Tag: vectors and transformations

Questions Related to vectors and transformations

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

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

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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. $-1$

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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

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

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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 $\overrightarrow{b}=\cfrac{(5i+2j+2k)}{3}$

$A$ vector $\vec V$ is inclined at equal angles to axes $OX,OY$ and $OZ$. If $\vec V$ is $6units$, then $\vec V$ is

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

  1. $2\sqrt 3\left( \hat i+\hat j+\hat k right )$

  2. $2\sqrt 3\left( \hat i-\hat j+\hat k right )$

  3. $\sqrt 2\left( \hat i+\hat j+\hat k right )$

  4. $2\sqrt 3\left( \hat i+\hat j-\hat k right )$

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Correct Option: A

$\sum _{ i=1 }^{ n }{ \vec { ai }  } =\vec { 0 } \quad where\quad |\vec { a\quad i\quad | } =1\forall i$ then the value of $\sum _{ 1\le i }^{  }{ \sum _{ <j\le n }^{  }{ \vec { { a } _{ i } }  }  } .\vec { { a } _{ j } } $ is 

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

  1. -n/2

  2. -n

  3. n/2

  4. n

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Correct Option: A

If $ \vec{a} $ and $ \vec{b} $ are two non-collinear unit vectors such that $ |\vec{a}+\vec{b}| = \sqrt{3}, $ find $(2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b}) $ 

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

  1. $ +\dfrac{11}{2} $

  2. $ -\dfrac{13}{2} $

  3. $ -\dfrac{11}{2} $

  4. $ +\dfrac{13}{2} $

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Correct Option: C
Explanation:

Given $ |\vec{a}+\vec{b}| = \sqrt{3}, $

Now squaring both sides we get,

$(\vec{a}+\vec{b}).(\vec{a}+\vec{b})=3$ [ Since$|\vec{a}|^2=\vec{a}.\vec{a}$ 
or, $|\vec{a}|^2+2\vec{a}.\vec{b}+|\vec{b}|^2=3$ [ Since 

$\vec{a}.\vec{b}=\vec{b}.\vec{a}$ ]
or, $\vec{a}.\vec{b}=\dfrac{1}{2}$.....(1). [ Since $\vec{a},\vec{b}$ are unit vectors then $|\vec{a}|=1=|\vec{b}|$ ]

Now,
$(2\vec{a}-5\vec{b}).(3\vec{a}+\vec{b}) $ 
$=6|\vec{a}|^2-13\vec{a}.\vec{b}-5|\vec{b}|^2$

$=6-\dfrac{13}{2}-5$ [ Using (1)]
$=-\dfrac{11}{2}$.

Which of the following can represent a vector?

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

  1. The length of the distance between the points $(0,0)$ and $(2,7)$

  2. A line segment beginning at $(2,7))$ and ending at $(0,0)$

  3. The length of the distance between the points $(2,7)$ and $(0,0)$

  4. A line segment beginning at $(0,0)$ and ending at $(2,7)$

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Correct Option: B,D
Explanation:
A vector is a quantity that can be described as having both magnitude and direction.
The length of the distance between any two points is a magnitude with no direction, so it can't represent a vector.
A line segment beginning at a certain point and ending at another can represent a vector. The magnitude of the vector is the distance between the points, and its direction is the direction from the initial point to the terminal point.
The following can represent a vector:
A line segment beginning at $(0,0)$ and ending at $(2,7)$.
A line segment beginning at $(2,7)$ and ending at $(0,0)$

Direction of zero vector

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

  1. does not exist

  2. towards origin

  3. indeterminate

  4. None of these

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Correct Option: C
Explanation:

As, zero vector represents a point.
Direction is indeterminate.
Hence, option C.