Mathematics

Straight Lines and Angles

162 Questions

Straight lines and angles are core components of coordinate geometry. This topic evaluates angle measures between intersecting lines, direction ratios, and perpendicular distances. Mastery of these mathematical concepts is necessary for high scores in quantitative exams.

Angle between linesDirection ratiosAngle bisectorsPerpendicular distanceSlope differences

Straight Lines and Angles Questions

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

If the intercepts made on the axes by the plane which bisects the line joining the points $(1, 2, 3)$ and $(-3, 4, 5)$ at right angles are $(a,0,0), (0,b,0)$ and $(0,0,c)$ then $(a,b,c)$ is 

  1. $\left (-\dfrac {9}{2}, 9, 9\right)$
  2. $\left (\dfrac {1}{2}, 1, 1\right)$
  3. $\left (1, -\dfrac {1}{2}, 1\right)$
  4. $\left (1, \dfrac {1}{2}, 1\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given points are (1,2,3) and (-3,4,5)
Mid point of this segment is, $(-1,3,4) = M$(say)
and direction ratio are, $(4,-2,-2)$
Therefore, normal vector perpendicular to required plane is $\vec{n} = 4\hat{i}-2\hat{j}-2\hat{k}$
Since required plane is bisecting given points perpendicularly, so point $M$ will lie in the plane.
Therefore equation of plane is given by,
$((x+1)\hat{i}+(y-3)\hat{j}+(z-4)\hat{k} ) \cdot \vec{n} = 0$
$\Rightarrow ((x+1)\hat{i}+(y-3)\hat{j}+(z-4)\hat{k} ) \cdot (4\hat{i}-2\hat{j}-2\hat{k}) = 0$
$\Rightarrow 2x-y-z+9=0$
Hence, intercepts made on the axes are $\left(-\cfrac{9}{2}, 9, 9\right)$

Multiple choice maths the plane equation of a plane in intercept form equation of a plane in different forms different forms of equation of a plane

If $5, 3, 2$ are the direction ratios of a normal to the plane passing through the point $(2, 3, 1)$, then the sum of the intercepts made by the plane on the $x$ -axis and $y$ - axis is

  1. $\displaystyle \dfrac{8}{21}$
  2. $56$
  3. $\displaystyle \dfrac{56}{5}$
  4. $\displaystyle \dfrac{217}{10}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The equation of the plane will be of the form $5x + 3y+2z = d$

Since, the plane passes through $(2,3,1)$

$d = 21$

Intercepts along the axes are $\displaystyle \dfrac{21}{5} , 7 , \dfrac{21}{2} $ respectively.

Their sum of $x$-axis and $y$-axis intercepts is $\displaystyle \dfrac{56}{5} $.

Multiple choice maths ellipse normal to an ellipse tangent and normal to an ellipse two dimensional analytical geometry-ii

The line $2x+y =3$ cuts the ellipse $4x^2+y^2 =5$ at P and Q . If $\theta$ be the angle between the normals  at these point then $tan \theta$ =

  1. $1/2$
  2. $3/4$
  3. $3/5$
  4. $5$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The line 2x+y=3 intersects the ellipse 4x^2+y^2=5 at P(1, 1) and Q(1/2, 2). The slopes of the normals at these points are calculated using the derivative dy/dx = -4x/y. The slope of the normal at P(1, 1) is 1/4 and at Q(1/2, 2) is 1. The angle theta between them satisfies tan(theta) = |(1 - 1/4) / (1 + 1*1/4)| = (3/4) / (5/4) = 3/5.

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

A line lies in $YZ-$plane and makes angle of $30^o$ with the $Y-$axis, then its inclination to the $Z-$axis is 

  1. $30^o$ or $60^o$
  2. $60^o$ or $90^o$
  3. $60^o$ or $120^o$
  4. $30^o$ or $150^o$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

since line lies on $y-z$ plane $\alpha ={ 90 }^{ 0 }$

$\beta ={ 30 }^{ 0 }$
$\therefore \cos ^{ 2 }{ \alpha  } +\cos ^{ 2 }{ \beta  } +\cos ^{ 2 }{ \gamma  } =1$
$\therefore \cos ^{ 2 }{ { 30 }^{ 0 } } +\cos ^{ 2 }{ { 30 }^{ 0 } } +\cos ^{ 2 }{ \gamma  } =1$
$\cos ^{ 2 }{ \gamma  } =\cfrac { 1 }{ 4 } \Rightarrow \cos { \gamma  } =\pm \cfrac { 1 }{ 2 } $
$\gamma ={ 60 }^{ 0 },{ 120 }^{ 0 }$
Ans: $C$

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The angle between the planes $\bar { r } \cdot \bar { n _{ 1 } } =\left| \bar { { d } _{ 1 } }  \right| $ and $\bar { r } \cdot \bar { n _{ 2 } } =\left| \bar { { d } _{ 2 } }  \right| $

  1. $\cos^{-1}\left(\displaystyle \frac{\bar{n _{1} }\cdot\bar{d} _{1}}{\left | \bar{d} _{1}\times \bar{d} _{2} \right |}\right)$
  2. $\cos^{-1}\left(\displaystyle \frac{\bar{n} _{1}.\bar{n} _{2}}{\left |\bar{n} _{1} \right |\left | \bar{n} _{2} \right |}\right)$
  3. $\cos^{-1}\left(\displaystyle \frac{\bar{n} _{1}\bar{n} _{2}}{\bar{n} _{1}\times \bar{n} _{2} }\right)$
  4. $\cos^{-1}\left(\displaystyle \frac{\bar{n} _{1}\cdot \left | \bar{d} _{2} \right |}{\left | \bar{n} _{1} \right |\left | \bar{n} _{2} \right |}\right)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Given planes are $\bar { r } \cdot \bar { n _{ 1 } } =\left| \bar { { d } _{ 1 } }  \right| $ and $\bar { r } \cdot \bar { n _{ 2 } } =\left| \bar { { d } _{ 2 } }  \right| $ 

Angle between the planes is same as the angle between the normal vectors.
Hence the angle  $\theta=\cos^{-1}\left(\dfrac{\bar{n} _1.\bar{n} _2}{|\bar{n} _1||\bar{n} _2|}\right)$

Multiple choice maths the plane angle between planes angle between two planes problems involving equation of plane

The angle between two planes $\displaystyle r.n=q$ and $\displaystyle r.n'=q'$ is

  1. $\displaystyle \sin ^{-1}\left ( \frac{n.n'}{nn'} \right )$
  2. $\displaystyle \cos ^{-1}\left ( \frac{n.n'}{nn'} \right )$
  3. $\displaystyle \tan ^{-1}\left ( \frac{n.n'}{nn'} \right )$

  4. None of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Given planes
$r\cdot n=q$------(1)
$r\cdot {n}'={q}'$-------(2)
Angle between two planes is between their normal vector 
$\left | n \right |\left | {n}' \right |cos\alpha=n \cdot {n}'$
$cos\alpha=\dfrac{n \cdot {n}'}{\left | n \right |\left | {n}' \right |}$
$\alpha=\cos^{-1}(\dfrac{n \cdot {n}'}{\left | n \right |\left | {n}' \right |})$
Multiple choice maths arithmetic progressions complete the a.p series with given information properties of an ap problems on ap

The line joining $A$ $\left( b\cos { \alpha ,\ b\sin { \alpha  }  }  \right)$ and $B$ $\left( a\cos { \beta ,\ a\sin { \beta  }  }  \right)$ is produced to the point $M$ $\left( x,y \right)$, so that $AM$ and $BM$ are in the ration $b:a$. Prove that
$x+y\ \tan { \left( \dfrac { \alpha +\beta  }{ 2 }  \right)  } =0$

  1. $-1$
  2. $0$
  3. $\sin (\alpha + \beta /2)$
  4. $\sin (\alpha - \beta /2)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Given $\dfrac{AM}{BM}=\dfrac{b}{a}$
$\Rightarrow M$ divides $AB$ externally in the ratio $b:a$
$\Rightarrow x=\dfrac{ba\cos \beta-ab\cos \alpha}{b-a}$ and $y=\dfrac{ba\sin \beta-ab\sin \alpha}{b-a}$
$\Rightarrow \dfrac{x}{y}=\dfrac{\cos \beta-\cos \alpha}{\sin \beta-\sin \alpha}$
$\cos \beta=\dfrac{1-\tan^2(\beta/2)}{1+\tan^2(\beta/2) }$, $\cos \alpha =\dfrac{1-\tan^2(\alpha /2)}{1+\tan^2(\alpha /2)}$, $\sin \beta=\dfrac{2\tan (\beta /2)}{1+\tan^2(\beta/2)}, \sin \alpha=\dfrac{2\tan \dfrac{\alpha}{2}}{1+\tan^2\dfrac{\alpha}{2}}$
$\Rightarrow \dfrac{x}{y}=\dfrac{\dfrac{1-\tan^2\dfrac{\beta}{2}}{1+\tan^2 \beta/2}-\dfrac{1-\tan^2\dfrac{\alpha}{2}}{1+\tan^2 \alpha/2}}{\dfrac{2\tan \beta/2}{1+\tan^2 \beta/2}-\dfrac{2\tan \alpha/2}{1+\tan^2 \alpha/2}}=\displaystyle \dfrac { 1+\tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } -\tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  } -\tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } \tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  } -1-\tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  } +\tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } +\tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } \tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  }  }{ 2\tan { \dfrac { \beta  }{ 2 }  } +2\tan { \dfrac { \beta  }{ 2 }  } \tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } -2\tan { \dfrac { \alpha  }{ 2 }  } -2\tan { \dfrac { \alpha  }{ 2 } \tan ^{ 2 }{ \dfrac { \beta  }{ 2 }  }  }  } $
$\Rightarrow \dfrac { x }{ y } =\dfrac { 2\left( \tan ^{ 2 }{ \dfrac { \alpha  }{ 2 }  } -\tan ^{ 2 } \beta /2 \right)  }{ 2\left( \tan  \beta /2-\tan  \dfrac { \alpha  }{ 2 }  \right) \left( 1-\tan  \dfrac { \alpha  }{ 2 } \tan { \beta /2 }  \right)  } =\dfrac { -\left( \tan  \dfrac { \alpha  }{ 2 } -\tan  \dfrac { \beta  }{ 2 }  \right) \left( \tan  \dfrac { \alpha  }{ 2 } +\tan  \dfrac { \beta  }{ 2 }  \right)  }{ \left( \tan  \dfrac { \alpha  }{ 2 } -\tan  \dfrac { \beta  }{ 2 }  \right) \left( 1-\tan  \dfrac { \alpha  }{ 2 } \tan  \dfrac { \beta  }{ 2 }  \right)  } $
$\Rightarrow x+y\dfrac{\tan \dfrac{\alpha}{2}+\tan \beta/2}{1-\tan \dfrac{\alpha}{2}\tan \dfrac{\beta}{2}}=0\Rightarrow x+y\tan \left(\dfrac{\alpha+\beta}{2}\right)=0$ Hence proved
Multiple choice

What is the bearing from point A to point B if the angle between the line connecting the two points and the north-south line is 120 degrees?

  1. N120E

  2. S120E

  3. N120W

  4. S120W
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The bearing from point A to point B is S120E because the angle between the line connecting the two points and the north-south line is 120 degrees east of south.

Multiple choice

What is the bearing from point A to point B if the angle between the line connecting the two points and the north-south line is 180 degrees?

  1. N180E

  2. S180E

  3. N180W

  4. S180W
Reveal answer Fill a bubble to check yourself
Correct answer
Explanation

The bearing from point A to point B is S0W because the angle between the line connecting the two points and the north-south line is 0 degrees west of south.

Multiple choice

What is the bearing from point A to point B if the angle between the line connecting the two points and the east-west line is 270 degrees?

  1. N270E

  2. S270E

  3. N270W

  4. S270W
Reveal answer Fill a bubble to check yourself
Correct answer
Explanation

The bearing from point A to point B is S90W because the angle between the line connecting the two points and the east-west line is 90 degrees west of south.

Multiple choice

Which of the following matrices represents a shear transformation that maps the line $y = x$ to the line $y = 2x$?

  1. $$\begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}$$
  2. $$\begin{bmatrix} 1 & 0 \\ 1 & 1 \end{bmatrix}$$
  3. $$\begin{bmatrix} 2 & 0 \\ 0 & 1 \end{bmatrix}$$
  4. $$\begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix}$$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The matrix $$\begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix}$$ represents a shear transformation that maps the line $y = x$ to the line $y = 2x$ because it satisfies the following equation: $$\begin{bmatrix} x' \ y' \end{bmatrix} = \begin{bmatrix} 1 & 1 \ 0 & 1 \end{bmatrix} \begin{bmatrix} x \ y \end{bmatrix}$$, where $$(x', y')$$ are the coordinates of a point after the shear transformation and $$(x, y)$$ are the coordinates of the point before the shear transformation.

Multiple choice

What is the angle between two lines in three-dimensional space?

  1. The angle between the two vectors that are parallel to the lines

  2. The angle between the two vectors that are perpendicular to the lines

  3. The angle between the two vectors that are parallel to the planes containing the lines

  4. The angle between the two vectors that are perpendicular to the planes containing the lines
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The angle between two lines in three-dimensional space is the angle between the two vectors that are parallel to the lines.