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

lf $\vec{a}=2\hat{i}+6n\hat{j}+m\hat{k}$ and $\vec{b}=\hat{i}+18\hat{j}+3\hat{k}$ are parallel to each other then the values of $m,n$ are:

  1. 6,6

  2. 6,1

  3. -1,6

  4. -1,-6
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since, the vector a is parallel to b, the corresponding coefficients of all the 3 components must bear the same ratio
i.e $ \dfrac{2}{1} = \dfrac{6n}{18} = \dfrac{m}{3} $
Or, $6n = 36, n = 6$
And $m = 6$

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

$\vec{A}$ and $\vec{B}$ are two vectors in a plane at an angle of $60^{0}$ with each other. $\vec{C}$ is another vector perpendicular to the plane containing vectors $\vec{A}$ and $\vec{B}$. Which of the following relations is possible?

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

Vector C is perpendicular to both vectors A and B. Hence, it can be equal to their cross product.
Options A and B make vector C in the plane of vectors A  and B which is not possible.
In option D, this is not possible as vector B is not perpendicular to  vector A

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

If $\vec{A} = 2\hat{i} + \hat{j}$ and $\vec{B} = \hat{i} - \hat{j}$, sketch vectors graphically and find the component of $\vec{A}$ along $\vec{B}$ and perpendicular to $\vec{B}$.

  1. Component of $A$ along $B$; $\dfrac{1}{2}(\hat{i}- \hat{j})$
    Component of $A$ perpendicular to $B$; $\dfrac{4}{2}(\hat{i}+\hat{j})$
  2. Component of $A$ along $B$; $\dfrac{1}{2}(\hat{i}- \hat{j})$
    Component of $A$ perpendicular to $B$; $\dfrac{3}{2}(\hat{i}+\hat{j})$
  3. Component of $A$ along $B$; $\dfrac{1}{2}(\hat{i}- \hat{j})$
    Component of $A$ perpendicular to $B$; $\dfrac{1}{2}(\hat{i}+\hat{j})$
  4. Component of $A$ along $B$; $\dfrac{1}{3}(\hat{i}- \hat{j})$
    Component of $A$ perpendicular to $B$; $\dfrac{3}{2}(\hat{i}+\hat{j})$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Component of A along B = (A dot B / |B|^2) * B. A dot B = 2-1 = 1. |B|^2 = 1^2 + (-1)^2 = 2. Component = (1/2)(i - j). Perpendicular component = A - (component along B) = (2i + j) - (0.5i - 0.5j) = 1.5i + 1.5j = (3/2)(i + j).

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

Given $\vec{A} = 2\hat{i} + p\hat{j} + q\hat{k}$ and $\vec{B}=5\hat{i}+7\hat{j} + 3\hat{k}$. If $\vec{A}|| \vec{B}$, then the values of $p$ and $q$ are, respectively,

  1. $\dfrac{14}{5}$ and $\dfrac{6}{5}$
  2. $\dfrac{14}{3}$ and $\dfrac{6}{5}$
  3. $\dfrac{6}{5}$ and $\dfrac{1}{3}$
  4. $\dfrac{3}{4}$ and $\dfrac{1}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given $\vec { A } =2\uparrow +p\hat { j } +q\hat { k } \quad \quad \vec { B } =5\uparrow +7\hat { j } +3\hat { k } $
$A\parallel B\Rightarrow A\times B=0$

Take $det{AB}$ and simplifying,

$\therefore$  $3p-7q=0$ and $6-5q=0$  and  $14-5p=0$
                               $\Rightarrow q=\dfrac { 6 }{ 5 } $                   $p=\dfrac { 14 }{ 5 } $
Multiple choice physics mathematical methods multiplication of vectors products of vectors scalar product

If the two vectors $\vec{A} = 2 \hat{i} + 3 \hat{j} + 4 \hat{k}$ and $\vec{B} = \hat{i} + 2 \hat{j} - n \hat{k}$ are perpendicular, then the value of $n$ is:-

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

Given,

$\begin{array}{l} \overrightarrow { A } =2\widehat { i } +3\widehat { j } +4\widehat { k }  \ \overrightarrow { B } =\widehat { i } +2\widehat { j } -n\widehat { k }  \end{array}$
They are perpendicular,
$\begin{array}{l} \overrightarrow { A } =2\widehat { i } +3\widehat { j } +4\widehat { k }  \ \overrightarrow { B } =\widehat { i } +2\widehat { j } -n\widehat { k }  \ \therefore \overrightarrow { A } .\overrightarrow { B } =0 \ \Rightarrow \left( { 2\widehat { i } +3\widehat { j } +4\widehat { k }  } \right) .\left( { \widehat { i } +2\widehat { j } -n\widehat { k }  } \right) =0 \ \Rightarrow 2\widehat { i } .\widehat { i } +3\widehat { j } .2\widehat { i } -4\widehat { k } .n\widehat { k } =0\, \, \, \, \, \, \left[ { \because \widehat { i } .\widehat { j } =j.\widehat { k } =\widehat { i } .\widehat { k } =0 } \right]  \ \Rightarrow 2+6-4n=0\, \, \, \, \, \, \, \, \, \, \, \, \, \, \left[ { \because \widehat { i } .\widehat { i } =\widehat { j } .\widehat { j } =\widehat { k } .\widehat { k } =1 } \right]  \ \Rightarrow 4n=8 \ \therefore n=2 \end{array}$
Hence, Option $B$ is correct.

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

Given $\overline { a } + \overline { b } + \vec { c } + \overline { d } = 0$ , which of the following statements is/are not a correct statement?

  1. $\vec { a } , \vec { b } , \vec { c }$ and $\vec { d }$ must be a null vector.
  2. The magnitude of $( \vec { a } + \vec { c } )$ equals the magnitude of $a( \vec { b } + \vec { d } )$
  3. The magnitude of $\vec { a }$ can never be greater than the sum of the magnitudes of $\vec { b } , \vec { c }$ and $\vec { d }$
  4. $\vec{b}$+$\vec{c}$ must He in the plane of $\vec{a}$ and $\vec{d}$ if $\vec{a}$ and $\vec{d}$ are not collinear and in the line of $\vec{a}$ and $\vec{d}$, if they are collinear.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In order to make vectors a + b + c + d = 0, it is not necessary to have all the four given vectors to be null vectors. There are many other combinations which can give the sum zero.

Simple example:- a=î,b=2î,c=–3î,d=0

(b) Correct
a + b + c + d = 0
a + c = – (b + d)
Taking modulus on both the sides, we get:
| a + c | = | –(b + d)| = | b + d |
Hence, the magnitude of (a + c) is the same as the magnitude of (b + d).

(c) Correct
a + b + c + d = 0
a = – (b + c + d)
Taking modulus both sides, we get:
| a | = | b + c + d |
| a |  ≤  | a | + | b | + | c |  …. (#)

Equation (#) shows that the magnitude of a is equal to or less than the sum of the magnitudes of b, c, and d.
Hence, the magnitude of vector a can never be greater than the sum of the magnitudes of b, c, and d.

(d) Correct
For a + b + c + d = 0
a + (b + c) + d = 0
The resultant sum of the three vectors a, (b + c), and d can be zero only if (b + c) lie in a plane containing a and d, assuming that these three vectors are represented by the three sides of a triangle.

If a and d are collinear, then it implies that the vector (b + c) is in the line of a and d. This implication holds only then the vector sum of all the vectors will be zero.

This is the explanation of correct solution.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

 The points with position vectors $\vec {a}=\hat {i}-2\hat {j}+3\hat {k}, \vec {b}=2\hat {i}+3\hat {j}-4\hat {k}$ & $-7\hat {j}+10\hat {k}$ are collinear.

  1. True

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

Three points with position vectors a, b, and c are collinear if (b-a) is a scalar multiple of (c-a). Given a = i - 2j + 3k, b = 2i + 3j - 4k, and c = 0i - 7j + 10k: b-a = i + 5j - 7k; c-a = -i - 5j + 7k. Since c-a = -1(b-a), the vectors are collinear.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The points $i + j + k, \, i + 2j, \, 2i+2j+k,\, 2i+3j+2k$ are

  1. collinear

  2. coplanar but not collinear

  3. non-coplanar

  4. none
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{matrix} A& B& C& D\i+j+k, &i+2j, &2i+2j+k,&2i+3j+2k \end{matrix}$
$\overline{AC} = (2-1)i + (2-1)j + k-k$
$=i+j$
$\overline{AB} = o + j - \overline{k} = j - \overline{k}$
$\overline{AD} = i + 2j + k$
$\begin{vmatrix} 1&1&0 \0&2 &1\end{vmatrix} = 1(1+2)-1(0+1)$
$=3-1 = 2 \neq 0$
Non coplanar.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If $\vec a, \, \vec b$ are two non-collinear vectors, then the position vector $\vec a + \vec b, \, \vec a - \vec b, \,and \, \vec a + \lambda {\vec b}$ are collinear for some real values of $\lambda$.

  1. True

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

For three points with position vectors to be collinear, the vectors connecting them must be parallel. The vectors a+b and a-b are not generally collinear with a+lambda*b unless specific conditions are met for lambda, and they are certainly not collinear for all real values of lambda.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If $\bar {a}, \bar {b}$ and $\bar {c}$ are non-zero non collinear vectors and $\theta(\neq 0 , \pi)$ is the angle between $\bar {b}$ and $\bar {c}$ if $(\bar {a}\times \bar {b}) \times \bar {c}=\dfrac {1}{2} |\bar {b}|\bar {c}|\bar {a}$. then $\sin \theta =$

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

We have

$\left( {\overrightarrow a  \times \overrightarrow b } \right) \times \overrightarrow c  = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $
$\overrightarrow c  \times \left( {\overrightarrow a  \times \overrightarrow b } \right) = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $
$ - \left[ {\left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a  - \left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b } \right] = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $
$\left( {\overrightarrow c .\overrightarrow a } \right)\overrightarrow b  - \left( {\overrightarrow c .\overrightarrow b } \right)\overrightarrow a  = \frac{1}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|\overrightarrow a $
$\overrightarrow c .\overrightarrow a  = 0$
$\overrightarrow c .\overrightarrow a  = \frac{{ - 1}}{2}\left| {\overrightarrow b } \right|\left| {\overrightarrow c } \right|$
$\cos \theta  = \frac{{ - 1}}{2}$
$ \Rightarrow \theta  = \frac{{2\pi }}{3}$
$\therefore \sin \theta  = \frac{{\sqrt 3 }}{2}$
Hence, $B$is the correct answer.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the three points  $A(\overline a),B(\overline  b),C(\overline c) $ are collinear ,the line passing through them is

$\overline r=\overline a+\lambda(\overline b-\overline a)$ then value of $\lambda $ is 

  1. $1$
  2. $2$
  3. $3$
  4. $4$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Given line
$\vec{r}=\vec{a}+\lambda(\vec{b}-\vec{a})$
$\vec{r}=(1-\lambda)\vec{a}+\lambda\vec{b}$
if $a$ and $b$ are collinear then 
$xa+yb=0$
$x=1-\lambda$
$y=\lambda$
if we pass line through c then 
$\vec{r}=1\neq0$
SO $\lambda=3$ to satisfy eq 
Multiple choice direction cosines and direction ratios three dimensional geometry maths

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

  1. $\vec {a}$
  2. $\vec {b}$
  3. $\vec {c}$
  4. $none\ of\ these$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
$\bar a+\bar b=K _{1}\bar c $

$\bar b+\bar c=K _{2}\bar a $

$\bar a- \bar c=K _{1}c-K _{2}\bar a$

$(k _{2}+1) \bar a-\bar c(1+k _{1})$=0

$k _{2}=-1 $ and $k _{1}=-1 $

$\bar{a}+\bar{b}+\bar{c}=0$
Multiple choice direction cosines and direction ratios three dimensional geometry maths

If the points with position vectors $60\hat{i}+3\hat{j}, 40\hat{i}-8\hat{j}$ and $a\hat{i}-52j$ are collinear, then $a=?$

  1. $-40$
  2. $-20$
  3. $20$
  4. $40$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Suppose, position vector $A=60\widehat i+3\widehat j$

position vector $B=40\widehat i-8\widehat j$

position vector $C=a\widehat i-52\widehat j$

Now, find vector AB and BC

$AB= -20\widehat i-11\widehat j$

$BC= (a-40)\widehat i-44\widehat j$

To be collinear,  angle between the vector AB and BC made by the given position vectors should be 0 or 180 degree.

That’s why the cross product of  the vectors should be zero

$ABXBC=(-20\widehat i-11\widehat j)X(a-40)\widehat i-44\widehat j$

$0\widehat i+0\widehat j+(880+11(a-40))=0$

$a-40= -80$

$a=-40$

Therefore, a should be $-40$ to be the given positions vectors collinear.


Multiple choice direction cosines and direction ratios three dimensional geometry maths

 Let $\overrightarrow{b}$ and  $\overrightarrow{c}$ be non collinear vectors.If $\overrightarrow{a}$ is a vector such that $\overrightarrow{a}.\left(\overrightarrow{b}+\overrightarrow{c}\right)=4$ and $\overrightarrow{a}\times\left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left({x}^{2}-2x+6\right)\overrightarrow{b}+\sin{y} .\overrightarrow{c}$ then $\left(x,y\right)$ lies on the line

  1. $x+y=0 $
  2. $x-y=0$
  3. $x=1$
  4. $y=\dfrac{\pi}{2}$
Reveal answer Fill a bubble to check yourself
C,D Correct answer
Explanation

$\overrightarrow{a}\times \left(\overrightarrow{b}\times \overrightarrow{c}\right)=\left(\overrightarrow{a}.\overrightarrow{c}\right).\overrightarrow{b}-\left(\overrightarrow{a}.\overrightarrow{b}\right).\overrightarrow{c}$
$\therefore \overrightarrow{a}.\overrightarrow{c}={x}^{2}-2x+6=-\sin y$
$\overrightarrow{a}.\left(\overrightarrow{b}+\overrightarrow{c}\right)=4 \Rightarrow -\sin y+{x}^{2}-2x+6=4$
$\Rightarrow {x}^{2}-2x+2=\sin y$
$\Rightarrow {\left(x-1\right)}^{2}+1=\sin y$
Left side $\ge 1$, right side $\le 1$
$\therefore $ they are equal if 
${\left(x-1\right)}^{2}+1=\sin y=1$
$\therefore y=\dfrac{\pi}{2},x=1$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Three points whose position vectors are $x\bar{i}+y\bar{j}+z\bar{k}$, $\bar{i}+2\bar{j}$ and $-\bar{i}-\bar{j}$ are collinear, then relation between $x, y, z$ is?

  1. $x-2y=1, z=0$
  2. $z+y=1, z=0$
  3. $x-y=1, z=0$

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

For three points to be collinear, the vectors connecting them must be proportional. The vectors are P1(x, y, z), P2(1, 2, 0), and P3(-1, -1, 0). The vector P2P3 is (-2, -3, 0). The vector P1P2 is (1-x, 2-y, -z). For these to be parallel, the ratios of components must be equal, implying z=0 and a specific linear relationship between x and y that is not listed in A, B, or C.