Tag: scalar product

Quizzes (0) Questions (48)

Questions Related to scalar product

Multiple choice physics mathematical methods multiplication of vectors products of vectors 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 :- 

  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. $6\hat{i}+2\hat{j}-2\hat{k}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $=(\vec{v}\hat{j})\hat{j}=2\sqrt{3}\dfrac{(\hat {  i}+\hat { j }+\hat { k })}{\sqrt{3}}$

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

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

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

  1. $\overline {C} \top \overline {A}$
  2. $\overline {C} \top \overline {B}$
  3. $\overline {C} \top \overline {A} \times \overline {B}$
  4. $\overline {C} \top \overline {A} + \overline {B}$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

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

  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. $\frac{\hat{i}+10\hat{j}+18k}{5\sqrt{17}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The vector perpendicular to two vectors is found by their cross product. Calculating (2i + 2j - k) x (6i - 3j + 2k) gives (4-3)i - (4+6)j + (-6-12)k = i - 10j - 18k. The magnitude is sqrt(1^2 + (-10)^2 + (-18)^2) = sqrt(1 + 100 + 324) = sqrt(425) = 5*sqrt(17).

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

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

  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} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

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

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

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

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

  1. $2\sqrt \frac {A}{B}\vec {B}$
  2. $2\vec {B}$
  3. $zero$
  4. $2 \vec A$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Velocity v = dP/dt = 2Bt. Acceleration a = dv/dt = 2B. Since P = A + Bt^2, at any time, the velocity is in the direction of B. The acceleration is also in the direction of B. The angle between them is 0, not 45 degrees. This question appears to have a conceptual error or typo.

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

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 }  $

  1. $ \frac { 5 }{ \sqrt { 6 } } $
  2. $ \frac { 7 }{ 10 } $
  3. $ \frac { 6 }{ \sqrt { 5 } } $
  4. $ \frac { 5 }{ \sqrt { 3 } }
Reveal answer Fill a bubble to check yourself
A Correct answer
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

Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

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

  1. $6$
  2. $3$
  3. $1$
  4. $0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The resultant magnitude R = 1 for vectors of magnitude 2 and 3 implies the vectors are in opposite directions (3 - 2 = 1). If they are in opposite directions, the angle between them is 180 degrees. The magnitude of the cross product is |A||B| sin(theta). Since sin(180) = 0, the cross product is 0.