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

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 $\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

If  $\bar { a }, \bar { b }, \bar { c }$ are non-coplaner vector , then the vectors $2\bar { a }- 4\bar { b }+ 4\bar { c }, \bar { a }- 2\bar { b }+ 4\bar { c }$ and $-\bar { a }+ 2\bar { b }+ 4\bar { c }$ are parellel.

  1. True

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

Three vectors are parallel if they are scalar multiples of each other. Here, the vectors are v1 = 2a - 4b + 4c, v2 = a - 2b + 4c, and v3 = -a + 2b + 4c. Since the coefficients of a and b are not proportional to the constant c across all three vectors, they cannot be parallel.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The position vectors of three points are $2\vec{a}-\vec{b}+3\vec{c}$, $\vec{a}-2\vec{b}+\lambda \vec{c}$ and $\mu \vec{a}-5\vec{b}$ where $\vec{a}, \vec{b}, \vec{c}$ are non coplanar vectors, then the points are collinear when

  1. $\displaystyle \lambda =-2, \mu =\dfrac{9}{4}$
  2. $\displaystyle \lambda =-\dfrac{9}{4}, \mu =2$
  3. $\displaystyle \lambda =\dfrac{9}{4}, \mu =-2$

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

When points $x, y, z$ are collinear, we have $\alpha x + \beta y = (\alpha + \beta)z$
Similarly, $x(2\vec{a} - \vec{b} + 3\vec{c}) + y(\vec{a} - 2\vec{b} + \lambda \vec{c}) = (x + y)(\mu \vec{a} - 5\vec{b})$
$\Rightarrow$ comparing the coefficients of $\vec{a} \rightarrow 2x + y = x\mu + y\mu $
$\vec{b} \rightarrow - x - 2y = - 5x - 5y$
$\vec{c} \rightarrow 3x + \lambda y = 0$
$\Rightarrow 4x = -3y$ and so $\lambda = \dfrac{9}{4}$
Also, $\mu = -2$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

$\bar a,\bar b,\bar c$ are three non-zero vectors such that any two of them are non-collinear. If  $\bar a+\bar b$ is collinear with  $\bar c$ and  $\bar b+\bar c$ is collinear with $\bar a$, then what is their sum?

  1. $-1$
  2. $0$
  3. $1$
  4. $2$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

We have

$\bar a+\bar b =t\bar c$ ----$(1)$
$\bar b+\bar c =s\bar a$ ----$(2)$
From $(1)$ and $(2)$
$\bar a+\bar b=t(s\bar a-\bar b)$
Since no two of them are collinear, comparing coeffficients gives
$st=1$ and $t=-1$
$\Rightarrow s=-1$ and $t=-1$
From $(1)$
$\therefore \bar a+\bar b+\bar c=0$
Hence, option $B$.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The vectors $2\hat i + 3\hat j, \ 5\hat i + 6\hat j$ and $8\hat i + \lambda \hat j$ have their initial points at $(1,1)$. The value of $\lambda$ so that the vectors terminate on one straight line is

  1. 9

  2. 6

  3. 3

  4. 0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Vectors starting from the same point (1,1) are collinear if their components are proportional. The vectors are v1 = (2, 3), v2 = (5, 6), and v3 = (8, lambda). The vector v2 - v1 = (3, 3). The vector v3 - v2 = (3, lambda - 6). For these to be collinear, the slopes must be equal, so (lambda - 6) / 3 = 3 / 3, which gives lambda - 6 = 3, so lambda = 9.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

If $\vec{a},\vec{b},\vec{c}$ are the position vectors of points lie on a line, then $\vec{a}\times \vec{b}+\vec{b}\times \vec{c}+\vec{c}\times \vec{a}=$

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

If points with position vectors a, b, c are collinear, then (b-a) is parallel to (c-b). This implies (b-a) x (c-b) = 0. Expanding this cross product gives b x c - b x b - a x c + a x b = 0. Since b x b = 0, we get b x c + a x b - a x c = 0, which rearranges to a x b + b x c + c x a = 0.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Assertion ($A$): The points with position vectors $\overline{a},\overline{b},\overline{c}$ are collinear if $2\overline{a}-7\overline{b}+5\overline{c}=0$.
Reason ($R$): The points with position vectors $\overline{a},\overline{b},\overline{c}$ are collinear if $l\overline{a}+m\overline{b}+n\overline{c}=\overline{0}$.

  1. Both $A$ and $R$ are true and $R$ is correct reason of $A$
  2. Both $A$ and $R$ are true and $R$ is not correct reason of $A$
  3. $A$ is true $R$ is false
  4. $A$ is false $R$ is true
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $ are collinear

$\Rightarrow \ni l,m,n$ all zeros such that 
$l\overrightarrow { a } +m\overrightarrow { b } +n\overrightarrow { c } =0$ if $2\overrightarrow { a } -7\overrightarrow { b } +5\overrightarrow { c } =0$
then $\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $ are collinear since $2-7+5=0$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The points with position vectors $\vec{a}+\vec{b},\vec{a}-\vec{b}$ and $\vec{a}+\lambda\vec{b}$ are collinear for

  1. Only integrals values of $\lambda$
  2. No value of $\lambda$
  3. All real values of $\lambda$
  4. Only rational values of $\lambda$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Three points with position vectors p1, p2, p3 are collinear if (p2-p1) is a multiple of (p3-p2). Here, p2-p1 = -2b and p3-p2 = (lambda-1)b. Since both vectors are multiples of b, they are parallel for any real value of lambda.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

The vectors $\bar {a}=x\hat {i}-2\hat {j}+5\hat {k}$ and $\bar {b}=\hat {i}+y\hat {j}-z\hat {k}$are collinear if 

  1. $x=1$, $y=-2$, $z=-5$
  2. $x=1/2$, $y=-4$, $z=-10$
  3. $x=-1/2$, $y=4$, $z=-10$
  4. $x=-1$, $y=2$, $z=5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
we have, $\vec{a} =x\hat{i}-x\hat{j}+5\hat{k}$ and $\vec{b} =\hat{i}+y\hat{j}-z\hat{k}$ 
Now, to have collinearity,
both vectors will be identical.
$\therefore x=1, y=-2,$ and $z=-5$ Ans
Multiple choice direction cosines and direction ratios three dimensional geometry maths

Three points whose position vectors are $\overrightarrow{a}$, $\overrightarrow{b}$, $\overrightarrow{c}$ will be collinear if

  1. $\lambda \overrightarrow{a}+\mu \overrightarrow{b}=\left ( \lambda +\mu \right )\overrightarrow{c}$
  2. $\overrightarrow{a}\times \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}+\overrightarrow{c}\times \overrightarrow{a}=\overrightarrow{0}$
  3. $\begin{bmatrix}

    \overrightarrow{a} & \overrightarrow{b} & \overrightarrow{c}

    \end{bmatrix}=0$

  4. None of these
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

If $\vec{a}$, $\vec{b}$ and $\vec{c}$ are collinear vectors, therefore they lie on the same line. Hence they are parallel. Hence there cross products will be 0.
Therefore
$(\vec{a}\times\vec{b})=(\vec{c}\times\vec{b})=(\vec{a}\times\vec{c})=0$
And application of the section formula gives us
$\vec{c}=\dfrac{\lambda \vec{a}+\mu\vec{b}}{\lambda+\mu}$
Or
$\vec{c}(\lambda+\mu)=\lambda \vec{a}+\mu\vec{b}$

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Assertion ($A$): 

Three points with position vectors $\vec{a},\vec{b},\ \vec{c}$ are collinear if $\vec{a}\times\vec{b}+\vec{b}\times\vec{c}+\vec{c}\times\vec{a}=\vec{0}$
Reason ($R$):
Three points ${A}, {B},\ {C}$ are collinear if $\vec{AB}={t}\ \vec{BC}$, where ${t}$ is a scalar quantity.

  1. Both $A$ and $R$ are individually true and $R$ is the correct explanation of $A$.
  2. Both $A$ and $R$ are individually true and $R$ is NOT the correct explanation of $A$.
  3. $A$ is true but $R$ is false.
  4. $A$ is false but $R$ is true.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let the position vectors of $A, B, C$ be $\vec a, \vec b, \vec c$ respectively.
Given, $\vec{AB} = t\ \vec{BC}$
$\Rightarrow \vec{AB}\times\vec{BC} = 0$
$\Rightarrow (\vec b - \vec a)\times(\vec c - \vec b) = 0$
$\Rightarrow (\vec a - \vec b)\times(\vec b - \vec c) = 0$
$\Rightarrow \vec a\times \vec b + \vec b\times \vec c + \vec c \times \vec a = 0$
Hence, $\vec a, \vec b, \vec c$ are collinear.

Hence, option A.