Mathematics · Physics

Vector Algebra and Calculus

214 Questions

Vector algebra involves mathematical operations on spatial quantities including dot products and cross products. These questions test the understanding of vector spaces and linear combinations. This topic is crucial for advanced mathematics and physics exams.

Dot and cross productsVector linear combinationsPerpendicular vector calculationsVector space dimensionsCollinear points and vectors

Vector Algebra and Calculus Questions

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

The value of $ (\bar { A } +\bar { B } )\times (\bar { A } -\bar { B } )$ is 

  1. $0$
  2. ${A}^{2}-{B}^{2}$
  3. $\bar { B } \times \bar { A }$
  4. $2(\bar { B } \times \bar { A })$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$(\overrightarrow A+\overrightarrow B)\times(\overrightarrow A-\overrightarrow B)=\overrightarrow A\times \overrightarrow A-\overrightarrow A\times \overrightarrow B+\overrightarrow B\times \overrightarrow A-\overrightarrow B\times \overrightarrow B=2(\overrightarrow B\times \overrightarrow A)$


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

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

Let $\vec {a}$ and $\vec {b}$ to two unit vectors. If the vectors $\vec {c}=\hat {a}+2\hat {b}$ and $\vec {d}=5\hat {a}-4\hat {b}$ are perpendicular to each other, then teh angle between $\vec {a}$ and $\vec {b}$ is

  1. $\dfrac {\pi}{3}$
  2. $\dfrac {\pi}{4}$
  3. $\dfrac {\pi}{6}$
  4. $\dfrac {\pi}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since c and d are perpendicular, their dot product is zero: (a + 2b) . (5a - 4b) = 0. Expanding this using a^2 = 1, b^2 = 1, and a . b = cos(theta) gives 5 + 6(a . b) - 8 = 0, so a . b = 1/2, making theta = pi/3.

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})$
$\Rightarrow \vec{A} \parallel (\vec{B} \times \vec{C})$