Tag: algebra of vectors

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Questions Related to algebra of vectors

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.