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

Let $\displaystyle \vec{A}=2\vec{i}+\vec{k},\,\vec{B}=\vec{i}+\vec{j}+\vec{k},$ and $\displaystyle \vec{C}=4\vec{i}-3\vec{j}+7\vec{k}$ Determine a vector $\displaystyle \vec{R}$satisfying $\displaystyle \vec{R}\times \vec{B}=\vec{C}\times \vec{B}$ and $\displaystyle \vec{R}.\vec{A}=0$

  1. $\displaystyle -\hat{i}-8\hat{j}+2\hat{k}$
  2. $\displaystyle -8\hat{i}-\hat{j}+2\hat{k}$
  3. $\displaystyle -2\hat{i}-\hat{j}+8\hat{k}$
  4. $\displaystyle -\hat{i}-2\hat{j}+8\hat{k}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given, $\displaystyle \vec{R}\times \vec{B}=\vec{C}\times \vec{B} \Rightarrow \vec{R}\times \vec{B}-\vec{C}\times \vec{B}=0$
$\Rightarrow (\vec{R}-\vec{C})\times \vec{B}=0\Rightarrow R = \vec{C}+k\vec{B}$
Also given, $\vec{R}\cdot \vec{A} =0\Rightarrow k =- \cfrac{\vec{C}\cdot \vec{A}}{\vec{B}\cdot\vec{A}}=-\cfrac{15}{3}=-5$
Hence $\vec{R} =\vec{C}-5\vec{B}=-\hat{i}-8\hat{j}+2\hat{k} $ 

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

Unit vector $\vec r$ which satisfies $\vec r \times \vec b = \vec r \times \vec c$ where $\vec b = \widehat i + 2 \widehat j + \widehat k $ & $ \vec c = 3 \widehat i + 2 \widehat k $, is

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

Given $\vec b = \widehat i + 2 \widehat j + \widehat k $ & $ \vec c = 3 \widehat i + 2 \widehat k $
$\vec r \times \vec b = \vec r \times \vec c$
$(\vec r \times \vec b) - (\vec r \times \vec c) = \vec 0 $
$\Rightarrow \vec r \times (\vec b - \vec c) = \vec 0$
$\Rightarrow \vec r = \lambda (\vec b - \vec c) = \lambda (- 2 \widehat i + 2 \widehat j - \widehat k)$
$\Rightarrow \displaystyle \widehat r = \pm \left ( \frac{2 \widehat i - 2 \widehat j + \widehat k}{3} \right )$

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

Let $\vec a = \widehat i + \widehat j$ and $\vec b = 2 \widehat i - \widehat k$, then the point of intersection of lines $\vec r \times \vec a = \vec b \times \vec a$ 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

$\vec r \times \vec a = \vec b \times \vec a\Rightarrow \vec{r} = \vec{b}+\lambda \vec{a}$
and $\vec r \times \vec b = \vec a \times \vec b\Rightarrow \vec{r} = \vec{a}+\mu \vec{b}$
For intersection of both the lines, $\vec{b}+\lambda \vec{a}=\vec{a}+\mu \vec{b}$
Comparing coefficients, $\lambda=\mu = 1$
Hence point of intersection is $\vec{r}=\vec{a}+\vec{b} = 3\hat{i}+\hat{j}-\hat{k}$

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

If three vectors $\overline{a},\overline{b},\ \overline{c}$ are such that $\overline{a}\neq 0$, $\overline{a}\times\overline{b}=2\overline{a}\times\overline{c},\ |\overline{a}|=|\overline{c}|=1,\ |\overline{b}|=4$ and the angle between $|\overline{b}|$ and $|\overline{c}|$ is $\displaystyle \cos^{-1}\frac{1}{4}$, then $\overline{b}-2\overline{c}=\lambda\overline{a}$ where $\lambda$ is equal to

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

Given a cross b = 2a cross c, we can write a cross (b - 2c) = 0. This implies (b - 2c) is parallel to a, so b - 2c = lambda a. Using the magnitude squared of (b - 2c) = lambda^2 * |a|^2, we expand |b|^2 + 4|c|^2 - 4(b dot c) = lambda^2. With |b|=4, |c|=1, and cos(theta)=1/4, we get 16 + 4 - 4(4*1*1/4) = 20 - 4 = 16 = lambda^2, so lambda = +/- 4.

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

If $\vec{a}\times\vec{b}=\vec{c}\times\vec{d}$ and $\vec{a}\times\vec{c}=\vec{b}\times\vec{d}$, then

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

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

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

If $\vec {a},\vec {b},\ \vec {c}$ are non-zero non-collinear vectors such that $\vec {a}\times\vec {b}=\vec {b}\times\vec {c}=\vec {c}\times\vec {a}$ , then $\vec {a}+\vec {b}+\vec {c}=$

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

$\vec{a}\times \vec{b}=\vec{b}\times \vec{c}$
$(\vec{a}+\vec{c})\times \vec{b}=o$
$(\vec{a+\vec{c}}+\vec{b})\times \vec{b}=\vec{b\times \vec{b}}$
$(\vec{a}+\vec{c}+\vec{b})\times \vec{b}=0$
$\vec{a}+\vec{c}+\vec{b}=\vec{0}$   hence $\vec{a},\vec{b},\vec{c}$ are non collinear

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

If $\vec {a}\times \vec {b}=\vec {c}\times \vec {d},\vec {a}\times \vec {c}=\vec {b}\times \vec {d}$, then

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

$\vec{a}\times \vec{b}=\vec{c}\times \vec{d}  -(1)$
$\vec{a}\times \vec{c}=\vec{b}\times \vec{d}  -(2)$
$(1) - (2)$
$\vec{a}\times \vec{b}+\vec{c}\times \vec{a}=\vec{c}\times \vec{d}+\vec{d}\times \vec{b}$
$(\vec{a}-\vec{d})\times \vec{b}=c\times (\vec{d}-\vec{a})$
$(\vec{a}-\vec{d})\times \vec{b}+(\vec{d}-\vec{a})\times \vec{c}=0$
$(\vec{a}-\vec{d})\times (\vec{b}-\vec{c})=0$
which mean $(\vec{a}-\vec{d})||(\vec{b}-\vec{c})$

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

If $\vec {a}$ and $\vec {b}$ are not perpendicular to each other and $\vec {r}\times\vec {a}=\vec {b}\times\vec {a},\ \vec {r}.\vec {c}=0$, then $\vec {r}$ is equal to

  1. $\vec {a}-\vec {c}$
  2. $\vec {b}+\lambda\vec {a}$, for all scalars $\lambda$
  3. $\displaystyle \vec {b}-\dfrac{(\vec {b}.\vec {c})}{(\vec {a}.\vec {c})}\vec {a}$
  4. $\vec {a}+\vec {c}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Let $r,a,b$ and $c$ be vectors.
It is given that
$r\times a=b\times a$
$r\times a-(b\times a)=0$
$(r-b)\times a=0$
Hence $r-b$ is a vector parallel to vector $a$.
$r=b+\mu a$ ...(i)
It is given that $r.c=0$.
Hence $r$ vector is perpendicular to $c$ vector.
$(b+\mu a).c=0$ ...(from i)
$b.c+\mu(a.c)=0$
$\mu(a.c)=-b.c$
$\mu=-\dfrac{b.c}{a.c}$
Hence $r=b-\dfrac{b.c}{a.c}a$

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

If $a$ and $b$ are two unit vectors inclined at an angle $\dfrac { \pi  }{ 3 }$, then $\left{ a\times \left( b+a\times b \right)  \right} \cdot b$ is equal to

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

Given, $\left| a \right| =\left| b \right| =1$ and $a\cdot b=\cos { \dfrac { \pi  }{ 3 }  } $
Consider,
$\left{ a\times \left( b+a\times b \right)  \right} \cdot b=\left{ a\times b+a\times \left( a\times b \right)  \right} \cdot b$
            $=\left( a\times b \right) \cdot b+\left{ \left( a\cdot b \right) \cdot a-\left( a\cdot a \right) \cdot b \right} \cdot b$
            $=\left[ \begin{matrix} a & b & b \end{matrix} \right] +{ \left( a\cdot b \right)  }^{ 2 }-{ \left| a \right|  }^{ 2 }{ \left| b \right|  }^{ 2 }$
            $=0+\cos ^{ 2 }{ \dfrac { \pi  }{ 3 }  } -1=\dfrac { 1 }{ 4 } -1=\dfrac { -3 }{ 4 } $

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

Let $\vec{\lambda }=\vec{a}\times \left ( \vec{b}+\vec{c} \right )$, $\vec{\mu }=\vec{b}\times \left ( \vec{c}+\vec{a} \right )$ and $\vec{\nu }=\vec{c}\times \left ( \vec{a}+\vec{b} \right )$, then

  1. $\vec{\lambda }+\vec{\mu }=\vec{\nu }$
  2. $\vec{\lambda }, \vec{\mu }, \vec{\nu }$ are coplanar
  3. $\vec{\lambda }+\vec{\nu }=2\vec{\mu }$

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

$\vec { \lambda  } +\vec { \mu  } =\vec { a } \times \left( \vec { b } +\vec { c }  \right) +\vec { b } \times \left( \vec { c } +\vec { a }  \right) $

$= \vec { a } \times \vec { b } +\vec { a } \times \vec { c } +\vec { b } \times \vec { c } +\vec { b } \times \vec { a }$
$ = \left( \vec { a } +\vec { b }  \right) \times \vec { c }$
$= -\vec { \nu  } \ \Rightarrow \quad \vec { \nu  }$
$ =-\left( \vec { \lambda  } +\vec { \mu  }  \right) $

one vector is expressed as linear combination of other two vectors
Hence,
$\vec { \lambda  } ,\vec { \mu  } ,\vec { \nu  } $ are coplanar vectors.

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

Let $\vec{r}\times \vec{a}=\vec{b}\times \vec{a}$ and $\vec{r}.\vec{c}=0$, where $\vec{a}\vec{b}\neq 0$, then $\vec{r}$ is equal to

  1. $\vec{b}+t\vec{a}$ where $t$ is a scalar
  2. $\displaystyle \vec{b}-\dfrac{\vec{b}.\vec{c}}{\vec{a}.\vec{c}}\vec{a}$
  3. $\vec{a}-\vec{c}$
  4. $None\ of\ these$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\vec{r}\times \vec{a}=\vec{b}\times \vec{a}$

$\Rightarrow \vec{r}\times \vec{a}-\vec{b}\times \vec{a}=0$
$\Rightarrow (\vec{r}-\vec{b}\times \vec{a} = 0\Rightarrow \vec{r} = \vec{b}+t\vec{a}$

Now taking dot product with $\vec{c}$ both side,
$\Rightarrow \vec{r}\cdot\vec{c} = \vec{b}\cdot\vec{c}+t \vec{a}\cdot\vec{c}=0$
$\Rightarrow t = -\dfrac{\vec{b}\cdot\vec{c}}{\vec{a}\cdot\vec{c}}$

Hence $\vec{r} = \vec{b}-\dfrac{\vec{b}\cdot\vec{c}}{\vec{a}\cdot\vec{c}}\vec{a}$

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

If $\overrightarrow{a}, \overrightarrow{b}, \overrightarrow{c}$ are any three vectors in space then $\left ( \overrightarrow{c}+\overrightarrow{b} \right )\times \left ( \overrightarrow{c}+\overrightarrow{a} \right ).\left ( \overrightarrow{c}+\overrightarrow{b}+\overrightarrow{a} \right )$ is equal to

  1. $3\begin{bmatrix}

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

    \end{bmatrix}$
  2. $0$
  3. $\begin{bmatrix}

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

    \end{bmatrix}$

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

$\left( \overrightarrow{c}+\overrightarrow{b} \right )\times \left ( \overrightarrow{c}+\overrightarrow{a} \right ).\left ( \overrightarrow{c}+\overrightarrow{b}+\overrightarrow{a} \right)$

$= \left(\overrightarrow{c} \times \overrightarrow{a}+\overrightarrow{b} \times \overrightarrow{c} + \overrightarrow{b}\times \overrightarrow{a}\right)\cdot \left ( \overrightarrow{c}+\overrightarrow{b}+\overrightarrow{a} \right)$
$= \overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{c}+\overrightarrow{b} \times \overrightarrow{c}\cdot \overrightarrow{c} + \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{c}$
$+\overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b} \times \overrightarrow{c}\cdot \overrightarrow{b} + \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{b}$
$+\overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{a}+\overrightarrow{b} \times \overrightarrow{c}\cdot \overrightarrow{a} + \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{a}$
$=0+0+ \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{c}+\overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{b}+0 + 0+0+\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a} + 0$
$= \overrightarrow{b}\times \overrightarrow{a}\cdot \overrightarrow{c}+\overrightarrow{c} \times \overrightarrow{a}\cdot \overrightarrow{b}+\overrightarrow{b}\times \overrightarrow{c}\cdot \overrightarrow{a}$

Multiple choice maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

The Cartesian equation of the plane $\vec r=(1+\lambda-\mu)\hat i+(2-\lambda)\hat j+(3-2\lambda+2\mu)\hat k$ is-

  1. $2x+y=5$
  2. $2x-y=5$
  3. $2x+z=5$
  4. $2x-z=5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given, $\vec{r} = (1+\lambda-\mu)\hat{i}+(2-\lambda)\hat{j}+(3-2\lambda+2\mu)\hat{k}$
$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k} = (1+\lambda-\mu)\hat{i}+(2-\lambda)\hat{j}+(3-2\lambda+2\mu)\hat{k}$
Comparing coefficient, we get
$ 1+\lambda-\mu = x, 2-\lambda=y, 3-2\lambda+2\mu=z$
$\Rightarrow\lambda = 2-y, \mu=1+\lambda - x = 3-y-x$
Eliminating $\mu$ and $\lambda$, we get
$2x+z=5$ which is required equation of plane in cartesian form.

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If A B C D E F is a regular hexagon with A B = a and B C = b, then CE equals

  1. b-a

  2. -b

  3. b-2a

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

In a regular hexagon ABCDEF, the vector CE can be found using vector addition. Since AB = a and BC = b, the vectors for the sides are related by the geometry of the hexagon, leading to CE = b - a.