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 maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

$A$ vector $\vec V$ is inclined at equal angles to axes $OX,OY$ and $OZ$. If $\vec V$ is $6units$, then $\vec V$ is

  1. $2\sqrt 3\left( \hat i+\hat j+\hat k right )$
  2. $2\sqrt 3\left( \hat i-\hat j+\hat k right )$
  3. $\sqrt 2\left( \hat i+\hat j+\hat k right )$
  4. $2\sqrt 3\left( \hat i+\hat j-\hat k right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a vector of magnitude 6 is inclined at equal angles to the axes, its components are equal (x=y=z). Thus, V = k(i + j + k). Since |V| = 6, k * sqrt(3) = 6, so k = 6/sqrt(3) = 2 * sqrt(3).

Multiple choice maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

$\sum _{ i=1 }^{ n }{ \vec { ai }  } =\vec { 0 } \quad where\quad |\vec { a\quad i\quad | } =1\forall i$ then the value of $\sum _{ 1\le i }^{  }{ \sum _{ <j\le n }^{  }{ \vec { { a } _{ i } }  }  } .\vec { { a } _{ j } } $ is 

  1. -n/2

  2. -n

  3. n/2

  4. n
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given sum(ai) = 0, we have |sum(ai)|^2 = 0. Expanding this, sum(|ai|^2) + 2 * sum(ai . aj) = 0. Since |ai| = 1, sum(1) + 2 * sum(ai . aj) = 0, so n + 2 * sum(ai . aj) = 0, which gives sum(ai . aj) = -n/2.

Multiple choice maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

If $ \vec{a} $ and $ \vec{b} $ are two non-collinear unit vectors such that $ |\vec{a}+\vec{b}| = \sqrt{3}, $ find $(2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b}) $ 

  1. $ +\dfrac{11}{2} $
  2. $ -\dfrac{13}{2} $
  3. $ -\dfrac{11}{2} $
  4. $ +\dfrac{13}{2} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given $ |\vec{a}+\vec{b}| = \sqrt{3}, $

Now squaring both sides we get,

$(\vec{a}+\vec{b}).(\vec{a}+\vec{b})=3$ [ Since$|\vec{a}|^2=\vec{a}.\vec{a}$ 
or, $|\vec{a}|^2+2\vec{a}.\vec{b}+|\vec{b}|^2=3$ [ Since 

$\vec{a}.\vec{b}=\vec{b}.\vec{a}$ ]
or, $\vec{a}.\vec{b}=\dfrac{1}{2}$.....(1). [ Since $\vec{a},\vec{b}$ are unit vectors then $|\vec{a}|=1=|\vec{b}|$ ]

Now,
$(2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b}) $ 
$=6|\vec{a}|^2-13\vec{a}.\vec{b}-5|\vec{b}|^2$

$=6-\dfrac{13}{2}-5$ [ Using (1)]
$=-\dfrac{11}{2}$.

Multiple choice maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

Which of the following can represent a vector?

  1. The length of the distance between the points $(0,0)$ and $(2,7)$
  2. A line segment beginning at $(2,7))$ and ending at $(0,0)$
  3. The length of the distance between the points $(2,7)$ and $(0,0)$
  4. A line segment beginning at $(0,0)$ and ending at $(2,7)$
Reveal answer Fill a bubble to check yourself
B,D Correct answer
Explanation
A vector is a quantity that can be described as having both magnitude and direction.
The length of the distance between any two points is a magnitude with no direction, so it can't represent a vector.
A line segment beginning at a certain point and ending at another can represent a vector. The magnitude of the vector is the distance between the points, and its direction is the direction from the initial point to the terminal point.
The following can represent a vector:
A line segment beginning at $(0,0)$ and ending at $(2,7)$.
A line segment beginning at $(2,7)$ and ending at $(0,0)$
Multiple choice maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

Which will result in a vector?

  1. Product of a scalar and a scalar.

  2. Product of a scalar and a vector.

  3. Addition of two vectors

  4. None of these
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation
Let two vectors
$\vec{a}=\hat{i}$ 
$\vec{b}=\hat{i}+\hat{j}$
Addition of both vector 
$\vec{a}+\vec{b}=\hat{i}+\hat{i}+\hat{j}$
$\vec{a}+\vec{b}=2\hat{i}+\hat{j}$
Here we get vector by addition of both vectors 
hence option C is correct

let two scalar $\lambda=2,\mu=1$
$\lambda\times\mu=2\times1=2$
SO from here we get a scalar quantity Hence 
Option A is not correct 

$\lambda\times\vec{a}=\lambda\hat{i}$
Here vector quantity is obtained 
hence option B is correct
Multiple choice maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

What is the value of $p$ for which the vector $p\left( 2\hat { i } -\hat { j } +2\hat { k }  \right)$ is of $ 3$ units length?

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

length of vector $a\hat { i } +b\hat { j } +c\hat { k } $ from origin is $\sqrt { a^2+b^2+c^2 } $ 

So $\sqrt { {(2p)}^2+{(-p)}^2+{(2p)}^2 } =\sqrt { 9p^2 }=3p $
Length is $3$ units given. 
$\therefore 3p=3\implies p=1$
Hence, A is correct.

Multiple choice maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

If $\vec{x}$ and $\vec{y}$ be unit vectors and $\displaystyle |\vec{z}| = \dfrac{2}{\sqrt 7}$ such that $\vec{z} + (\vec{z} \times \vec{x}) = \vec{y}$ and $\theta$ is the angle between $\vec{x}$ and $\vec{z}$, then the value of sin $\theta$ is

  1. $\displaystyle \dfrac{1}{2}$
  2. $1$
  3. $\displaystyle \dfrac{\sqrt 3}{2}$
  4. $\displaystyle \dfrac{\sqrt 3 -1}{2 \sqrt 2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$|\vec{z} + (\vec{z} \times \vec{x}) | = | \vec{y}|^2 \,\,\,\,\,\Rightarrow$
$|\vec{z}|^2+|\vec{z}|^2 |\vec{x}|^2 \,sin^2\,\theta =1$
$\displaystyle \Rightarrow \,|z| = \frac{1}{\sqrt {1 + sin^2\,\theta}} = \frac{2}{\sqrt 7} \Rightarrow sin\,\theta = \frac{\sqrt 3}{2}$

Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

A unit vector parallel to the intersection of the planes $\vec r\cdot (\hat i-\hat j+\hat k)=5$ and $\vec r\cdot (2\hat i+\hat j-3\hat k)=4$ can be

  1. $\dfrac {2\hat i+5\hat j+3\hat k}{\sqrt {38}}$
  2. $\dfrac {2\hat i-5\hat j+3\hat k}{\sqrt {38}}$
  3. $\dfrac {-2\hat i-5\hat j-3\hat k}{\sqrt {38}}$
  4. $\dfrac {-2\hat i+5\hat j-3\hat k}{\sqrt {38}}$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation

Vector parallel to intersection of planes 


$\vec p= \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 1 & -1 & 1 \ 2 & 1 & -3 \end{vmatrix}$


$= \hat{i} (3 - 1) - \hat{j} (-3 - 2) + \hat{k} (1 + 2)$

$= 2 \hat{i} + 5 \hat{j} + 3 \hat{k}$

$|\vec p|=\sqrt{2^2 + 5^2 + 3^2}=\sqrt{38}$

unit vector parallel to intersection of planes

$= \pm \dfrac{(2 \hat{i} + 5 \hat{j} + 3 \hat{k} )}{\sqrt{2^2 + 5^2 + 3^2}}$

$= \pm \dfrac{(2 \hat{i} + 5 \hat{j} + 3 \hat{k})}{\sqrt{38}}$

Multiple choice line of intersection of two planes line and a plane vectors, lines and planes three dimensional geometry maths

A non-zero vector $\vec{a}$ is parallel to the line of intersection of the plane determined  by the vectors $\hat{i},\hat{i}+\hat{j}$ and the plane determined by the vectors $\hat { i } -\hat { j } ,\hat { i } -\hat { k }$. The angle between $\vec{a}$ and $\hat { i } -2\hat { j } +2\hat { k } $ is

  1. $\pi/3$
  2. $\pi/4$
  3. $\pi/6$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The line of intersection is parallel to the cross product of the normals of the two planes. Calculating these vectors and their cross product, then finding the angle with the given vector, yields pi/3.

Multiple choice maths addition of vectors vectors from a geometric viewpoint basic concepts of vector introduction to vectors

If $\overrightarrow A ,\overrightarrow B $ and $\overrightarrow C $ are vectors such that $\left| {\overrightarrow B } \right| = \left| {\overrightarrow C } \right|$ , then  $\left{ {\left( {\overrightarrow A  + \overrightarrow B } \right)} \right. \times \left. {\left( {\overrightarrow A  + \overrightarrow C } \right)} \right} \times \left( {\overrightarrow B  \times \overrightarrow C } \right).\left( {\overrightarrow B  + \overrightarrow C } \right) = 1 $  these relation is ?

  1. True

  2. False
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The expression involves a complex vector product. Given the properties of cross products and dot products, the result of such a complex vector identity is generally not equal to 1 unless specific conditions are met. This is a false statement.

Multiple choice maths addition of vectors vectors from a geometric viewpoint basic concepts of vector introduction to vectors

If the vectors $\overrightarrow a  = \left( {2,{{\log } _3}x,\;a} \right)$ $and\;\overrightarrow b  = \left( { - 3,a{{\log } _3}x,{{\log } _3}x} \right)$ are included at an acute angle then-

  1. a=0

  2. a<0

  3. a>0

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} \overrightarrow { a } =2\left( { { { \log   } _{ 3 } }x,a } \right) ,\overrightarrow { b } =\left( { -3,a{ { \log   } _{ 3 } }x,{ { \log   } _{ 3 } }x } \right)  \ \overrightarrow { a } \overrightarrow { b } =\left| a \right| \left| b \right| \cos  \theta  \ \cos  \theta =\frac { { \overrightarrow { a } .\overrightarrow { b }  } }{ { \left| a \right| \left| b \right|  } }  \ =\frac { { -6+a{ { \left( { { { \log   } _{ 3 } }x } \right)  }^{ 2 } }+a.{ { \log   } _{ 3 } }x } }{ { \sqrt { 4+{ { \left( { { { \log   } _{ 3 } }x } \right)  }^{ 2 } }+{ a^{ 2 } } } .\sqrt { 9+{ a^{ 2 } }{ { \left( { { { \log   } _{ 3 } }x } \right)  }^{ 2 } }+{ { \left( { { { \log   } _{ 3 } }x } \right)  }^{ 2 } } }  } }  \ for\, \, acute\, \, angle\, \, a>0 \ Option\, \, \, C\, \, is\, \, correct. \end{array}$

Multiple choice maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

If $\overline{a},\overline{b},\overline{c}$ are three non-zero vectors and $\overline{a}\neq\overline{b}$, $\overline{a}\times\overline{c}=\overline{b}\times\overline{c}$, then

  1. $\overline{a}-\overline{b}$ is parallel to $\overline{c}$
  2. $\overline{a}-\overline{b}$ is perpendicular to $\overline{c}$
  3. $\overline{a}+\overline{b}$ is parallel to $\overline{c}$
  4. $\overline{a}+\overline{b}$ is perpendicular to $\overline{c}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We get 

$(\vec{a}-\vec{b})\times\vec{c}=0$
So    $(\vec{a}-\vec{b})\parallel \vec{c}$

Multiple choice maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors


If $\left| \vec { a }  \right| =1,\ \left| \vec { b }  \right| =2,\ (\vec { a },\vec { b })=\dfrac{2\pi}{3}$ then $\left{(\vec { a } +3\vec { b } )\times \left( 3\vec { a } -\vec { b }  \right) \right}^{2}=$


  1. $425$
  2. $\dfrac{147}{2}$
  3. $325$
  4. $300$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$|\bar{a}|=1, |\bar{b}|=2, (\bar{a}, \bar{b})=\dfrac{2z}{3}$


$[|\bar{a}|+3|\bar{b}|]=1+3\times 2=1+6=7$ ......... $(1)$


Also, $3|\bar{a}|-|\bar{b}|=3\times 1-2=3-2=1$ ......... $(2)$

$|(\bar{a}+3\bar{b})\times (3\bar{a}-\bar{b})|=|(\bar{a}+3\bar{b})||3\bar{a}-\bar{b}|\sin \theta$

$=|\bar{a}+3\bar{b}||3\bar{a}-\bar{b}|\sin (\bar{a}, \bar{b})$

$=|(\bar{a}+3\bar{b}||3\bar{a}-\bar{b}|\sin\left(\dfrac{2z}{3}\right)$

$=|\bar{a}+3\bar{b}||3\bar{a}-\bar{b}|\sin 60^{o}$

$|\bar{a}+3\bar{b}||3\bar{a}-\bar{b}|\left(\dfrac{\sqrt{3}}{2}\right)$

from $(1)$ & $(2)$

$=7\times 1\dfrac{\surd{3}}{2}=\dfrac{7\sqrt{3}}{2}$

$[(\bar{a}+3\bar{b}\times (3\bar{a}-\bar{b})]^{2}=\left(7\dfrac{\sqrt{3}}{2}\right)^{2}$

$=\dfrac{49\times 3}{4}=\dfrac{147}{2}$

Multiple choice maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

If $\vec a = \hat i + \hat j + \hat k,\,\vec b = \hat i + \hat j,\,\,\hat c = \hat i$ and $\left( {\vec a \times \vec b} \right) \times \vec c = \lambda \vec a \times \mu \vec b$ then $\lambda  + \mu $

  1. $0$
  2. $1$
  3. $2$
  4. $3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Few things to consider 
$\hat i\times \hat i = \hat j\times \hat j=\hat k\times  \hat k =0$
$\hat i\times \hat j = \hat k , \hat j \times \hat k=\hat i , \hat k \times \hat i =\hat k$
$\hat j\times \hat i = -\hat k , \hat i \times \hat k=-\hat j , \hat k \times \hat j =-\hat i$
Now, $(\vec a \times \vec b) \times \vec c \Rightarrow $
$\vec a =(\hat i+\hat j+\hat k), \vec b=\hat i+\hat j, \vec c=\hat i$
$\Rightarrow \ (\hat i+\hat j+\hat k) \times (\hat i+\hat j)\times (\hat i)$
$\Rightarrow \ (\hat k-\hat k+\hat j-\hat i)\times (\hat i)$
$\Rightarrow \ -\hat k$
Which does not involve any term like $\lambda \vec a\times \mu \vec b \ \therefore \ \lambda +\mu =0$


Multiple choice maths vectors and transformations vectors from a geometric viewpoint basic concepts of vector introduction to vectors

Let $\vec{a} = \widehat{i} + \widehat{j}$, $\vec{b} = 2 \widehat{i} - \widehat{k}$, then vector $\vec{r}$ satisfying the equations $\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$ and $\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$ is

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

$\vec{r} \times \vec{a} = \vec{b} \times \vec{a}$ or $(\vec{r} - \vec{b}) \times \vec{a} = 0$
$\vec{r} \times \vec{b} = \vec{a} \times \vec{b}$ or $(\vec{r} - \vec{a}) \times \vec{b} = 0$
If $\vec{r} = x \widehat{i} + y \widehat{j} + z\widehat{k}$, then 


$\begin{vmatrix}\widehat{i} & \widehat{j} & \widehat{k}\ x-2 & y & z+1\ 1 & 1 & 0\end{vmatrix} = 0$ and $\begin{vmatrix}\widehat{i} & \widehat{j} & \widehat{k}\ x-1 & y-1 & z\ 2 & 0 & -1 \end{vmatrix}= 0$

$\Rightarrow z + 1 = 0, x - y = 2$
and $y -1 = 0, x - 1 + 2z = 0$
$\Rightarrow x = 3, y = 1, z = -1$