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

Which of the following symbols represent vectors?

  1. $v$
  2. $\vec{AB}$
  3. $\displaystyle \left| v \right| $
  4. $\displaystyle u+\left| v \right| $
Reveal answer Fill a bubble to check yourself
B Correct answer
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.

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

Which of the following expressions represent vectors ?

  1. $v$
  2. $AB$
  3. $\displaystyle \left| v \right| $
  4. $\displaystyle \vec { AB } $
Reveal answer Fill a bubble to check yourself
D Correct answer
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.

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

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

  1. Equilateral triangle

  2. Isosceles triangle

  3. Right angled isosceles triangle

  4. Right angled triangle
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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    

  1. $1$
  2. $2$
  3. $0$
  4. $-1$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

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

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

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 $\left | \overrightarrow{b} \right |$

  1. 6
  2. $\sqrt{22}$
  3. $\sqrt{32}$

  4. 11
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given the conditions that (a+b) is perpendicular to c and the projection of b on a is a, one can solve for the components of b. The calculation leads to |b| = sqrt(22).

Multiple choice maths vectors and transformations introduction to vector algebra algebra of vectors operations on vectors 
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.
  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. $\displaystyle -2\overrightarrow{a}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $\overrightarrow{b}=\cfrac{(5i+2j+2k)}{3}$