Tag: scalar product

Questions Related to scalar product

The velocity of a particle is $\vec{v}=6\hat{i}+2\hat{j}-2\hat{k}.$ The component of the velocity parallel to vector $\vec{a}=\hat{i}+\hat{j}+\hat{k}$ is :- 

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $6\hat{i}+2\hat{j}+2\hat{k}$

  2. $2\hat{i}+2\hat{j}+2\hat{k}$

  3. $\hat{i}+\hat{j}+\hat{k}$

  4. $6\hat{i}+2\hat{j}-2\hat{k}$

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Correct Option: B
Explanation:

$\vec{v}=6\hat{i}+2\hat{j}-2\hat{k}$


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

component $\vec {a  }$ is $(\vec { v } \cdot\hat {  a})$
$\vec {  v}.\hat{a}=\vec{v}.\dfrac{\vec{a}}{|a|}=\vec{v}.\dfrac{(\hat { i }+\hat { j }+\hat { k })}{\sqrt{1+1+1}}$

$\Rightarrow \vec{v}.\hat{a}=\dfrac{(6\hat { i }+2\hat { j }-2\hat { k }).(\hat { i }+\hat { j }+\hat { k })}{\sqrt{3}}$

$\Rightarrow \vec{v}.\hat{a}=\dfrac{6+2-2}{\sqrt{3}}=2\sqrt{3}$

$\Rightarrow$ component $=(\vec{v}\hat{j})\hat{j}=2\sqrt{3}\dfrac{(\hat {  i}+\hat { j }+\hat { k })}{\sqrt{3}}$

$\boxed{component=2\hat { i }+2\hat { i }+2\hat { k }}$

If $\overline {A} \times\overline {B} =\overline {C}$ which of the following statement is not correct?

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $\overline {C} \top \overline {A}$

  2. $\overline {C} \top \overline {B}$

  3. $\overline {C} \top \overline {A} \times \overline {B}$

  4. $\overline {C} \top \overline {A} + \overline {B}$

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Correct Option: C
Explanation:

The cross product $\vec{A}\times\vec{B}$  is defined as a vector $\vec{C}$ that is perpendicular (orthogonal) to both A and B, with a direction given by the right-hand rule and a magnitude equal to the area of the parallelogram that the vectors span.

In C option $\vec{C}$ is not perpendicular to  $\vec{A}\times\vec{B}$  
Hence C option is correct 

Two forces of magnitude 20N and 20N act along the adjacent sides of the parallelogram and the magnitude of the resultant force of these two forces is $20\sqrt{3}$. Then the angle between these forces is:

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $30^0$

  2. $45^0$

  3. $90^0$

  4. $60^0$

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Correct Option: A

What is the unit vector perpendicular to the following vectors $ 2\hat{i} + 2\hat{j}- k$ and $6\hat{i}-3\hat{j}+2k$ 

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $\frac{\hat{i}+10\hat{j}-18k}{5\sqrt{17}}$

  2. $\frac{\hat{i}-10\hat{j}+18k}{5\sqrt{17}}$

  3. $\frac{\hat{i}-10\hat{j}-18k}{5\sqrt{17}}$

  4. $\frac{\hat{i}+10\hat{j}+18k}{5\sqrt{17}}$

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Correct Option: C

$(\overline{A} + \overline{B} )\times ( \overline{A} - \overline{B} )$ is

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $(\overline{A} ^2- \overline{B} ^2)$

  2. $2\overline{A} \overline{B} $

  3. $(\overline{A} \times \overline{B} )$

  4. $( \overline{B} \times \overline{A} $

Show answer
Correct Option: B
Explanation:

$(\bar A+\bar B)\times (\bar A-\bar B)$

$=\bar A \bar A-\bar A\bar B+\bar A\bar B-\bar B \bar B$
$=2\bar A\bar B$ .

The vectors $\vec{A}=4\hat{i}+3\hat{j}+\hat{k}$ and $\vec{B}=12\hat{i}+9\hat{j}+3\hat{k}$ are parallel to each other.

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. True

  2. False

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Correct Option: A

The momentum of a particle is $\vec { P } =\vec { A } +\vec { B } { t }^{ 2 }$, where $\vec { A }$ and $\vec { B }$ are constant perpendicular vectors. The force acting on the particle when its acceleration is at ${45}^{o}$ with its velocity is

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $2\sqrt \frac {A}{B}\vec {B}$

  2. $2\vec {B}$

  3. $zero$

  4. $2 \vec A$

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Correct Option: A

Find the projection of $ \vec A =2\hat { i } -\hat { j } +\hat { k } \quad on\quad \vec  B  =\quad \hat { i } -2\hat { j } +\hat { k }  $

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $ \frac { 5 }{ \sqrt { 6 } } $

  2. $ \frac { 7 }{ 10 } $

  3. $ \frac { 6 }{ \sqrt { 5 } } $

  4. $ \frac { 5 }{ \sqrt { 3 } }

Show answer
Correct Option: A
Explanation:

Given that,

  $ \vec{A}=2\hat{i}-\hat{j}+\hat{k} $

 $ \vec{B}=\hat{i}-2\hat{j}+\hat{k} $

Now, the projection  $\vec{A}$ on $\vec{B}$

  $ =\dfrac{\vec{A}\centerdot \vec{B}}{|\vec{B}|} $

 $ =\dfrac{5}{\sqrt{6}} $

Hence, this is the required solution

The resultant of the two vector is having magnitude 2 and 3 is 1. What is their cross product 

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $6$

  2. $3$

  3. $1$

  4. $0$

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Correct Option: D

The vector of  magnitude 18 which is perpendicular to both vectors $4\hat i-\hat j+3\hat k \,and  -2\hat i+\hat j-2\hat k$  is

physics mathematical methods multiplication of vectors products of vectors scalar product

  1. $12\hat i+ 12 \hat j-6\hat k$

  2. $6\hat i-12\hat j-12\hat k$

  3. $12\hat i+6\hat j+12 \hat k$

  4. $-6\hat i+12\hat j+12\hat k$

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Correct Option: A