Mathematics · Physics

Vector Algebra and Calculus

192 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 direction cosines and direction ratios three dimensional geometry maths

Let $\vec{a}, \vec{b}$ and $\vec{c}$ be three non-zero vectors, no two of which are collinear. If the vector $\vec{a}+2\vec{b}$ is collinear with $\vec{c}$ and $\vec{b}+3\vec{c}$ is collinear with $\vec{a}$, then $\vec{a}+2\vec{b}+6\vec{c}$ is equal to.

  1. $\lambda \vec{a}$
  2. $\lambda \vec{b}$
  3. $\lambda \vec{c}$
  4. $\vec{0}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since, $\vec{a}+2\vec{b}$ is collinear with $\vec{c}$.
$\therefore \vec{a}+2\vec{b}=x\vec{c}, x\epsilon R$ and $\vec{b}+3\vec{c}$ is collinear with $\vec{a}$.
$\therefore \vec{b}+3\vec{c}=y\vec{a}, y\epsilon R$
$\Rightarrow \vec{a}+2\vec{b}+6\vec{c}=(1+2y)\vec{a}$
Also, $\vec{a}+2\vec{b}+6\vec{c}=(x+6)\vec{c}$
$\therefore (x+6)\vec{c}=(1+2y)\vec{a}$
$\Rightarrow x+6=0$
and $1+2y=0$
$\Rightarrow x=-6$ and $y=-1/2$
$\therefore \vec{a}+2\vec{b}+6\vec{c}=\vec{0}$

Multiple choice physics turning effects of forces centre of gravity forces - vectors and moments acceleration due to gravity

The position vector of three particles of masses $m _1\, =\,1kg,\, m _2\, =\, 2\, kg$ and $m _3\, =\, 3\, kg$ are $\vec{r} _1\, =\, (\hat{i}\, +\, 4\hat{j}\, +\, \hat{k})\, m,\, \vec{r} _2\, =\, (\hat{i}\, +\, \hat{j}\, +\, \hat{k}) m$ and $\vec{r} _3\, =\, (2\hat{i}\, -\, \hat{j}\, -\, 2\hat{k})$ m respectively. Find the position vector of their center of mass.

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

The position vector of COM of the.three particles will be given by
$\vec{r} _{COM}\, =\, \displaystyle \frac {m _1\vec{r} _1\, +\, m _2\vec{r} _2\, +\, m _3\vec{r} _3}{m _1\, +\, m _2\, +\, m _3}$
Substituting the values, we get
$\vec{r} _{COM}\, =\, \displaystyle \frac {(1) (\hat{i}\, +\, 4\hat{j}\, +\, \hat{k})\, +\, (2) (\hat{i}\, +\, \hat{j}\, +\, \hat{k})\, +\, (3) (2\hat{i}\, -\, \hat{j}\, -\, 2\hat{k})}{1+2+3}\, =\, \displaystyle \frac {1}{2}\, (3\hat{i}\, +\, \hat{j}\, -\, \hat{k})\, m$.
Hence, the position vector of their center of mass is $\, \displaystyle \frac {1}{2}\, (3\hat{i}\, +\, \hat{j}\, -\, \hat{k})\, m$.

Multiple choice

A vector is an arrow that shows

  1. magnitude

  2. direction

  3. magnitude and direction

  4. the same as an equal (=) sign in a chemical equation (→)
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A vector is a mathematical or physical quantity that possesses both magnitude (size) and direction. Scalars only have magnitude, while vectors require both to be fully defined.

Multiple choice

The adding of two vectors is called what?

  1. Scalar quantity

  2. Vector

  3. Resultant
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

A resultant is the vector sum of two or more vectors. It represents the single vector that has the same effect as the individual vectors combined.

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

If vectors $\bar{b}=\left(\tan\alpha, -1 2\sqrt{\sin \dfrac{\alpha}{2}}\right)$ and $\bar{c}=\left(\tan \alpha , \tan\alpha -\dfrac{3}{\sqrt{\sin \alpha/2}}\right)$ are orthogonal and vector $\bar{a}=(1, 3, \sin 2\alpha)$ make an obtuse angle with the z-axis, then?

  1. $\alpha =\tan^{-1}(-2)$
  2. $\alpha =\tan^{-1}(-3)$
  3. $\alpha =\tan^{-1}(2)$
  4. $-2 < \alpha < 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The orthogonality condition (dot product = 0) and the obtuse angle condition (dot product with z-axis < 0) constrain the value of alpha. Solving the equations leads to the specified interval.

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

Let $\overrightarrow{A}$ be vector parallel to the line of intersection of planes ${p} _{1}$ and ${p} _{2}$ through the origin. ${p} _{1}$ is parallel to the vectors $\overrightarrow{a}=2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=4\hat{j}-3\hat{k}$ and ${p} _{2}$ is parallel to the vectors $\overrightarrow{c}=\hat{j}-\hat{k}$ and $\overrightarrow{d}=3\hat{i}+3\hat{j}$. The angle between $\overrightarrow{A}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is 

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

Plane ${P} _{1}$ is parallel to $\overrightarrow{a}$ and $\overrightarrow{b}$.
The normal to ${P} _{1}$ is along $\overrightarrow{a}\times \overrightarrow{b}$.
Plane ${P} _{2}$ is parallel to $\overrightarrow{c}$ and $\overrightarrow{d}$.
The normal to ${P} _{2}$ is along $\overrightarrow{c}\times \overrightarrow{d}$.
$\overrightarrow{A}$ is along the line of intersection of planes ${P} _{1}$ and ${P} _{2}$.
$\therefore \overrightarrow{A}$ is along $\left(\overrightarrow{a}\times\overrightarrow{b}\right)\times\left(\overrightarrow{c}\times\overrightarrow{d}\right)$
$\overrightarrow{a}\times\overrightarrow{b}=\left|\begin{matrix} \hat{i} &\hat{j}  &\hat{k}  \ 0 & 2 & 3 \ 0 &4  &-3  \end{matrix}\right|$
$=\left(-6-12\right)\hat{i}-0.\hat{j}+0.\hat{k}$ on simplification
$=-18\hat{i}$
$\left(\overrightarrow{a}\times\overrightarrow{b}\right)\times\left(\overrightarrow{c}\times\overrightarrow{d}\right)$
$\overrightarrow{c}\times\overrightarrow{d}=\left|\begin{matrix} \hat{i} &\hat{j}  &\hat{k}  \ 0 & 1 & -1 \ 3 &3  &0 \end{matrix}\right|$
$=\left(0+3\right)\hat{i}-\left(0+3\right)\hat{j}+\left(0+3\right)\hat{k}$ on simplification
$=3\hat{i}-3\hat{j}-3\hat{k}$
$=3\left(\hat{i}-\hat{j}-\hat{k}\right)$
The angle between $\overrightarrow{A}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is  $\theta$
$\cos{\theta}=\dfrac{\overrightarrow{A}}{\left|\overrightarrow{A}\right|}.\dfrac{\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3\sqrt{2}}$
$\pm\dfrac{\left(\hat{j}-\hat{k}\right).\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3\sqrt{2}}$
$\pm\dfrac{1}{\sqrt{2}}$
and $\cos{\theta}=\pm\dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta=\dfrac{\pi}{4},\dfrac{3\pi}{4}$

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

Let $\overrightarrow{A}$ be vector parallel to the line of intersection of planes ${p} _{1}$ and ${p} _{2}$ through the origin. ${p} _{1}$ is parallel to the vectors $\overrightarrow{a}=2\hat{j}+3\hat{k}$ and $\overrightarrow{b}=4\hat{j}-3\hat{k}$ and ${p} _{2}$ is parallel to the vectors $\overrightarrow{c}=\hat{j}-\hat{k}$ and $\overrightarrow{d}=3\hat{i}+3\hat{j}$. The angle between $\overrightarrow{A}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is:

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

Plane ${p} _{1}$ is parallel to $\overrightarrow{a}$ and $\overrightarrow{b}$ the normal to ${p} _{1}$ is along $\overrightarrow{a}\times \overrightarrow{b}$ 
Plane ${p} _{2}$ is parallel to $\overrightarrow{c}$ and $\overrightarrow{d}$ the normal to ${p} _{2}$ is along $\overrightarrow{c}\times \overrightarrow{d}$
$\overrightarrow{A}$ is along the line of intersection of planes ${p} _{1}$ and ${p} _{2}$ 
$\therefore \overrightarrow{A}$ is along $\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{c}\times \overrightarrow{d}\right)$
$\overrightarrow{a}\times \overrightarrow{b}=\left[\begin{matrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 2 &  3\ 0 & 4 & -3 \end{matrix}\right]$
$=\hat{i}\left(-6-12\right)-\hat{j}\left(0-0\right)+\hat{k}\left(0\right)$
$=-18\hat{i}$
$\overrightarrow{c}\times \overrightarrow{d}=\left|\begin{matrix} \hat{i} & \hat{j} & \hat{k} \ 0  & 1 & -1 \ 3 & 3 & 0 \end{matrix}\right|$
$=\hat{i}\left(0+3\right)-\hat{j}\left(0+3\right)+\hat{k}\left(0-3\right)$
$=3\hat{i}-3\hat{j}-3\hat{k}$
$=3\left(\hat{i}-\hat{j}-\hat{k}\right)$
$ \therefore \overrightarrow{A}$ is along $\hat{i}\times \left(\hat{i}-\hat{j}-\hat{k}\right)=\hat{j}-\hat{k}$
The angle between $\overrightarrow{A}$ and $2\hat{i}+\hat{j}-2\hat{k}$ is $\theta$
$\cos{\theta}=\dfrac{\overrightarrow{A}}{\left|\overrightarrow{A}\right|}.\dfrac{\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3}$
$   =\pm \dfrac{\left(\hat{j}-\hat{k}\right)\left(2\hat{i}+\hat{j}-2\hat{k}\right)}{3\sqrt{2}}$
$=\pm\dfrac{\left(1+2\right)}{3\sqrt{2}} = \pm \dfrac{1}{\sqrt{2}}$
and $\cos{\theta}=\pm \dfrac{1}{\sqrt{2}}$
$\Rightarrow \theta=\dfrac{\pi}{4},\dfrac{3\pi}{4}$

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

Let $\overrightarrow{a},\overrightarrow{b},\overrightarrow{c},\overrightarrow{d}$ are such that $\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{c}\times \overrightarrow{d}\right)=0$.Let ${p} _{1}$ and ${p} _{2}$ be the planes determined by the pairs of vectors $\overrightarrow{a},\overrightarrow{b}$ and $\overrightarrow{c},\overrightarrow{d}$ respectively . The angle between the planes ${p} _{1}$ and ${p} _{2}$ is

  1. $0$
  2. $\dfrac{\pi}{4}$
  3. $\dfrac{\pi}{3}$
  4. $\dfrac{\pi}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The plane ${p} _{1}$ contains the vectors $\overrightarrow{a}$ and $\overrightarrow{b}$ into normal is along $\overrightarrow{a}\times \overrightarrow{b}$
The normal to plane ${p} _{2}$ is along  $\overrightarrow{c}\times \overrightarrow{d}$.
$\left(\overrightarrow{a}\times \overrightarrow{b}\right)\times \left(\overrightarrow{c}\times \overrightarrow{d}\right)=0$
$\Rightarrow$ two normals are parallel
$\therefore$ the angle between the planes is zero

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

Two planes are prependicular  to one another. One of them contains vector $\vec{a}, \vec{b}$ and the other contains $\vec{c}, \vec{d}$ then $(\vec{a} \times \vec{b}) . (\vec{c}\times \vec{d}) = $

  1. $1$
  2. $0$
  3. $[\vec{a} \vec{b} \vec{c} ]$
  4. $[ \vec{b} \vec{c} \vec{d} ]$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let plane $P$, contains $a,b$ vector
$\vec{n} _{1}=\ \vec{a}\times \vec{b}$
Plane $P _{2}$ contain $\vec{c},\vec{d}$ vector
$\vec{n} _{2}=\vec{c}\times \vec{d}$
If $ P _{1}\perp P _{2}$ than $ n _{1}\perp\ n _{2}$
$(\vec{a}\times \vec{b}).(\vec{c}\times \vec{d})=0$

Multiple choice

In an inner product space, the inner product of two vectors (x) and (y) is denoted by:

  1. $\langle x, y \rangle$
  2. $\lVert x \rVert \cdot \lVert y \rVert$
  3. $\lVert x - y \rVert$
  4. $\lVert x + y \rVert$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The inner product of two vectors (x) and (y) in an inner product space is denoted by (\langle x, y \rangle).

Multiple choice

The norm of a vector (x) in an inner product space is defined as:

  1. $\lVert x \rVert = \sqrt{\langle x, x \rangle}$
  2. $\lVert x \rVert = \langle x, x \rangle$
  3. $\lVert x \rVert = \lVert x \rVert^2$
  4. $\lVert x \rVert = \lVert x - 0 \rVert$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The norm of a vector (x) in an inner product space is defined as (\lVert x \rVert = \sqrt{\langle x, x \rangle}).

Multiple choice

In an inner product space, the angle between two vectors (x) and (y) is given by:

  1. $\theta = \arccos\left(\frac{\langle x, y \rangle}{\lVert x \rVert \lVert y \rVert}\right)$
  2. $\theta = \arcsin\left(\frac{\langle x, y \rangle}{\lVert x \rVert \lVert y \rVert}\right)$
  3. $\theta = \arctan\left(\frac{\langle x, y \rangle}{\lVert x \rVert \lVert y \rVert}\right)$
  4. $\theta = \arccot\left(\frac{\langle x, y \rangle}{\lVert x \rVert \lVert y \rVert}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In an inner product space, the angle between two vectors (x) and (y) is given by (\theta = \arccos\left(\frac{\langle x, y \rangle}{\lVert x \rVert \lVert y \rVert}\right)).

Multiple choice

Let (V) be a vector space and (\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3) be vectors in (V). Which of the following statements is true about linear independence?

  1. If \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\) are linearly independent, then \(\mathbf{v}_1 + \mathbf{v}_2 + \mathbf{v}_3 = \mathbf{0}\).
  2. If \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\) are linearly independent, then no vector in the set can be expressed as a linear combination of the others.
  3. If \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\) are linearly independent, then they span the entire vector space \(V\).
  4. If \(\mathbf{v}_1, \mathbf{v}_2, \mathbf{v}_3\) are linearly independent, then they form a basis for \(V\).
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Linear independence means that no vector in the set can be written as a linear combination of the other vectors in the set. This implies that none of the vectors can be expressed as a multiple of the others.