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Quizzes (598) Questions (13441)

Questions Related to physics

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.

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

The component of vector $2\ \hat {i}+3\hat {j}$ along vector $-\hat {j}+5\hat {i}$ is:

  1. $\dfrac{7}{\sqrt{13}}$
  2. $\dfrac{7}{\sqrt{26}}$
  3. $\dfrac{13}{\sqrt{13}}$
  4. $none\ of\ these$.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The component of A along B is (A dot B) / |B|. A = 2i + 3j, B = 5i - j. A dot B = (2*5) + (3*-1) = 10 - 3 = 7. |B| = sqrt(5^2 + (-1)^2) = sqrt(26). Component = 7/sqrt(26).

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

If $\vec {u},\vec {v}$ and $\vec {w}$ are three non-coplanar vectors, then
$(\vec {u}+\vec {v}-\vec {w}).(\vec {u}-\vec {v})\times (\vec {v}-\vec {w})$ equals

  1. $3\vec {u}.\vec {v} \times \vec {w}$
  2. $0$
  3. $\vec {u}.\vec {v}\times \vec {w}$
  4. $\vec {u}.\vec {w} \times \vec {v}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The scalar triple product (A+B-C) dot ((A-B) x (B-C)) expands using properties of cross products. (A-B) x (B-C) = A x B - A x C - B x B + B x C = A x B - A x C + B x C. Dotting this with (A+B-C) results in terms like A dot (A x B) which are zero. The result is 0.

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

Which of the following vector is perpendicular to the vector $A=2\hat{i}+3\hat{j}+4\hat{k}$?

  1. $\hat{i}+\hat{j}+\hat{k}$
  2. $4\hat{i}+3\hat{j}-2\hat{k}$
  3. $\hat{i}-3\hat{j}+\hat{k}$
  4. $\hat{i}+2\hat{j}-2\hat{k}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

A vector is perpendicular to A if their dot product is zero. For A = 2i + 3j + 4k, check option D: (2*1) + (3*2) + (4*-2) = 2 + 6 - 8 = 0.

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

Which of the following vector is perpendicular to the vector $\vec { A } =\hat { 2i } +\hat { 3j } +\hat { 4k } $?

  1. $\hat { i } +\hat { j } +\hat { k } $
  2. $\hat { 4i } +\hat { 3j } -\hat { 2k } $
  3. $\hat { i } -\hat {3 j } +\hat { k } $
  4. $\hat { i } +\hat { 2j } -2\hat { k } $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

This is an identical duplicate of question 443269. Two vectors are perpendicular if their dot product equals zero. For A = 2i+3j+4k and option D (i+2j-2k), we calculate: 2*1 + 3*2 + 4*(-2) = 2 + 6 - 8 = 0. This confirms they are perpendicular.

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

Find a vector $\vec {x}$ which is perpendicular to both $\vec {A}$ and $\vec {B}$ but has magnitude equal to that of $\vec {B}$. Vector $\vec {A}=3\hat{i}-2 \hat {j} +\hat {k}$ and $\vec {B}=4\hat{i}+3 \hat {j} -2\hat {k}$

  1. $\displaystyle \frac{1}{\sqrt{10}}(\hat{i}+10\hat{j}+17\hat{k})$
  2. $\displaystyle \frac{1}{\sqrt{10}}(\hat{i}-10\hat{j}+17\hat{k})$
  3. $\sqrt {\displaystyle \frac{29}{390}}(\hat{i}-10\hat{j}+17\hat{k})$
  4. $\sqrt {\displaystyle \frac{29}{390}}(\hat{i}+10\hat{j}+17\hat{k})$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\vec {A} \times \vec {B}=\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 3 & -2 & 1\ 4 & 3 & -2\end{vmatrix}$

$\hat {n}=\displaystyle \frac{\vec{A} \times \vec{B}}{|\vec{A} \times \vec{B}|}=\displaystyle \frac{\hat{i}+10\hat{j}+17\hat{k}}{\sqrt{390}}$

$\vec{x}=|\vec{B}|\hat{n}=\displaystyle \frac{\sqrt{29}(\hat{i}+10\hat{j}+17\hat{k})}{\sqrt{390}}$

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

Three vectors $\vec A, \vec B$ and $\vec C$ satisfy the relation $\vec {A}\cdot \vec {B}=0$ and $\vec{A}\cdot \vec{C}=0$. The vector $A$ is parallel to :

  1. $\vec {B}. \vec {C}$
  2. $\vec {B}$
  3. $\vec {C}$
  4. $\vec {B} \times \vec {C}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

$\vec A$ is perpendicular to $\vec B$ and $\vec A$ is perpendicular to $\vec C$. Thus $\vec A$ must lie along the direction of the cross product of $\vec B$ and $\vec C$.

Alternatively:
Given,
$\vec {A}.\vec {B}=0$
$\vec{A}.\vec{C}=0$
$ \Rightarrow \vec {A}.\vec {B} - \vec {A}.\vec {C}=\vec{A}( \vec{B} -\vec{C}) =0$
$ \Rightarrow \vec{A} \perp (\vec{B} -\vec{C})$
$\Rightarrow \vec{A} \parallel (\vec{B} \times \vec{C})$