Tag: maths

Questions Related to maths

$\displaystyle a\times \left ( b+c \right )+b\times \left ( c+a \right )+c\times \left ( a+b \right )$ is equal to

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $\displaystyle 2\left [ a:b:c \right ]$

  2. $0$

  3. $3$

  4. None of these

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Correct Option: B
Explanation:

$\displaystyle a\times \left ( b+c \right )+b\times \left ( c+a \right )+c\times \left ( a+b \right )$
$=a\times b+a\times c+b\times c+b\times a+c\times a+c\times b$
$=a\times b-c\times a+b\times c-a\times b+c\times a-b\times c = 0$

Let $\displaystyle a=i+j$ and $\displaystyle b=2i-k,$ the point of intersection of the lines $\displaystyle r\times a=b\times a $ and $\displaystyle r\times b=a\times b $ is

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $\displaystyle -i+j+k$

  2. $\displaystyle 3i-j+k$

  3. $\displaystyle 3i+j-k$

  4. $\displaystyle i-j-k$

Show answer
Correct Option: C
Explanation:

$\vec r \times \vec a = \vec b \times \vec a\Rightarrow \vec{r} = \vec{b}+\lambda \vec{a}$
and $\vec r \times \vec b = \vec a \times \vec b\Rightarrow \vec{r} = \vec{a}+\mu \vec{b}$
For intersection of both the lines, $\vec{b}+\lambda \vec{a}=\vec{a}+\mu \vec{b}$
Comparing coefficients, $\lambda=\mu = 1$
Hence point of intersection is $\vec{r}=\vec{a}+\vec{b} = 3\hat{i}+\hat{j}-\hat{k}$

If $\overline{a},\overline{b},\overline{c}$ are three non-zero vectors and $\overline{a}\neq\overline{b}$, $\overline{a}\times\overline{c}=\overline{b}\times\overline{c}$, then

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $\overline{a}-\overline{b}$ is parallel to $\overline{c}$

  2. $\overline{a}-\overline{b}$ is perpendicular to $\overline{c}$

  3. $\overline{a}+\overline{b}$ is parallel to $\overline{c}$

  4. $\overline{a}+\overline{b}$ is perpendicular to $\overline{c}$

Show answer
Correct Option: A
Explanation:

We get 

$(\vec{a}-\vec{b})\times\vec{c}=0$
So    $(\vec{a}-\vec{b})\parallel \vec{c}$

If $a +2b +3c = 0$, then $a \times b + b\times c + c\times a = ka\times b,$
Where $k$ is equal to ?

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: C
Explanation:

Given,

$\begin{array}{l} a+2b+3c=0 \ a=-\left( { 2b+3c } \right)  \ a\times a=-\left( { 2\left( { b\times a } \right) +3\left( { c\times a } \right)  } \right)  \ \Rightarrow 0=-\left( { 2b\times a+3\left( { c\times a } \right)  } \right) ..........\left( i \right)  \end{array}$
Again,
$\begin{array}{l} 2b=-\left( { a+3c } \right)  \ 2b\times b=-\left( { a\times b+3c\times b } \right)  \ \Rightarrow 0=-\left( { a\times b+3c\times b } \right) ..........\left( { ii } \right)  \end{array}$
equation (i) and (ii)
$\begin{array}{l} -\left( { 2b\times a+3c\times a } \right) +\left( { a\times b } \right) +3c\times b=0 \ \Rightarrow 3a\times b-3c\times a-3b\times c=0 \end{array}$
We know that,
$a \times b =  - b \times a$
$a \times b = c \times a + b \times c...............\left( {iii} \right)$
Now,$ATQ;$
$a \times b + b \times c + c \times a = K\,\,\,a \times b$
$1 = 2a \times b = k\,a \times b$         {from equation (iii)}
$\therefore k=2$
Option $C$ is correct answer.


If $\left| \vec { a }  \right| =1,\ \left| \vec { b }  \right| =2,\ (\vec { a },\vec { b })=\dfrac{2\pi}{3}$ then $\left{(\vec { a } +3\vec { b } )\times \left( 3\vec { a } -\vec { b }  \right) \right}^{2}=$


maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $425$

  2. $\dfrac{147}{2}$

  3. $325$

  4. $300$

Show answer
Correct Option: B
Explanation:

$|\bar{a}|=1, |\bar{b}|=2, (\bar{a}, \bar{b})=\dfrac{2z}{3}$


$[|\bar{a}|+3|\bar{b}|]=1+3\times 2=1+6=7$ ......... $(1)$


Also, $3|\bar{a}|-|\bar{b}|=3\times 1-2=3-2=1$ ......... $(2)$

$|(\bar{a}+3\bar{b})\times (3\bar{a}-\bar{b})|=|(\bar{a}+3\bar{b})||3\bar{a}-\bar{b}|\sin \theta$

$=|\bar{a}+3\bar{b}||3\bar{a}-\bar{b}|\sin (\bar{a}, \bar{b})$

$=|(\bar{a}+3\bar{b}||3\bar{a}-\bar{b}|\sin\left(\dfrac{2z}{3}\right)$

$=|\bar{a}+3\bar{b}||3\bar{a}-\bar{b}|\sin 60^{o}$

$|\bar{a}+3\bar{b}||3\bar{a}-\bar{b}|\left(\dfrac{\sqrt{3}}{2}\right)$

from $(1)$ & $(2)$

$=7\times 1\dfrac{\surd{3}}{2}=\dfrac{7\sqrt{3}}{2}$

$[(\bar{a}+3\bar{b}\times (3\bar{a}-\bar{b})]^{2}=\left(7\dfrac{\sqrt{3}}{2}\right)^{2}$

$=\dfrac{49\times 3}{4}=\dfrac{147}{2}$

If $\vec a = \hat i + \hat j + \hat k,\,\vec b = \hat i + \hat j,\,\,\hat c = \hat i$ and $\left( {\vec a \times \vec b} \right) \times \vec c = \lambda \vec a \times \mu \vec b$ then $\lambda  + \mu $

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $0$

  2. $1$

  3. $2$

  4. $3$

Show answer
Correct Option: A
Explanation:
Few things to consider 
$\hat i\times \hat i = \hat j\times \hat j=\hat k\times  \hat k =0$
$\hat i\times \hat j = \hat k , \hat j \times \hat k=\hat i , \hat k \times \hat i =\hat k$
$\hat j\times \hat i = -\hat k , \hat i \times \hat k=-\hat j , \hat k \times \hat j =-\hat i$
Now, $(\vec a \times \vec b) \times \vec c \Rightarrow $
$\vec a =(\hat i+\hat j+\hat k), \vec b=\hat i+\hat j, \vec c=\hat i$
$\Rightarrow \ (\hat i+\hat j+\hat k) \times (\hat i+\hat j)\times (\hat i)$
$\Rightarrow \ (\hat k-\hat k+\hat j-\hat i)\times (\hat i)$
$\Rightarrow \ -\hat k$
Which does not involve any term like $\lambda \vec a\times \mu \vec b \ \therefore \ \lambda +\mu =0$


Let $\vec{a} = \widehat{i} + \widehat{j}$, $\vec{b} = 2 \widehat{i} - \widehat{k}$, then vector $\vec{r}$ satisfying the equations $\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$ and $\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$ is

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $\widehat{i} - \widehat{j} + \widehat{k}$

  2. $3\widehat{i} - \widehat{j} + \widehat{k}$

  3. $3\widehat{i} + \widehat{j} - \widehat{k}$

  4. $\widehat{i} - \widehat{j} - \widehat{k}$

Show answer
Correct Option: C
Explanation:

$\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$ or $(\vec{r} - \vec{b}) \times \vec{a} = 0$
$\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$ or $(\vec{r} - \vec{a}) \times \vec{b} = 0$
If $\vec{r} = x \widehat{i} + y \widehat{j} + z\widehat{k}$, then 


$\begin{vmatrix}\widehat{i} & \widehat{j} & \widehat{k}\ x-2 & y & z+1\ 1 & 1 & 0\end{vmatrix} = 0$ and $\begin{vmatrix}\widehat{i} & \widehat{j} & \widehat{k}\ x-1 & y-1 & z\ 2 & 0 & -1 \end{vmatrix}= 0$

$\Rightarrow z + 1 = 0, x - y = 2$
and $y -1 = 0, x - 1 + 2z = 0$
$\Rightarrow x = 3, y = 1, z = -1$

If the vector $\bar{c}, \bar{a} = x\bar{i}+y\bar{j}+ z\bar{k}, \bar{b}= \bar{j}$ are such that $\bar{a}, \bar{c}, \bar{b}$ from R.H.S then $\bar{c}$ = 

maths addition of vectors vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $z\bar{i} -x\bar{k}$

  2. $z\bar{i} -3\bar{k}$

  3. $x\bar{j} -y\bar{k}$

  4. $y\bar{j} -x\bar{k}$

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Correct Option: A

If $a,b,c$ are unit vectors, then the maximum value of $|a+2b|^{2}+|b+3c|^{2}+|c+4a|^{2}$ is 

maths addition of vectors vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $50$

  2. $21$

  3. $48$

  4. $58$

Show answer
Correct Option: A
Explanation:
Given,

$\left|a+2b\right|^2+\left|b+3c\right|^2+\left|c+4a\right|^2$

$=\left(a+2b\right)^2+\left(b+3c\right)^2+\left(4a+c\right)^2$

$=a^2+4ab+4b^2+b^2+6bc+9c^2+c^2+8ac+16a^2$

$=17a^2+4ab+8ac+5b^2+10c^2+6bc$

$=50$

If $\displaystyle \bar{a}+p\bar{b}+q\bar{c}=0 $ then

maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

  1. $\displaystyle p(\bar{a}\times\bar{b})=pq(\bar{b}\times\bar{c})=q(\bar{c}\times\bar{a})$

  2. $\displaystyle \bar{a}\times\bar{b}=pq(\bar{c}\times\bar{a})$

  3. $\displaystyle \bar{c}\times\bar{a}=p(\bar{a}\times\bar{b})$

  4. $\displaystyle \bar{a}\times\bar{c}=q(\bar{b}\times\bar{c})$

Show answer
Correct Option: A
Explanation:

$\bar{a}+p\bar{b}+q\bar{c}=0$
$\bar{b}=\dfrac{-q\bar{c}-\bar{a}}{p}$

Then
$p(\bar{a}\times b)$
$=p(\bar{a}\times (\dfrac{-q\bar{c}-\bar{a}}{p}))$

$=((q\bar{c}+\bar{a})\times \bar{a})$
$=q(\bar{c}\times \bar{a})$
$=q(\bar{c}\times (-p\bar{b}-q\bar{c}))$
$=q((p\bar{b}+q\bar{c})\times \bar{c})$
$=pq(\bar{b}\times \bar{c})$.