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 direction cosines and direction ratios three dimensional geometry maths

If the points $\bar a + \bar b,\bar a - \bar b,\bar a + k\bar b$ are collinear, then  

  1. $k$ has only one real value
  2. $k$ has two real value
  3. $k$ has no real values
  4. $k$ has infinite number of real values
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
As the $3$ point should be collinear area of triangle termed by then should be zero

considering $2$ direction to be $(\bar{a}+\bar{b})-(\bar{a}-\bar{b})=2\bar{b}$
& $(\bar{a}+\bar{b})-(\bar{a}+k\bar{b})=(1-k)\bar{b}$

$(2\bar{b})\times (1-k)\bar{b}=0$

this will be for any value of $k$ as cross produced of $2$ linear vector $=0$


Multiple choice direction cosines and direction ratios three dimensional geometry maths

If points $\hat i + \hat j, \hat i - \hat j$ and $p \hat i + q \hat j + r \hat k$ are collinear, then

  1. $p = 1$
  2. $r = 0$
  3. $q \in R$
  4. $q \neq 1$
Reveal answer Fill a bubble to check yourself
A,B,D Correct answer
Explanation

Points $A(\hat i + \hat j), B (\hat i - \hat j)$ and $C(p \hat i + q \hat j + r \hat k)$ are collinear
Now $\vec{AB} = - 2 \hat j$ and $\vec{BC} = (p -1) \hat i + (q - 1) \hat j + r k$
Vectors $\vec{AB}$ and $\vec{BC}$ must be collinear
$\Rightarrow p = 1,  r= 0 $ and $q \neq 1$

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

Given $A(1,-1,0)$; $B(3,1,2)$;$C(2,-2,4)$ and $D(-1,1,-1)$ which of the following points neither lie on $AB$ nor on $CD$

  1. $(2,2,4)$
  2. $(2,-2,4)$
  3. $(2,0,1)$
  4. $(0,-2,-1)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given:- $A(1,-1,0)$; $B(3,1,2)$;$C(2,-2,4)$ and $D(-1,1,-1)$ 

The equation of line $AB$ is given by $r=i-j+t(2i+2j+2k)$ or $\dfrac{x-1}{2} = \dfrac{y+1}{2} = \dfrac{z}{2} $ 
The points $(2,0,1)$ and $(0,-2,1)$ lies on $AB$. 
The equation of line $CD$ is given by $r=2i-2j+4k+t(3i-3j+5k)$ or $ \dfrac{x-2}{3} = \dfrac{y+2}{-3} = \dfrac{z-4}{5}$
$(2,-2,4)$ lie on $CD$. 
$(2,2,4)$ does not lie on any of the lines AB or CD.
Hence, option A is correct.

Multiple choice direction cosines and direction ratios three dimensional geometry maths

Given $A(1,-1,0)$; $B(3,1,2)$; $C(2,-2,4)$ and $D(-1,1,-1)$ which of the following points neither lie on $AB$ nor on $CD$?

  1. $(2,2,4)$
  2. $(2,-2,4)$
  3. $(2,0,1)$
  4. $(0,-2,-1)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$A(1,-1,0) , B(3,1,2), C(2,-2,4), D(-1,1,-1)\ \vec { AB } = <3-1, 1-(-1), 2-0>$ 

       $= <2,2,2>$ 
$\therefore \quad Equation\quad of\quad line\quad AB:\ \dfrac { x-1 }{ 2 } =\dfrac { y-(-1) }{ 2 } =\dfrac { z-0 }{ 2 } \quad \Longrightarrow \quad \dfrac { x-1 }{ 2 } =\dfrac { y+1 }{ 2 } =\dfrac { z-0 }{ 2 }$ 
$\vec { CD } = <-1-2, 1-(-2), -1-4>$ 
        $= <-3,3,-5>$ 
$\therefore \quad Equation\quad of\quad line\quad CD:\ \dfrac { x-2 }{ -3 } =\dfrac { y-(-2) }{ 3 } =\dfrac { z-4 }{ -5 } \quad \Longrightarrow \quad \dfrac { x-2 }{ -3 } =\dfrac { y+2 }{ 3 } =\dfrac { z-4 }{ -5 }$ 
By putting the values of points given in the choices in the equation of lines $AB$ and $CD$, $(2,2,4)$ is the point which neither lie on $AB$ nor on $CD$.

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

If the three points with position vectors $\displaystyle \bar{a}-2\bar{b}+3\bar{c}, \ 2\bar{a}+\lambda \bar{b}-4\bar{c}, \ -7\bar{b}+10\bar{c} $ are collinear, then $\displaystyle \lambda= $

  1. <font color="#888888">$1$</font>

  2. <span class="MathJax_Preview"><span class="MJXp-math"><span class="MJXp-mn">2
  3. $3$

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

The given vectors are collinear, so $l(\bar{a} - 2\bar{b} + 3\bar{c}) + k(2\bar{a} + \lambda\bar{b} - 4\bar{c}) = (l + k)(-7\bar{b} + 10\bar{c})$
Comparing the coefficients of $\bar{a} \rightarrow l + 2k = 0 $
$\bar{b} \rightarrow -2l + \lambda k = -7l -7k$
$\bar{c} \rightarrow 3l - 4k = 10l + 10k$
$\Rightarrow l = -2k$ and so $\lambda = 3$

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

A point $P$ lies on a line whose ends are $A(1,2,3)$ and $B(2,10,1).$ If $z$ component of $P$ is $7,$ then the coordinates of $P$ are

  1. $(-1,-14,7)$
  2. $(1,-14,7)$
  3. $(-1,14,7)$
  4. $(1,14,7)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The equation of line passing through $A$ and $B$ is $\displaystyle\frac { x-1 }{ 2-1 } =\frac { y-2 }{ 10-2 } =\frac { z-3 }{ 1-3 } \Rightarrow \frac { x-1 }{ 1 } =\frac { y-2 }{ 8 } =\frac { z-3 }{ -2 } =r$

Substitute $z=7$, we get
$\Rightarrow \displaystyle \frac { x-1 }{ 1 } =\frac { y-2 }{ 8 } =\frac { 7-3 }{ -2 } =-2$
Solve it to get $x=1-2=-1$, $y=2-16=-14$
So $P:$ $\left( -1,-14,7 \right) $

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

If the points whose position vectors are $2i+j+k, 6i-j+2k$ and $14i-5j+pk$ are collinear, then the value of p is?

  1. $2$
  2. $4$
  3. $6$
  4. $8$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$Positive\, \, vector\, \, are\, \, \, 2i+j+k,6\hat { i } +\hat { j } +2\hat { k }  \ and\, \, 4i-5j+pk\, \, are\, \, collinear\, \, then\, \, P=2 \ if\, \, three\, \, position\, \, vectors\, \, are\, \, collinear\, \, then,\, \, its\, \, { { determinantsis } }\left( 0 \right)  \ \Rightarrow \left| \begin{matrix} 2\, \, \, \, \, \, \, 1\, \, \, \, \, \, \, 1 \ 6\, \, \, \, -1\, \, \, \, \, \, 2 \ 14\, \, -5\, \, \, \, P \  \end{matrix} \right| =0 \ \Rightarrow 2\left( { -P+10 } \right) -1\left( { 6P-28 } \right) +\left( { -30+14 } \right) =0 \ \Rightarrow -2P+20-6P+28-16=0 \ \Rightarrow -8P+32=0 \ \therefore P=4\, $