Tag: maths

Questions Related to maths

Which of the following symbols represent vectors?

maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

  1. $v$

  2. $\vec{AB}$

  3. $\displaystyle \left| v \right| $

  4. $\displaystyle u+\left| v \right| $

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Correct Option: B
Explanation:

The vector is visual representation of a physical quantity that which has the both direction and the magnitude.

And out of these option,
Only option $B$ have the both direction and magnitude.
So, the correct answer is $\vec {AB}$.
Because it has a particular direction from $A$ to $B$ and also have the magnitude.

Which of the following expressions represent vectors ?

maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

  1. $v$

  2. $AB$

  3. $\displaystyle \left| v \right| $

  4. $\displaystyle \vec { AB }  $

Show answer
Correct Option: D
Explanation:

The vector is visual representation of a physical quantity that which has the both direction and the magnitude.

And out of these option,
Only option $D$ have the both direction and magnitude.
So, the correct answer is $\vec {AB}$.
Because it has a particular direction from $A$ to $B$ and also have the magnitude.

Which of the following expressions represent vectors ?

maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

  1. $\displaystyle -v$

  2. $\displaystyle u+v$

  3. $\displaystyle \vec{u}+ \vec{v} $

  4. $\displaystyle -\left| \vec { A  B} \right| $

Show answer
Correct Option: C
Explanation:

$\vec u + \vec v$ represents vector.


Hence, the option $C$ is correct.

State the following statement is true or false

In vector Algebra displacement is a vector quantity.

maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

  1. True

  2. False

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Correct Option: A
Explanation:
In vector algebra, displacemant is a vector quantity. since displacement has both value and direction. 
Distance is a scalar quantity.
Therefore, given statement is $TRUE$. 

On ethe rectangularcomponent of a viewer of 20m is 10 m .The component is

maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

  1. 20m

  2. $\eqalign{
    & \cr
    & 20\sqrt 3 \cr} $

  3. $\eqalign{
    & \cr
    & 10\sqrt 3 \cr} $

  4. None

Show answer
Correct Option: A

$[ a \times b \times c ] =  2[ a b c ] ^ { 2 }$

maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

  1. True

  2. False

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Correct Option: B

If the position vectors of the vertices of atriangle are $2 \overline { i } - \overline { j } + \overline { k } , \overline { i } - 3 \vec { j } - 5 \overline { k }$ and $3 \vec { i } - 4 \overline { j } - 4 \overline { k }$ then the triangle is

maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

  1. Equilateral triangle

  2. Isosceles triangle

  3. Right angled isosceles triangle

  4. Right angled triangle

Show answer
Correct Option: D
Explanation:

$\\Let\>A\>(2\hat{i}-\hat{j}+\hat{k}),\>B(\hat{i}-3\hat{j}-5\hat{k})\>and\\C(3\hat{i}-4\hat{j}-4\hat{k})\\then\\\overrightarrow{AB}=-\hat{i}-2\hat{j}-6\hat{k}\\\therefore\>|\overrightarrow{AB}|=\sqrt{1+4+36}=\sqrt{41}\\\overrightarrow{BC}=2\hat{i}-\hat{j}+\hat{k}\\\therefore\>|\overrightarrow{BC}|=\sqrt{4+1+1}=\sqrt{6}\\\overrightarrow{CA}=-\hat{i}+3\hat{j}+5\hat{k}\\\therefore\>|\overrightarrow{CA}|=\sqrt{1+9+25}=\sqrt{35}\\clearly\>\>\>|\overrightarrow{BC}|^2+|\overrightarrow{CA}|^2=|\overrightarrow{AB}|^2\\\therefore\>Triangle\>is\>a\>right\>angled\>triangle$

If $\displaystyle {\sec}^{2}A\hat{i}+\hat{j}+\hat{k}$, $\displaystyle \hat{i}+{\sec}^{2}B\hat{j}+\hat{k}$,and $\displaystyle \hat{i}+\hat{j}+{\sec}^{2}C\hat{k}$, are coplanar then $\displaystyle {\cot}^{2}A+{\cot}^{2}B+{\cot}^{2}{C}$ is    

maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

  1. $1$

  2. $2$

  3. $0$

  4. $-1$

Show answer
Correct Option: D
Explanation:

$\begin{array}{l} \left| { \begin{array} { *{ 20 }{ c } }{ { { \sec   }^{ 2 } }A } & 1 & 1 \ 1 & { { { \sec   }^{ 2 } }B } & 1 \ 1 & 1 & { { { \sec   }^{ 2 } }C } \end{array} } \right| =0 \\ { C _{ 1 } }\to { C _{ 1 } }-{ C _{ 2 } }\, \, \, \, \, \, \, \, \, { C _{ 2 } }\to { C _{ 2 } }-{ C _{ 3 } } \\ \left| { \begin{array} { *{ 20 }{ c } }{ { { \tan   }^{ 2 } }A } & 0 & 1 \ { -{ { \tan   }^{ 2 } }B } & { { { \tan   }^{ 2 } }B } & 1 \ 0 & { -{ { \tan   }^{ 2 } }C } & { { { \sec   }^{ 2 } }C } \end{array} } \right| =0 \\ { \tan ^{ 2 }  }A\left[ { { { \tan   }^{ 2 } }B{ { \sec   }^{ 2 } }C+{ { \tan   }^{ 2 } }C } \right] +{ \tan ^{ 2 }  }B{ \tan ^{ 2 }  }C=0 \ \\dfrac { { { { \sec   }^{ 2 } }C } }{ { { { \tan   }^{ 2 } }C } } +{ \cot ^{ 2 }  }B+{ \cot ^{ 2 }  }A=0 \ \\cos  e{ c^{ 2 } }C+{ \cot ^{ 2 }  }B+{ \cot ^{ 2 }  }A=0 \\ { \cot ^{ 2 }  }A+{ \cot ^{ 2 }  }B+{ \cot ^{ 2 }  }C=-1 \\ Hence,\, the\, option\, D\, is\, \, the\, correct\, answer. \end{array}$

Let        $\dot{a}$ = $\hat{i}$ + $\hat{j}$ + $\sqrt{2}\hat{k}$
              $\dot{b}$ = $b _1\hat{i}$ + $b _2\hat{j}$ + $\sqrt{2}\hat{k}$
              $\dot{c}$ = $5\hat{i}$ + $\hat{j}$ + $\sqrt{2}\hat{k}$
& ($\dot{a}$ + $\dot{b}$) is perpendicular to \overrightarrow{c} and projection vector of $\dot{b}$ on $\overrightarrow{a}$ is $\overrightarrow{a}$ then find $\left | \overrightarrow{b} \right |$

maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

  1. 6

  2. $\sqrt{22}$

  3. $\sqrt{32}$

  4. 11

Show answer
Correct Option: B 
Let a=i+j+k and c=j-k . If b is a vector satisfying a×b=c and a.b=3, then find b.
maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors

  1. $\dfrac{1}{3}(5\hat{i}+2\hat{j}+2\hat{k})$

  2. $\dfrac{1}{3}(2\hat{i}+3\hat{j}+\hat{k}$

  3. $\displaystyle 2\overrightarrow{a}$

  4. $\displaystyle -2\overrightarrow{a}$

Show answer
Correct Option: A
Explanation:

Let$b=xi+yj+zk$

Now 
$a.b=x+y+z=3$
Now $a\times b=(i+j+k)(xi+yj+zk)\ \quad=(z-y)i+(x-z)j+(y-x)k=j-k$
Now $z.y=0$ hence $k=y$
Now  $x-z=1$ and $y-x=-1$
Now
$x+y+z=3\1+z+z+z=3\z=\cfrac{2}{3}$
Hence $y=\cfrac{2}{3}\x=\cfrac{5}{3}$
Now $\overrightarrow{b}=\cfrac{(5i+2j+2k)}{3}$