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Let the volume of a parallelopiped whose coterminous edges are given by \vec{u}=\widehat{i}+\widehat{j}+\lambda \widehat{k},\vec{v}=\widehat{i}+\widehat{j}+3\widehat{k} and \vec{w}=2\widehat{i}+\widehat{j}+\widehat{k} be 1 cu. unit. If \theta be the angle between the edges \vec{u} and \vec{w}, then \cos \theta can be :
Option: 1 \frac{7}{6\sqrt{6}}
Option: 2 \frac{5}{7}
Option: 3 \frac{7}{6\sqrt{3}}
Option: 4 \frac{5}{3\sqrt{3}}
 

 

 

Dot (Scalar) Product in Terms of Components -


Angle between two vectors

\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;}\vec{\mathbf a} \cdot \vec{\mathbf b}=|\vec{\mathbf a}||\vec{\mathbf b}| \cos \theta\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;}\cos \theta=\frac{\vec{\mathbf a} \cdot \vec{\mathbf b}}{|\vec{\mathbf a}||\vec{\mathbf b}| }\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \theta=\cos^{-1}\left (\frac{\vec{\mathbf a} \cdot \vec{\mathbf b}}{|\vec{\mathbf a}||\vec{\mathbf b}| } \right )\\\\\text{If }\;\;\vec {\mathbf a}=a_{1} \hat{\mathbf{i}}+a_{2} \hat{\mathbf{j}}+a_{3} \hat{\mathbf{k}}\;\;\text{and }\;\;\vec{\mathbf b}=b_{1} \hat{\mathbf{i}}+b_{2} \hat{\mathbf{j}}+b_{3} \hat{\mathbf{k}}\\\\\mathrm{\Rightarrow \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;} \theta=\cos ^{-1}\left(\frac{a_{1} b_{1}+a_{2} b_{2}+a_{3} b_{3}}{\sqrt{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}} \sqrt{b_{1}^{2}+b_{2}^{2}+b_{3}^{2}}}\right)

 

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Geometrical Interpretation of Scalar Triple Product -

Let vectors \vec {\mathbf a},\;\;\vec {\mathbf b} and \vec {\mathbf c} represented the sides of a parallelepiped OA, OB and OC respectively. Then, \vec {\mathbf b}\times \vec {\mathbf c} is a vector perpendicular to the plane of \vec {\mathbf b} and \vec {\mathbf c}. Let ? be the angle between vectors \vec{\mathbf b} and \vec {\mathbf c} and α be the angle between \vec {\mathbf a} and \vec {\mathbf b}\times\vec {\mathbf c}.

If \hat {\mathbf n} is a unit vector along \vec {\mathbf b}\times\vec {\mathbf c}, then α is the angle between \hat {\mathbf n} and \vec {\mathbf a}.

\\\left [ \vec{\mathbf a}\;\;\vec{\mathbf b}\;\;\vec{\mathbf c} \right ]= \vec{\mathbf c}\cdot \left ( \vec{\mathbf b}\times \vec{\mathbf c} \right )\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\vec{\mathbf a} \cdot(\mathbf {b} \mathbf {c} \sin \theta\; \hat{\mathbf n})\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=(\mathbf {b} \mathbf {c} \sin \theta)(\vec{\mathbf a} \cdot \hat{\mathbf n})\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=(\mathbf {b} \mathbf {c} \sin \theta)({\mathbf a} \cdot {\mathbf 1}\cdot \cos \alpha)\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=({\mathbf a} \cdot \cos \alpha)(\mathbf {b} \mathbf {c} \sin \theta)\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=(\text{Height} )\cdot \text{(Area of Base)}\\\mathrm{\;\;\;\;\;\;\;\;\;\;\;\;\;\;}=\text{Volume of parallelepiped}

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\\\pm 1=\left|\begin{array}{lll}{1} & {1} & {\lambda} \\ {1} & {1} & {3} \\ {2} & {1} & {1}\end{array}\right| \Rightarrow=-\lambda+3=\pm 1 \Rightarrow \lambda=2 \text { or } \lambda=4 \\ {\text { For } \lambda=4 \frac{1}{90}} \\\\ {\cos \theta=\frac{2+1+4}{\sqrt{6} \sqrt{18}}=\frac{7}{6 \sqrt{3}}}

Correct Option (3)

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Kuldeep Maurya

Let \vec{a}=\hat{i}-2\hat{j}+\hat{k} and \vec{b}=\hat{i}-\hat{j}+\hat{k} be two vectors. If \vec{c} is a vector such that \vec{b}\times \vec{c}=\vec{b}\times \vec{a} and \vec{c}\cdot \vec{a}=0, then \vec{c}\cdot \vec{b} is equal to :
Option: 1 \frac{1}{2}
Option: 2-\frac{3}{2}
Option: 3 -\frac{1}{2}
Option: 4 -1

 

 

Vector Triple Product -

For three  vectors \vec {\mathbf a},\;\vec{\mathbf b} and \vec {\mathbf c} vector triple product is defined as \vec {\mathbf a}\times\left ( \vec{\mathbf b}\times \vec{\mathbf c} \right ).
\vec {\mathbf a}\times\left ( \vec{\mathbf b}\times \vec{\mathbf c} \right )=\left ( \vec {\mathbf a}\cdot \vec {\mathbf c} \right )\cdot \vec {\mathbf b}\;-\;\left ( \vec {\mathbf a}\cdot \vec {\mathbf b} \right )\cdot \vec {\mathbf c}

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\\\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{c}})=\overrightarrow{\mathrm{a}} \times(\overrightarrow{\mathrm{b}} \times \overrightarrow{\mathrm{a}}) \\ {\quad-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{c}}=(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{a}}) \overrightarrow{\mathrm{b}}-(\overrightarrow{\mathrm{a}} \cdot \overrightarrow{\mathrm{b}}) \overrightarrow{\mathrm{a}}} \\ {\quad\quad\quad\quad-4 \overrightarrow{\mathrm{c}}=6(\hat{\mathrm{i}}-\hat{j}+\hat{\mathrm{k}})-4(\hat{\mathrm{i}}-2 \hat{\mathrm{j}}+\hat{\mathrm{k}})} \\ {\quad\quad\quad\quad-4 \overrightarrow{\mathrm{c}}=2 \hat{\mathrm{i}}+2 \hat{\mathrm{j}}+2 \hat{\mathrm{k}}} \\ {\quad\quad\quad\quad\quad\overrightarrow{\mathrm{c}}=-\frac{1}{2}(\hat{\mathrm{i}}+\hat{\mathrm{j}}+\hat{\mathrm{k}})} \\ {\quad\quad\quad\overrightarrow{\mathrm{b}} \cdot \overrightarrow{\mathrm{c}}=-\frac{1}{2}}

Correct Option (3)

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vishal kumar

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Let \vec{a},\vec{b} and \vec{c} be three unit vectors such that \vec{a}+\vec{b}+\vec{c}=\vec{0}.If \lambda =\vec{a}\cdot \vec{b}+\vec{b}\cdot \vec{c}+\vec{c}\cdot \vec{a} and \vec{d} =\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}, then the ordered pair,  \left ( \lambda ,\vec{d} \right ) is equal to :
Option: 1 \left [ \frac{3}{2},3\vec{a}\times \vec{c} \right ]      
      
Option:2  \left [ -\frac{3}{2},3\vec{c}\times \vec{b} \right ]

Option: 3 \left [ -\frac{3}{2},3\vec{a}\times \vec{b} \right ]

Option: 4  \left [ \frac{3}{2},3\vec{b}\times \vec{c} \right ]
 

 

 

Addition and subtraction of Vectors -

 

Properties of vector addition

The sum of two vectors is always a vector.

\\1.\;\;\;\;\vec a+\vec b =\vec b +\vec a \quad \quad\quad\;\;\;\;\;\;\;\;\;\;\;\;\; \text { (Commutative property) }\\2.\;\;\;(\vec{a}+\vec{b})+\vec{c}=\vec{a}+(\vec{b}+\vec{c})\quad \quad \text { (Associative property) }\\3.\;\;\;\;\vec{a}+\overrightarrow{0}=\overrightarrow{0}+\vec{a}=\vec{a}\quad \quad\quad\;\; \text { (additive identity) }\\4.\;\;\;\;\vec{a}+{\left (-\vec a \right )}={\left ( -\vec a \right )}+\vec{a}=\vec{0}\quad \quad \text { (additive inverse) }

 

Properties of vector Subtraction

\\1.\;\;\;\;\vec a-\vec b\neq \vec b-\vec a\\2.\;\;\;\;(\vec a-\vec b )-\vec c\neq \vec a-(\vec b-\vec c)\\3.\;\;\;\;\mathrm{For \;any \;two\;vectors\;\;\ \overrightarrow{a}\;\;and\;\; \overrightarrow{b}}\\\mathrm{\;\;\;\;\;\;\;\;\;\;(a)}\;\; {|a+b| \leq|a|+|b|} \\\mathrm{\;\;\;\;\;\;\;\;\;\;(b)}\;\; {|a+b| \geq|a|-|b|} \\\mathrmlk{\;\;\;\;\;\;\;\;\;\;(c)}\;\; {|a-b| \leq|a|+|b|} \\\mathrm{\;\;\;\;\;\;\;\;\;\;(d)}\;\; {|a-b| \geq|a|-|b|}

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{|\vec{a}+\vec{b}+\vec{c}|^{2}=0} \\ {3+2(\vec{a} \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=0} \\ {(\vec{a} \cdot \vec{b}+\vec{b} \cdot \vec{c}+\vec{c} \cdot \vec{a})=\frac{-3}{2}} \\ {\Rightarrow \lambda=\frac{-3}{2}}

\overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}=\overrightarrow{0}

\\ {\vec{d}=\vec{a} \times \vec{b}+\vec{b} \times(-\vec{a}-\vec{b})+(-\vec{a}-\vec{b}) \times \vec{a}} \\ {=\vec{a} \times \vec{b}+\vec{a} \times \vec{b}+\vec{a} \times \vec{b}} \\ {\vec{d}=3(\vec{a} \times \vec{b})}

Correct Option (3)

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Ritika Jonwal

Let \overrightarrow{\mathrm{a}}, \overrightarrow{\mathrm{b}}, \overrightarrow{\mathrm{c}} be three mutually perpendicular vectors of the same magnitude and equally inclined at an angle \theta , with the vector \overrightarrow{\mathrm{a}}+\overrightarrow{\mathrm{b}}+\overrightarrow{\mathrm{c}}. Then 36 \cos ^{2} 2 \theta 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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Kuldeep Maurya

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Let \overrightarrow{\mathrm{a}}=2 \hat{i}-\hat{j}+2 \hat{k}$ and $\overrightarrow{\mathrm{b}}=\hat{i}+2 \hat{j}-\hat{k}. Let a vector \vec{v}be in the plane containing \vec{a}$ and $\vec{b}. If \vec{v} is perpendicular to the vector 3 \hat{i}+2 \hat{j}-\hat{k} and its projection on \vec{a} is 19 units, then |2 \vec{v}|^{2} 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

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Kuldeep Maurya

Let \vec{a}, \vec{b}, \vec{c} be three vectors mutually perpendicular to each other and have same magnitude. If a vector \vec{r} satisfies
\vec{a} \times\{(\vec{r}-\vec{b}) \times \vec{a}\}+\vec{b} \times\{(\vec{r}-\vec{c}) \times \vec{b}\}+\vec{c} \times\{(\vec{r}-\vec{a}) \times \vec{c}\}=\overrightarrow{0},then \vec{r} is equal to :
Option: 1 \begin{aligned} &\frac{1}{3}(\vec{a}+\vec{b}+\vec{c}) \\ \end{aligned}
Option: 2 \frac{1}{3}(2 \vec{a}+\vec{b}-\vec{c}) \\
Option: 3 \frac{1}{2}(\vec{a}+\vec{b}+\vec{c}) \\
Option: 4 \frac{1}{2}(\vec{a}+\vec{b}+2 \vec{c})

|\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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Kuldeep Maurya

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Let \vec{a}$ and $\vec{b} be two vectors such that |2 \vec{a}+3 \vec{b}|=|3 \vec{a}+\vec{b}| and the angle between \vec{a} and \overrightarrow{\mathrm{b}} is 60^{\circ}. If \frac{1}{8} \overrightarrow{\mathrm{a}} is a unit vector, then |\overrightarrow{\mathrm{b}}| is equal to :
Option: 1 8
Option: 2 4
Option: 3 6
Option: 4 5

\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)

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Kuldeep Maurya

The angle between the straight lines, whose direction cosines are given by the equations 2 l+2 m-n=0$ and $m n+n l+l m=0, is :
Option: 1 \pi-\cos ^{-1}\left(\frac{4}{9}\right)
Option: 2 \frac{\pi}{2}
Option: 3 \cos ^{-1}\left(\frac{8}{9}\right)
Option: 4 \frac{\pi}{3}

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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Kuldeep Maurya

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Let \overrightarrow{\mathrm{a}}=\hat{i}+5 \hat{j}+\alpha \hat{k}, \vec{b}=\hat{i}+3 \hat{j}+\beta \hat{k}$ and $\vec{c}=-\hat{i}+2 \hat{j}-3 \hat{k} be three vectors such that, |\vec{b} \times \vec{c}|=5 \sqrt{3}$ and $\vec{a} is perpendicular to \vec{b}. Then the greatest amongst the values of |\vec{a}|^{2} 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

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Kuldeep Maurya

If the projection of the vector \hat{i}+2\hat{j}+\hat{k} on the sum of the two vectors 2\hat{i}+4\hat{j}-5\hat{k} and -\lambda \hat{i}+2\hat{j}+3\hat{k} is 1, then \lambda 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

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Kuldeep Maurya

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