Tag: maths

Questions Related to maths

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

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

  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 $\vec {c}+\vec {d}$

Show answer
Correct Option: A
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 $(\vec{a}-\vec{d})||(\vec{b}-\vec{c})$

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

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

  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. $\vec {a}+\vec {c}$

Show answer
Correct Option: C
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 $r=b-\dfrac{b.c}{a.c}a$

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

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

  1. $\dfrac { 1 }{ 4 } $

  2. $\dfrac { -3 }{ 4 } $

  3. $\dfrac { 3 }{ 4 } $

  4. $\dfrac { 1 }{ 2 } $

Show answer
Correct Option: B
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 } $

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

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

  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

Show answer
Correct Option: B
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.

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

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

  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. $None\ of\ these$

Show answer
Correct Option: B
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 $\vec{r} = \vec{b}-\dfrac{\vec{b}\cdot\vec{c}}{\vec{a}\cdot\vec{c}}\vec{a}$

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

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

  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

Show answer
Correct Option: C
Explanation:

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

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

LCM of numbers 1, 2, 3 is equal to their

maths hcf-lcm introduction to multiples multiples lcm

  1. product

  2. division

  3. sum

  4. difference

Show answer
Correct Option: A,C
Explanation:

$2, 3$ are primes.
$\therefore$ Each number has no factor other than $1$ and itself.
$\therefore$ Their LCM is the product of the numbers.
$\therefore$ LCM of $1,2,3=2\times 3=6$.
Also here $1+2+3=6$.
Answer- Option A and Option C.

L.C.M. of two co-prime numbers is their

maths hcf-lcm introduction to multiples multiples lcm

  1. sum

  2. difference

  3. product

  4. quotient

Show answer
Correct Option: C
Explanation:

The two numbers which have only 1 as their common factor are called co-primes.

For example, Factors of $ 5 $  are $ 1, 5 $
Factors of $ 3 $ are $ 1, 3 $

Common factors is $ 1 $.
So they are co-prime numbers.

To find their LCM, we

then choose each prime number with the greatest power and multiply them to get the LCM.
$ => LCM = 3 \times 5 = 15 $

Hence, LCM of two co-prime numbers is their product.

What are the three common multiples of $18$ and $6$?

maths hcf-lcm introduction to multiples multiples lcm

  1. $18, 6, 9$

  2. $18,36,6$

  3. $36, 54, 72$

  4. none of these

Show answer
Correct Option: C
Explanation:

Multiples of $ 18 = 18, 36, 54, 72..... $

Multiples of $ 6 = 6, 12, 18, 24.... $
The first common multiple will be $ 18 $

And the next common multiples will be multiples of $ 18 $
Hence, the common multiples of $ 18, 6 $ are $ 18, 36, 54, 72 $...

Bhushan counted to $60$ using multiples of $6.$ Which statement is true about multiples of $6?$

maths hcf-lcm introduction to multiples multiples lcm

  1. They are all odd numbers.

  2. They all have $6$ in the ones place.

  3. They can all be divided evenly by $3.$

  4. They can all be divided evenly by $12.$

Show answer
Correct Option: C
Explanation:

Multiple of $6$ like $6, 12, 18,24,30$ and they can all be divided evenly by $3$.
So option $C$ is correct.