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Introduction to vector algebra - class-XI

Description: introduction to vector algebra
Number of Questions: 26
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Tags: vectors vectors and transformations maths vector algebra addition of vectors vector algebra - i
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In a square matrix $A$ of order $3$ the elements, $a _{i\ i}s^{'}$ are the sum of the roots of the equation $x^{2}-(a+b)x+ab=0;\ a _{1,\ i+1}s^{'}$ are the product of the roots, $a _{1,\ i-i}s^{'}$ are all unity and the rest of the elements all zero. The value of the det. $(A)$ is equal to

  1. $0$

  2. $(a+b)^{3}$

  3. $a^{3}-b^{3}$

  4. $(a^{2}+b^{2})$


Correct Option: B
Explanation:
Given, square matrix $A$ of order $3$ with $a _{ii}= (a+b) $ (sum of roots)

$a _{1 i+1}=a _{12}=ab$  (product of roots)

$a _{1i-1}= a _{10}=1$

$|A|= \begin{vmatrix} (a+b) & 0 & 0 \\ 1 & (a+b) & ab \\ 0 & 0 & (a+b) \end{vmatrix}= (a+b)[(a+b)(a+b)-0]$

$=(a+b)^{3}$

$\therefore $ Option $B$ is correct.


If $(\vec{a}\times\vec{b})^{2}+(\vec{a}.\vec{b})^{2}=144$ and $|\vec{a}|=4,\ |\vec{b}|=$

  1. $16$

  2. $8$

  3. $3$

  4. $5$


Correct Option: C

One of the following is not a vector

  1. displacement

  2. work

  3. centrifugal force

  4. gravitational field


Correct Option: B
Explanation:

Displacement,Centrifugal force and Gravitation are vectors.
Work $W=\overrightarrow { F } \cdot \overrightarrow { S } $ a scalar.
Hence, option B.

Which one of the following is not a scalar quantity?

  1. Temperature

  2. Density

  3. Mass

  4. Weight


Correct Option: D
Explanation:

Temperature,Density,Mass are clearly scalars.
Whereas in case of weight, it acts always downwards beacuse of
accelration due to gravity.
Hence, option D.

If a be an unit vector then

  1. direction of a is constant

  2. magnitude of a is constant

  3. direction and magnitude of a is constant

  4. any one of direction or magnitude is constant.


Correct Option: C
Explanation:

For a given unit vector , Magnitude and direction are constant (fixed).
since, a vector cannot have two directions.
Hence, option C.

If two vectors have the same magnitude and direction regardless of the positions of their initial points, then they are

  1. Co-initial vectors

  2. Colinear vectors

  3. Equal Vectors

  4. Unit Vector


Correct Option: C
Explanation:

 A quantity that has magnitude as well as direction is called a vector.  

or we can say ,
Two vectors are said to be equal if they have the same magnitude and directionregardless of the positions of their initial points.

A line PQ is symbolically written as

  1. PQ

  2. $\overrightarrow{PQ}$

  3. $\overleftrightarrow{PQ}$

  4. None of these


Correct Option: C
Explanation:

We know that the symbolically is  $\leftrightarrow$

now,  option $C$ is correct.

A ray has ................ end point/points.

  1. 1

  2. 2

  3. 3

  4. no


Correct Option: A
Explanation:

In geometry, a ray is usually taken as a half-infinite line (also known as a half-line) with one of the two points and taken to be at infinity. 

Which of the following symbols represent vectors?

  1. $v$

  2. $\vec{AB}$

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

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


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 ?

  1. $v$

  2. $AB$

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

  4. $\displaystyle \vec { AB }  $


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 ?

  1. $\displaystyle -v$

  2. $\displaystyle u+v$

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

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


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.

  1. True

  2. False


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

  1. 20m

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

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

  4. None


Correct Option: A

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

  1. True

  2. False


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

  1. Equilateral triangle

  2. Isosceles triangle

  3. Right angled isosceles triangle

  4. Right angled triangle


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    

  1. $1$

  2. $2$

  3. $0$

  4. $-1$


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

  1. 6

  2. $\sqrt{22}$

  3. $\sqrt{32}$

  4. 11


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


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

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


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 

  1. -n/2

  2. -n

  3. n/2

  4. n


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}) $ 

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

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

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

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


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?

  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 $(2,7)$


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

  1. does not exist

  2. towards origin

  3. indeterminate

  4. None of these


Correct Option: C
Explanation:

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

Which will result in a vector?

  1. Product of a scalar and a scalar.

  2. Product of a scalar and a vector.

  3. Addition of two vectors

  4. None of these


Correct Option: B,C
Explanation:
Let two vectors
$\vec{a}=\hat{i}$ 
$\vec{b}=\hat{i}+\hat{j}$
Addition of both vector 
$\vec{a}+\vec{b}=\hat{i}+\hat{i}+\hat{j}$
$\vec{a}+\vec{b}=2\hat{i}+\hat{j}$
Here we get vector by addition of both vectors 
hence option C is correct

let two scalar $\lambda=2,\mu=1$
$\lambda\times\mu=2\times1=2$
SO from here we get a scalar quantity Hence 
Option A is not correct 

$\lambda\times\vec{a}=\lambda\hat{i}$
Here vector quantity is obtained 
hence option B is correct

What is the value of $p$ for which the vector $p\left( 2\hat { i } -\hat { j } +2\hat { k }  \right)$ is of $ 3$ units length?

  1. $1$

  2. $2$

  3. $3$

  4. $6$


Correct Option: A
Explanation:

length of vector $a\hat { i } +b\hat { j } +c\hat { k } $ from origin is $\sqrt { a^2+b^2+c^2 } $ 

So $\sqrt { {(2p)}^2+{(-p)}^2+{(2p)}^2 } =\sqrt { 9p^2 }=3p $
Length is $3$ units given. 
$\therefore 3p=3\implies p=1$
Hence, A is correct.

If $\vec{x}$ and $\vec{y}$ be unit vectors and $\displaystyle |\vec{z}| = \dfrac{2}{\sqrt 7}$ such that $\vec{z} + (\vec{z} \times \vec{x}) = \vec{y}$ and $\theta$ is the angle between $\vec{x}$ and $\vec{z}$, then the value of sin $\theta$ is

  1. $\displaystyle \dfrac{1}{2}$

  2. $1$

  3. $\displaystyle \dfrac{\sqrt 3}{2}$

  4. $\displaystyle \dfrac{\sqrt 3 -1}{2 \sqrt 2}$


Correct Option: C
Explanation:

$|\vec{z} + (\vec{z} \times \vec{x}) | = | \vec{y}|^2 \,\,\,\,\,\Rightarrow$
$|\vec{z}|^2+|\vec{z}|^2 |\vec{x}|^2 \,sin^2\,\theta =1$
$\displaystyle \Rightarrow \,|z| = \frac{1}{\sqrt {1 + sin^2\,\theta}} = \frac{2}{\sqrt 7} \Rightarrow sin\,\theta = \frac{\sqrt 3}{2}$

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