = \frac{19\times 3}{28+12+36}= \frac{19\times 3}{76}= \frac{3}{4}$
$so\: \left | 2\vartheta ^{2} \right |= \left | 2\times \frac{3}{4}\left ( 14\hat{i}-12\hat{j}+18\hat{k} \right ) \right |^{2}$
$= 9\left ( 7^{2}+6^{2}+9^{2} \right )$
$= 1494$

#### Let be three vectors mutually perpendicular to each other and have same magnitude. If a vector satisfiesthen is equal to :Option: 1Option: 2Option: 3Option: 4

$|\vec{a}|=|\vec{b}|=|\vec{c}|\: and \: \vec{a} \cdot \vec{b}=\vec{b} \cdot \vec{c}=\vec{c} \cdot \vec{a}=0\\$

$Let\: \vec{r}= x\vec{a}+y \vec{b}+z \vec{c} \\$

$where \:\: \vec{r}\cdot \vec{a}=x|\vec{a}|^{2}, \vec{r} \cdot \vec{b}=y|\vec{b}|^{2}, \vec{r} \cdot \vec{c}=z|\vec{c}|^{2}$

Give expression is

$(\vec{a} \times(\vec{r} \times \vec{a}))-(\vec{a} \times(\vec{b} \times \vec{a}))+\vec{b} \times(\vec{r} \times \vec{b})-\vec{b} \times(\vec{c} \times \vec{b})+\\$
$\vec{c} \times(\vec{r} \times \vec{c})-(\vec{c} \times(\vec{a}\times \vec{c))=0} \\$

$\Rightarrow(\vec{a} \cdot \vec{r}) \vec{a}-|\vec{a}|^{2} \vec{r}-(\vec{a} \cdot \vec{b}) \vec{a}+|\vec{a}|^{2} \vec{b}+(\vec{b} \cdot \vec{r}) \vec{b}-|\vec{b}|^{2} \vec{r}-\\$
$(\vec{b} \cdot \vec{c}) \vec{b}+|\vec{b}|^{2}\vec{c}+(\vec{c}\cdot \vec{r})\vec{c}-|\vec{c}|^{2}\vec{r}-(\vec{c}\cdot \vec{a})\vec{a}+|\vec{c}|^{2}\vec{a}=0\\$

$\Rightarrow x|\vec{a}|^{2}\vec{a}+y|\vec{b}|^{2}\vec{b}+z|\vec{c}|^{2}\vec{c}-\vec{r}(|\vec{a}|^{2}+|\vec{b}|^{2}+|\vec{c}|^{2})+\\$
$|\vec{a}|^{2}\vec{b}+|\vec{b}|^{2}\vec{c}+|\vec{c}|^{2}\vec{a}=0\\$

$\Rightarrow |\vec{a}|^{2}(x\vec{a}+y\vec{b}+z\vec{c})-3|\vec{a}|^{2}\vec{r}+|\vec{a}|^{2}(\vec{a}+\vec{b}+\vec{c})=0\\$

$\Rightarrow 3\vec{r}-\vec{r}=\vec{a}+\vec{b}+\vec{c}\\$

$\Rightarrow \vec{r}=\frac{1}{2}\left ( \vec{a}+\vec{b}+\vec{c} \right )$

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#### Let be two vectors such that and the angle between and is . If is a unit vector, then is equal to :Option: 1Option: 2Option: 3Option: 4

$\left | \frac{\vec{a}}{8} \right |= 1\Rightarrow \left | \vec{a} \right |= 8$
$\left | 2\vec{a}+3\vec{b} \right |^{2}= \left | 3\vec{a}+\vec{b} \right |^{2}$
$\Rightarrow 4\left | \vec{a} \right |^{2}+9\left | \vec{b} \right |^{2}+ 12 \, \vec{a}\cdot \; \vec{b}= 9\left | \vec{a} \right |^{2}+\left | \vec{b} \right |^{2}+6\, \vec{a}\,\;\cdot \vec{b}$
$\Rightarrow 5\left | \vec{a} \right |^{2}-8\left | \vec{b} \right |^{2}-6\; \left | \vec{a} \right |\left | \vec{b} \right |\cos 60^{\circ}= 0$
$\Rightarrow 5\times 8^{2}-8\left | \vec{b} \right |^{2}-6\times 8\times \left | \vec{b} \right |\times \frac{1}{2}= 0$
$\Rightarrow \left | \vec{b} \right |^{2}+3 \left | \vec{b} \right |-40= 0$
$\Rightarrow \left | \vec{b} \right |^{2}+8 \left | \vec{b} \right |-5\left | \vec{b} \right |-40= 0$
$\Rightarrow \left | \vec{b} \right |^{2}= -8 \times \left ( Not\: possible \right )$
OR $\left | \vec{b} \right |= 5$
option (4)

#### The angle between the straight lines, whose direction cosines are given by the equations , is :Option: 1Option: 2Option: 3Option: 4

$2l+2m-n= 0\\$

$\Rightarrow n= 2l+2m\\$            ...............(1)

$mn+nl+lm= 0\\$

$\Rightarrow n\left ( m+l \right )+lm= 0\\$

$\Rightarrow 2\left ( l+m \right )\left ( l+m \right )+lm= 0\\$

$\Rightarrow 2l^{2}+4ml+2m^{2}+lm= 0\\$

$\Rightarrow 2l\left ( l+2m \right )+m\left ( l+2m \right )= 0\\$

$\Rightarrow \left ( 2l+m \right )\left ( l+2m \right )= 0\\$

$\Rightarrow m= -2l\: or\: l= -2m\\$               ..........(2)

From (1) and (2)

$n= 2l+2m\\$

$= 2l-4l= -2l\: or\: -4m+2m= -2m\\$

So $m= -2l= n\:\: OR\: \: l= -2m= n\\$

Also $l^{2}+m^{2}+n^{2}= 1\\$

$\Rightarrow l^{2}+4l^{2}+4l^{2}-1\Rightarrow l= \pm \frac{1}{3}\\$

$OR\: \: 4m^{2}+m^{2}+4m^{2}= 1\Rightarrow m= \frac{1}{3}\\$

So $L_{1}:\pm \left ( \frac{1}{3} ,\frac{-2}{3},\frac{-2}{3}\right )\: L_{2}:\pm \left ( \frac{-2}{3},\frac{1}{3}, \frac{-2}{3}\right )\\$

Angle $\cos\theta= \left | l_{1}l_{2}+m_{1}m_{2}+n_{1}n_{2} \right |\\$

$= \left | \frac{-2}{9}-\frac{2}{9}+\frac{4}{9} \right |= 0\\$

$\Rightarrow \theta= 90^{\circ}= \frac{\pi}{2}$

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#### Let be three vectors such that, is perpendicular to . Then the greatest amongst the values of is__________

$\vec{b}\times \vec{c}= \begin{vmatrix} \hat{i} & \hat{j} &\hat{k} \\ 1& 3 &\beta \\ -1& 2 & -3 \end{vmatrix}$
$= \left ( -9-2\beta \right )\hat{i}+\left ( 3-\beta \right )\hat{j}+5\hat{k}$
$\left | \vec{b}\times \vec{c} \right |^{2}= 75$
$\Rightarrow 4\beta ^{2}+36\beta +81+\beta ^{2}-6\beta +9+25= 75$
$\Rightarrow 5\beta ^{2}+30\beta +40= 0$
$\Rightarrow 5\beta ^{2}+20\beta +10\beta +40= 0$
$\Rightarrow \beta = -4 \: \: OR\, \,2$
$\vec{a}\cdot \vec{b}= 0\Rightarrow 1+15+\alpha \beta = 0$
$\Rightarrow \alpha \beta = -16$
$if\, \beta = -4;\alpha = 4,if \: \beta = -2,\alpha = 8$
$\left | \vec{a}\right |^{2}_{max}= 1^{2}+5^{2}+8^{2}$
$=90$

#### If the projection of the vector on the sum of the two vectors and is 1, then is equal to ________.

Sum of vectors ( $\vec{b}$ ) is
$\left ( 2-\lambda \right )i+6j-2k$

Projection of $\vec{a}= i+2j+k$ on $\vec{b}$ is
$\Rightarrow \frac{\vec{a}\cdot \vec{b}}{\left | \vec{b} \right |}= 1$
$\Rightarrow \left ( 2-\lambda \right )+12-2= \sqrt{\left ( 2-\lambda \right )^{2}+36+4}$
$\Rightarrow \left ( 12-\lambda \right )^{2}= \left ( 2-\lambda \right )^{2}+40$
$\Rightarrow \lambda ^{2}+144-24\lambda = \lambda ^{2}+4-4\lambda +40$
$\Rightarrow \lambda = 5$