Tag: vectors and transformations

Quizzes (7) Questions (68)

Questions Related to vectors and transformations

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

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

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

  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. $2\sqrt 3\left( \hat i+\hat j-\hat k right )$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If a vector of magnitude 6 is inclined at equal angles to the axes, its components are equal (x=y=z). Thus, V = k(i + j + k). Since |V| = 6, k * sqrt(3) = 6, so k = 6/sqrt(3) = 2 * sqrt(3).

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

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

  1. -n/2

  2. -n

  3. n/2

  4. n
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given sum(ai) = 0, we have |sum(ai)|^2 = 0. Expanding this, sum(|ai|^2) + 2 * sum(ai . aj) = 0. Since |ai| = 1, sum(1) + 2 * sum(ai . aj) = 0, so n + 2 * sum(ai . aj) = 0, which gives sum(ai . aj) = -n/2.

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

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}) $ 

  1. <span>$ +\dfrac{11}{2} $</span>

  2. <span>$ -\dfrac{13}{2} $</span>

  3. <span>$ -\dfrac{11}{2} $</span>

  4. <span>$ +\dfrac{13}{2} $</span>
Reveal answer Fill a bubble to check yourself
C Correct answer
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)]
$=-\dfrac{11}{2}$.

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

Which of the following can represent a vector?

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

  2. <span>A line segment beginning at $(2,7))$ and ending at $(0,0)$</span>

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

  4. <span>A line segment beginning at $(0,0)$ and ending at $(2,7)$</span>
Reveal answer Fill a bubble to check yourself
B,D Correct answer
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 $(0,0)$