Tag: maths

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Questions Related to maths

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

If the vector $a, b$ and $c$ form the sides $BC, CA $ and $AB $ and equal magnitute respectively of a triangle $ABC,$ then

  1. $ a \cdot b + b\cdot c + c \cdot a = 0$
  2. $a \times b = b \times c = c \times a$
  3. $a \cdot b = b\cdot c = c \cdot a$
  4. $a \times b + b \times c + c \times a = O$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

By triangle law, $\overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c } =\overrightarrow { 0 } $
Taking cross product by $\overrightarrow { a } ,\overrightarrow { b } ,\overrightarrow { c } $ respectively 
$\overrightarrow { a } \times \left( \overrightarrow { a } +\overrightarrow { b } +\overrightarrow { c }  \right) =\overrightarrow { a } \times \overrightarrow { 0 } =\overrightarrow { 0 } $
$\Rightarrow \overrightarrow { a } \times \overrightarrow { a } \times \overrightarrow { a } \times \overrightarrow { b } +\overrightarrow { a } \times \overrightarrow { c } =\overrightarrow { a } $
$\Rightarrow \overrightarrow { a } \times \overrightarrow { b } =\overrightarrow { c } \times \overrightarrow { a } \quad \left[ \because \overrightarrow { a } \times \overrightarrow { a } =\overrightarrow { 0 }  \right] $
Similarly, $\overrightarrow { a } \times \overrightarrow { b } =\overrightarrow { b } \times \overrightarrow { c. } $
$\therefore \overrightarrow { a } \times \overrightarrow { b } =\overrightarrow { b } \times \overrightarrow { c } =\overrightarrow { c } \times \overrightarrow { a. } $

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

If $\displaystyle a\cdot b=a\cdot c$ and $\displaystyle a\times b=a\times c,$ then

  1. either $\displaystyle a=0$ or $\displaystyle b=c$
  2. $a$ is parallel to $\displaystyle \left ( b-c \right )$
  3. $a$ is perpendicular to $\displaystyle \left ( b-c \right )$

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

Given,  $\displaystyle a\cdot b=a\cdot c$ and $\displaystyle a\times b=a\times c,$
$\Rightarrow a\cdot (b-c) = 0$ and $a\times (b-c) = 0$
Hence  either $a=0$ or $b=c$

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

Let $\displaystyle \vec{a}=\hat{i}+\hat{j}$ & $\displaystyle \vec{b}=2\hat{i}+\hat{j}$ The point of intersection of the lines $\displaystyle \vec{r}\times \vec{a}=\vec{b}\times \vec{a}& \vec{r}\times \vec{b}=\vec{a}\times \vec{b}$ is

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

$\displaystyle \overrightarrow{a}=\hat{i}+\hat{j}$ & $\displaystyle \hat{b}=\hat{2i}-\hat{k}:\overrightarrow{r}\times \overrightarrow{a}=\overrightarrow{b}\times \overrightarrow{a}\Rightarrow \left ( \overrightarrow{r}-\overrightarrow{b} \right )\times \overrightarrow{a}=0
\Rightarrow \overrightarrow{r}=\overrightarrow{b}+\lambda \overrightarrow{a}$ 

Similarly $\displaystyle \overrightarrow{r}\times \overrightarrow{b}=\overrightarrow{a}\times \overrightarrow{b}\Rightarrow \overrightarrow{r}=\overrightarrow{a}+\mu \overrightarrow{b}$ 
Substitute the vector $\displaystyle \overrightarrow{a}$ & $\displaystyle \overrightarrow{b}$ in (i) & (ii) and equating we get 
$\displaystyle 2\hat{i}-\hat{k}+\lambda \left ( \hat{i}+\hat{j} \right )=\hat{i}+\hat{j}+\mu \left ( 2\hat{i}-\hat{k} \right )
\Rightarrow 2+\lambda =1+2\mu ,\lambda =1,\mu =1
\therefore $ Point of intersection is $\displaystyle 3\hat{i}+\hat{j}-\hat{k}$

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

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 $\overline{a}\times\overline{b}=\overline{b}\times\overline{c}$, then

  1. $\overline{b}=\overline{a}\times\overline{c}$
  2. $\overline{b}||\overline{a}-\overline{c}$
  3. $\overline{b}\Vert(\overline{a}+\overline{c})$
  4. $\overline{b}=\overline{a}-\overline{c}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\overrightarrow{a}\times\overrightarrow{b}=\overrightarrow{b}\times\overrightarrow{c}$
$(\overrightarrow{a}+\overrightarrow{c})\times\overrightarrow{b}=0$
$\Rightarrow (\overrightarrow{a}+\overrightarrow{c})\parallel \overrightarrow{b}$

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