Tag: mathematical methods

Questions Related to mathematical methods

If $\overrightarrow a  + b + \overrightarrow c  = 0$ The angle between $\overrightarrow a \,\,and\,\,\overrightarrow b \,,b\,and\,\overrightarrow c \,and\,{150^0}\,\,and\,\,{120^0}$ respectively.The the magnitude of vectors $\overrightarrow a ,\overrightarrow b \,\,and\,\,\overrightarrow c $ are in ratio of .

  1. $1:2:2$

  2. $1:2$:$\sqrt 3 $

  3. $\sqrt 3 $:2:1

  4. $2$:$\sqrt 3 $:1


Correct Option: B

If  $\vec { A } = 4 \vec { i } + 5 \vec { j } - 6 \vec { k }$  and  $\vec { B } = 2 \vec { i }  - 3 \vec { j } + 4 \vec { k }$  then  $( \vec { A } + \vec { B } ) \cdot (\vec { A } - \vec { B } )$  is

  1. $6$

  2. $48$

  3. $67$

  4. $13$


Correct Option: B

If the two given vectors $ 2 \hat i + 3 \hat j + 4 \hat k $ and $ 6 \hat i +  \alpha \hat j + \beta \hat k $ are parallel , the value of $ \alpha $ and $ \beta $ will be

  1. $9$ and $12$

  2. $3$ and $14$

  3. $6$ and $8$

  4. $4$ and $12$


Correct Option: A

If $\overrightarrow{A}=4\widehat{i}+6\widehat{j} $  and  $\overrightarrow{B}=2\widehat{i}+3\widehat{j}$ .Then :

  1. $\overrightarrow{A}.\overrightarrow{B} =29$

  2. $\overrightarrow{A}\times \overrightarrow{B}=0$

  3. $\dfrac{|\overrightarrow{A}|}{|\overrightarrow{B}|}=\dfrac{2}{1} $

  4. angles between $ \overrightarrow{A}$ and $\overrightarrow{B} $ is $ 30^{\circ}$


Correct Option: B
Explanation:
(A) $\vec A.\vec B=(4\hat i+6\hat j).(2\hat i+3\hat j)=8+18=26$

(B) $\overrightarrow{A} \times \overrightarrow{B}= \begin{vmatrix} \widehat{i}& \widehat{j} &\widehat{k}\\  4& 6  & 0\\ 2 & 3 & 0\end{vmatrix} = \widehat{i}(0-0)-\widehat{j}(0-0)+\widehat{k}(12-12)=0 $

(C) $|\vec A|=\sqrt{4^2+6^2}=7.21$
     $|\vec B|=\sqrt{2^2+3^2}=3.60$
     $\dfrac{|\vec A|}{|\vec B|}=2$

(D) Angle between $\vec A$ and $\vec B$ $= cos ^{-1}\dfrac{\vec A.\vec B}{|\vec A|.|\vec B|}=cos^{-1}1=0^0$

If $  \overrightarrow{A} \times \overrightarrow{B}=0,$ $  \overrightarrow{B} \times \overrightarrow{C}=0, $then $  \overrightarrow{A} \times \overrightarrow{C}= $

  1. $AC$

  2. $\dfrac{AB^2}{C} $

  3. $Zero$

  4. $None  of  these$


Correct Option: C
Explanation:

$  \overrightarrow{A} \times \overrightarrow{B}=0   \overrightarrow{A}  \ and\  \overrightarrow{B} $ are parallel.
$  \overrightarrow{B} \times \overrightarrow{C}=0  \Rightarrow  \overrightarrow{B}  \ and \  \overrightarrow{C} $ are parallel 
 $\Rightarrow \vec{A} \parallel \vec{C}$

Consider a vector $F=4\hat{i}-3\hat{j} $. Another vector which is perpendicular to $\vec F$ is:

  1. $ 4\hat{i}+3\hat{j}$

  2. $ 6\hat{i}$

  3. $7\hat{k} $

  4. $ 3\hat{i}-4\hat{j}$


Correct Option: C
Explanation:

The vector perpendicular to the $i$ & $j$ plane would be along the unit vector $k$.
Another vector in $i$ & $j$ plane can be perpendicular to $\vec{F}$
$(3\widehat{i} +4\widehat{j}) \perp (4\widehat{i} -3\widehat{j}) $
But,  from the options only $7\widehat{k} \perp (4\widehat{i} -3\widehat{j})$

Show that the vector is parallel to a vector $\displaystyle \vec{A}=\hat{i}-\hat{j}+2\hat{k}$ is parallel to a vector $\displaystyle \vec{B}=3\hat{i}-3\hat{j}+6\hat{k}.$

  1. $\displaystyle \frac{1}{3}$ times the magnitude of $\displaystyle \vec{B}.$

  2. $\displaystyle \frac{1}{4}$ times the magnitude of $\displaystyle \vec{B}.$

  3. $\displaystyle \frac{1}{2}$ times the magnitude of $\displaystyle \vec{B}.$

  4. None of these


Correct Option: A
Explanation:

A vector $\displaystyle \vec{A}$ is parallel to an another vector $\displaystyle \vec{B}$ if it can be written as
$\displaystyle \vec{A}= m\vec{B}$ where $m$ is a constant.
Here, $\displaystyle \vec{A}=\left ( \hat{i}-\hat{j}+2\hat{k} \right )=\frac{1}{3}\left ( 3\hat{i}-3\hat{j}+6\hat{k} \right )$
or $\displaystyle \vec{A}=\frac{1}{3}\vec{B}$
This implies that $\vec A || \displaystyle \vec{B}$ and magnitude of $\displaystyle \vec{A}$ is $\displaystyle \frac{1}{3}$ times the magnitude of $\displaystyle \vec{B}.$

If $\vec{a}=x _1\hat {i}+y _1\hat {j}$ and $\vec{b}=x _2\hat {i}+y _2\hat {j}$. The condition that would make $\vec{a}$ and $\vec{b}$ parallel to each other is........... .

  1. $x _1y _2=x _2y _1$

  2. $x _1/y _1=x2y2$

  3. $x _1y _1=x _2/y _2$

  4. $x _1y _1=y _2/x _2$


Correct Option: A
Explanation:

$\vec{a}\times\vec{b}=\begin{pmatrix}x _1\hat {i}+y _1\hat {j}\end{pmatrix}\times\begin{pmatrix}x _2\hat {i}+y _2\hat {j}\end{pmatrix}$

$\;\;\;\;\;\;\;\;\;\;\;=x _1y _2\hat {k}-x _2y _1\hat {k}=\vec{0}\Rightarrow x _1y _2=x _2y _1$

A vector $\bar{P} _{1}$ is along the positive x- axis. If its cross product with another vector $\bar{P} _{2}$ is zero, then $\bar{P} _{2}$ could be:

  1. $4\hat{j}$

  2. $-4\hat{i}$

  3. $(\hat{i}+\hat{k})$

  4. $-(\hat{i}+\hat{j})$


Correct Option: B
Explanation:

The vector product of two vectors $\vec{A}$ and $\vec{B}$ is defined by $\vec{A} \times \vec{B} = \hat{n} |A| |B| \sin x $, where  $\hat{n}$ is the unit vector perpendicular to both A and B vectors and x is the angle between them.
Here in this question the vector product of $\vec{P1}$ and $\vec{P2}$ vectors is zero. This is only possible in two cases: 1) Any of the vectors is zero itself or 2) the $\sin$ of the angle between them is zero.
From the given options, the vector parallel to the given vector  $\hat{i}$ is  $-4\hat{i}$. 

If three vectors satisfy the relation $ \overrightarrow A . \overrightarrow B = 0 $ and $ \overrightarrow A . \overrightarrow C = 0 $ , then $ \overrightarrow A $ can be parallel to

  1. $ \overrightarrow C $

  2. $ \overrightarrow B $

  3. $ \overrightarrow B \times \overrightarrow C $

  4. $ \overrightarrow B . \overrightarrow C $


Correct Option: C
Explanation:

$\displaystyle \displaystyle \overrightarrow A .  \overrightarrow B = 0 \Rightarrow  \overrightarrow A \bot  \overrightarrow B$
and, $\overrightarrow A .  \overrightarrow C = 0 \Rightarrow  \overrightarrow A \bot  \overrightarrow C$
Also, $\overrightarrow B \times \overrightarrow C $ is perpendicular to both $\vec { B } \ and  \ \vec { C }$
$Thus, \overrightarrow { A } ||\, (\overrightarrow { B } \times \overrightarrow { C } )$