Tag: vectors and transformations

Questions Related to vectors and transformations

The area of triangle formed by the lines $x+y-3=0$, $x-3y+9=0$ and $3x-2y+1=0$ is:

combining transformations transformations vectors and transformations maths

  1. $\cfrac { 16 }{ 7 } $ sq. units

  2. $\cfrac { 10 }{ 7 } $ sq. units

  3. $4$ sq. units

  4. $9$ sq. units

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Correct Option: B
Explanation:

$L _1:x+y-3=0$

$L _2:x-3y+9=0$
$L _3:3x-2y+1=0$

$A$ is intersection point of $L _1$ and $L _2$ which is $(0,3)$ on solving $L _1$ and $L _2$.

$B$ is intersection point of $L _2$ and $L _3$ which is $(1,2)$ on solving $L _2$ and $L _3$.

$C$ is intersection point of $L _3$ and $L _1$ which is $(\dfrac{15}{7},\dfrac{26}{7})$ on solving $L _3$ and $L _1$.

Now, area of $\Delta ABC=\dfrac{1}{2}|(x _1y _2-x _2y _1)+(x _2y _3-x _3y _2)+(x _3y _1-x _1y _3)|$

$=\dfrac{1}{2}|(0-3)+(\dfrac{26}{7}-\dfrac{30}{7})+(\dfrac{45}{7}-0)|$

$=\dfrac{1}{2}|-3-\dfrac{4}{7}+\dfrac{45}{7}|=\dfrac{1}{2}\times \dfrac{20}{7}$

$=\dfrac{10}{7} sq. units$

if the equation $4{x^2} + 2xy + 2{y^2} - 1 = 0$ becomes $5{x^2} + {y^2} = 1,$ when the axes are rotate through an angle ${45^ \circ }\,$  , then the original equation of the curve is :

combining transformations transformations vectors and transformations maths

  1. $\,{15^ \circ }$

  2. $\,{30^ \circ }\,$

  3. ${45^ \circ }$

  4. ${60^ \circ }$

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Correct Option: D

If the axes are shifted to $(-2, -3)$ and rotated $\dfrac{\pi}{4}$ then Transformed equation of $2x^{2}+4xy-5y^{2}+20x-22y-14=0$ is 

combining transformations transformations vectors and transformations maths

  1. $X^{2}-14XY-7Y^{2}=2$

  2. $X^{2}-14XY-7Y^{2}=4$

  3. $X^{2}-14XY+7Y^{2}=2$

  4. $X^{2}+14XY+7Y^{2}=2$

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Correct Option: A

The point $A(2, 1)$ is translated parallel to the line $x- y = 3$ by a distance $4$ units. If the new position $A'$ is in third quadrant, then the coordinates of $A'$ are

combining transformations transformations vectors and transformations maths

  1. $(2 + 2 \sqrt{2}, 1 + 2\sqrt{2})$

  2. $(-2 + \sqrt{2}, -1 -2 \sqrt{2})$

  3. $(2 - 2 \sqrt{2}, 1 - 2 \sqrt{2})$

  4. none of these

Show answer
Correct Option: C
Explanation:

Since the point $A(2, 1)$ is translated parallel to $x-y =3$, therefore $AA'$ has the same slope as that of $x-y =3$. Therefore, $AA'$ passes through $(2, 1)$ and has the slope of $1$. 

Here, $\tan  \theta = 1 $
$\Rightarrow \cos \theta = \displaystyle\frac{1 }{\sqrt{2}}, \sin \theta = \displaystyle \frac {1 }{ \sqrt{2}}$
Using the parametric form of the line, the coordinates of $A'$ can be written as $\left(2\pm \displaystyle\frac {4}{\sqrt 2}, 1\pm \displaystyle\frac{4}{\sqrt 2}\right) $. 
Now, $A'$ is in the third quadrant. 
Hence, the coordinates of $A'$ are $(2- 2 \sqrt{2}, 1 - 2 \sqrt{2})$.

If the axes are rotated through an angle of ${30}^{o}$ in the anti-clockwise direction, the coordinates of point $(4,-2\sqrt{3})$ with respect to new axes are-

combining transformations transformations vectors and transformations maths

  1. $(2,\sqrt{3})$

  2. $(\sqrt{3}, -5)$

  3. $(2,3)$

  4. $(\sqrt{3},2)$

Show answer
Correct Option: B
Explanation:

$x=r\cos \alpha=4$

$y=r\sin\alpha=-2\sqrt3$
$x^1=r\cos(\alpha-30^o)$
$y^1=r\sin(\alpha-30^o)$
$x^1=r[\cos\alpha\cos 30^o+\sin\alpha\sin 30^o]$
$x^1=\dfrac{4\sqrt{3}}{2}-2\sqrt{3}\times \dfrac{1}{2}$
$x^1=2\sqrt{3}-\sqrt{3}$
$\therefore x^1=\sqrt{3}$
$y^1=r[\sin\alpha\cos 30^o-\cos \alpha\sin 30^o]$
      $=-2\sqrt{3}\times \dfrac{\sqrt{3}}{2}-4\times \dfrac{1}{2}$
$y^1=-5$
$\therefore \alpha(\sqrt{3}, -5)$

Let $\displaystyle A=(1,0)$ and $\displaystyle B=(2,1).$ The line $AB$ turns about $A$ through an angle $ \dfrac{\pi}6$ in the clockwise sense, and the new position of $B$ is $B'$. Then $B'$ has the coordinates

combining transformations transformations vectors and transformations maths

  1. $\displaystyle \left ( \frac{3+\sqrt{3}}{2},\frac{\sqrt{3}-1}{2} \right )$

  2. $\displaystyle \left ( \frac{3\sqrt{3}}{2},\frac{\sqrt{3}+1}{2} \right )$

  3. $\displaystyle \left ( \frac{1-\sqrt{3}}{2},\frac{1+\sqrt{3}}{2} \right )$

  4. none of these

Show answer
Correct Option: A
Explanation:

Given, points are $A=(1,0)$ and $B=(2,1)$

Slope of $AB=\cfrac { 1-0 }{ 2-1 } =1$
Then angle of $AB$ with $x$-axis is 
$\angle BAX={ 45 }^{ 0 }$
Hence, $\angle B'AX={ 45 }^{ 0 }-{ 30 }^{ 0 }={ 15 }^{ 0 }$
Therefore for $B'\left( h,k \right) $
$h=1+\sqrt{2}\cos{ 15 }^{ 0 },k= \sqrt{2}\sin{ 15 }^{ 0 }\$
We have, $\sin \left(15^{0} \right) = \dfrac{\sqrt 6 -\sqrt 2 }{4} $
$\Rightarrow \cos \left (15^{0} \right) = \dfrac{\sqrt6 + \sqrt 2 }{4}$
$ \Rightarrow h=\cfrac { 3+\sqrt { 3 }  }{ 2 } ,k=\cfrac { \sqrt { 3 } -1 }{ 2 } $

The transformed equation of $3{ x }^{ 2 }+3{ y }^{ 2 }+2xy=2$. When the coordinate axes are rotated through an angle of $45$, is

combining transformations transformations vectors and transformations maths

  1. ${ x }^{ 2 }+2{ y }^{ 2 }=1$

  2. $2{ x }^{ 2 }+{ y }^{ 2 }=1$

  3. ${ x }^{ 2 }+{ y }^{ 2 }=1$

  4. ${ x }^{ 2 }+3{ y }^{ 2 }=1$

Show answer
Correct Option: B
Explanation:

Since, the axes are rotated through an angle $45$, then we replace $\left( x,y \right) $ by
$\left( x\cos { 45 } -y\sin { 45 } ,x\sin { 45 } +y\cos { 45 }  \right) $
Given equation is $3{ x }^{ 2 }+3{ y }^{ 2 }+2xy=2$
Therefore, $ 3{ \left( \dfrac { x }{ \sqrt { 2 }  } -\dfrac { y }{ \sqrt { 2 }  }  \right)  }^{ 2 }+3\dfrac { x+y }{ \sqrt { 2 }  } +2\dfrac { x-y }{ \sqrt { 2 }  } \dfrac { x+y }{ \sqrt { 2 }  } =2$
$\Rightarrow \dfrac { 3 }{ 2 } \left( { x }^{ 2 }+{ y }^{ 2 }+2xy \right) +\dfrac { 3 }{ 2 } \left( { x }^{ 2 }+{ y }^{ 2 }-2xy \right) +\dfrac { 2 }{ 2 } \left( { x }^{ 2 }-{ y }^{ 2 } \right) =2$
$\Rightarrow  4{ x }^{ 2 }=2{ y }^{ 2 }=2$
$\Rightarrow  2{ x }^{ 2 }+{ y }^{ 2 }=1$

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

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

  1. $0$

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

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

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

Show answer
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 $|\vec{a}|=4,\ |\vec{b}|=$

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

  1. $16$

  2. $8$

  3. $3$

  4. $5$

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Correct Option: C

One of the following is not a vector

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

  1. displacement

  2. work

  3. centrifugal force

  4. gravitational field

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Correct Option: B
Explanation:

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