Tag: vectors and transformations

Questions Related to vectors and transformations

When the axes are rotated through an angle $\dfrac{\pi}{6}$ , find the new coordinate for $(1,0)$

  1. $(\dfrac{\sqrt3}{2},\dfrac{-1}{2})$

  2. $(\dfrac{\sqrt4}{2},\dfrac{-1}{2})$

  3. $(\dfrac{\sqrt5}{2},\dfrac{-1}{2})$

  4. $(\dfrac{\sqrt3}{2},\dfrac{-1}{3})$


Correct Option: A
Explanation:

$\\ sin(\frac{\pi}{6})=(\frac{b}{1})\\\therefore b= (\frac{1}{2})\\ so new y-coordinate wil be  = (\frac{-1}{2})\\cos(\frac{\pi}{6})=(\frac{a}{1})\\\therefore a=(\frac{\sqrt3}{2})\\\therefore new x-coordinate =(\frac{\sqrt3}{2})$

The point to which is shifted in order to remove the first degree terms in $ 2x^{ 2 }+5xy+3y^{ 2 }+6x+7y+1=0 $ is

  1. (2,1)

  2. (1,-2)

  3. (2,-1)

  4. (1,2)


Correct Option: A

If the transformed equation of a curve is $9x^{2}+16y^{2}=144$ when the axes rotated through an angle of $45^{o}$ then the original equation of a curve is:

  1. $25x^{2}+14yxy+25y^{2}=228$

  2. $25x^{2}-14yxy+25y^{2}=228$

  3. $25x^{2}+14yxy-25y^{2}=228$

  4. $25x^{2}-14yxy-25y^{2}=228$


Correct Option: A

By translating the axes the equation $xy-x+2y=6$ has changed to $XY=C$, then $C=$

  1. 4

  2. 5

  3. 6

  4. 7


Correct Option: A
Explanation:

We have

$xy-x+2y=6$


Now,

$ xy-x+2y=6 $

$ \Rightarrow xy-x+2y-6=0 $

$ \Rightarrow xy-x+2y-2=4 $

$ \Rightarrow x\left( y-1 \right)+2\left( y-1 \right)=4 $

$ \Rightarrow \left( x+2 \right)\left( y-1 \right)=4 $


This equation is the form of $XY=C$

Then,

$C=4$


Hence, this is the answer.

lf the axes are translated to the point $(-2, -3)$ , then the equation $\mathrm{x}^{2}+3\mathrm{y}^{2}+4\mathrm{x}+18\mathrm{y}+30=0$ transforms to

  1. $\mathrm{X}^{2}+\mathrm{Y}^{2}=4$

  2. $\mathrm{X}^{2}+3\mathrm{Y}^{2}=1$

  3. $\mathrm{X}^{2}-\mathrm{Y}^{2}=4$

  4. $\mathrm{X}^{2} - 3 \mathrm{Y}^{2}=1$


Correct Option: B
Explanation:

When the point $(x,y)$ changes to $(X,Y)$ on shifting the origin to $(h,k)$
Then, $x=X+h,y=Y+k$
$x=X-2 , y=Y-3$
So, the equation transform to 
$(X-2)^2+3(Y-3)^2+4(X-2)+18(Y-3)+30=0$
$\Rightarrow X^2+3Y^2-1=0$
So, the transformed eqn is $X^2+3Y^2-1=0$

lf the origin is shifted to the point $(-1, 2)$ without changing the direction of axes, the equation ${x}^{2} -{y}^{2}+2{x}+4{y}=0$ becomes 

  1. ${X}^{2}+{Y}^{2}+3=0$

  2. ${X}^{2}+{Y}^{2}-3=0$

  3. ${X}^{2}-{Y}^{2}+3=0$

  4. ${X}^{2}-{Y}^{2}-3=0$


Correct Option: C
Explanation:

Let the point $(x,y)$ on the line changes to $(X,Y)$ on shifting the origin to $(h,k)$
Then, $x=X+h,y=Y+k$
$x=X-1 , y=Y+2$
So, the equation transform to 
$(X-1)^2-(Y+2)^2+2(X-1)+4(Y+2)=0$
$\Rightarrow X^2-Y^2+3=0$
So, the transformed eqn is $X^2-Y^2+3=0$

lf the axes are rotated through an angle $60^{\mathrm{o}}$, then the transformed equation of  $\mathrm{x}^{2}+\mathrm{y}^{2}=25$ is 

  1. $\mathrm{X}^{2}+\mathrm{Y}^{2}=1$

  2. $\mathrm{X}^{2}+\mathrm{Y}^{2}=9$

  3. $\mathrm{X}^{2}+\mathrm{Y}^{2}=16$

  4. $\mathrm{X}^{2}+\mathrm{Y}^{2}=25$


Correct Option: D
Explanation:

The radius of the circle is $5$. Despite rotating the reference frame, the origin still remains the same. Hence, the equation of the circle remains unchanged. 
Hence the new equation is:
$ X^2 + Y^2 =25 $

The transformed equation of $\mathrm{x}\mathrm{c}\mathrm{o}\mathrm{s}\alpha+\mathrm{y}\mathrm{s}\mathrm{i}\mathrm{n}\alpha = \mathrm{P}$ when the axes are rotated through an angle $\alpha$ is

  1. $\mathrm{X}=\mathrm{P}$

  2. $\mathrm{X}+\mathrm{P}=0$

  3. $\mathrm{Y}=\mathrm{P}$

  4. $\mathrm{Y}+\mathrm{P}=0$


Correct Option: A
Explanation:

By  rotation  of  axes,
$ x = x _{1} \cos\alpha -y _{1}\sin\alpha $
$ y = x _{1} \sin \alpha + y _{1} \cos \alpha $
$ x \cos  \alpha  + y \sin  \alpha = P$..........(given)
$ \Rightarrow  x _{1} (\cos^2 \alpha + \sin^{2}\alpha ) + y (\sin  \alpha  \cos  \alpha - \cos  \alpha  \sin  \alpha ) = P$..............(substitute the values of x and y) 
$x _{1} = P$ => X=P.

When axes are rotated by an angle of $135^{0}$, initial coordinates of the new coordinate $(4, -3)$ are             

  1. $\left(\displaystyle \frac{1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$

  2. $\left(\displaystyle \frac{1}{\sqrt{2}}, \frac{-7}{\sqrt{2}}\right)$

  3. $\left(\displaystyle \frac{-1}{\sqrt{2}}, \frac{-7}{\sqrt{2}}\right)$

  4. $\left(\displaystyle \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$


Correct Option: D
Explanation:

When axes are rotated through an angle $\theta $, then 

$x=X \cos\theta -Y \sin\theta$

$y=Y \cos\theta -X \sin\theta$

where(x,y) are initial coordinates and (X,Y) are new coordinates

$\therefore x=4\left(\dfrac{-1}{\sqrt{2}}\right)-\left(-3\right)\left(\dfrac{1}{\sqrt{2}}\right)=\dfrac{-1}{\sqrt{2}}$

$y=4\left(\dfrac{1}{\sqrt{2}}\right)-3\left(\dfrac{-1}{\sqrt{2}}\right)=\dfrac{7}{\sqrt{2}}$


$\therefore (x,y)=\left(\displaystyle \frac{-1}{\sqrt{2}}, \frac{7}{\sqrt{2}}\right)$

The point $(4,3)$ is translated to the point $(3,1)$ and then axes are rotated through $30^{\mathrm{o}}$ about the origin, then the new position of the point is 

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

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

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

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


Correct Option: C
Explanation:

When axes are rotated through an angle $\theta $, then

$x=\alpha +x^{1}\cos\theta -y^{1}\sin\theta  ; y=\beta +x^{1}\sin\theta +y^{1}\cos\theta $

$\Rightarrow 4=3+\dfrac{x^{1}\sqrt{3}}{2}-\dfrac{y^{1}}{2} ;3=1+\dfrac{x^{1}}{2}+\dfrac{y^{1}\sqrt{3}}{2}$

$\Rightarrow x^{1}\sqrt{3}-y^{1}=2$

$ x^{1}+y^{1}\sqrt{3}=4$

$\Rightarrow x=\dfrac{\sqrt{3}+2}{2}$

$y=\dfrac{2\sqrt{3}-1}{2}$