Tag: maths

Questions Related to maths

If the angle between the line $x=\cfrac{y-1}{2}=\cfrac{z-3}{\lambda}$ and the plane $x+2y+3z=4$ is $\cos ^{ -1 }{ \left( \sqrt { \cfrac { 5 }{ 14 }  }  \right)  } $, then $\lambda$ equals

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\cfrac{15}{2}$

  2. $\cfrac{3}{2}$

  3. $\cfrac{2}{5}$

  4. $\cfrac{5}{3}$

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Correct Option: A
Explanation:

equation of line $\dfrac{x-0}{1}=\dfrac{y-1}{2}=\dfrac{2-3}{\lambda}$ where $a _1=1, a _2=2, a _3=\lambda$

equation of plane $n+2y+32=4$
where $b _1=1, b _2=2, b _3=3$
angle between Line and plane
$\cos\theta=\dfrac{|a _1b _1+a _2b _2+a _3b _3|}{\sqrt{a _1^2+a _2^2+a _3^2}.\sqrt{b _1^2+b _2^2+b _3^2}}$
$\cos \theta=\dfrac{|1.1+2.2+3.\lambda|}{\sqrt{1+4+\lambda^2}.\sqrt{1+4+9}}$
$\cos \theta =\dfrac{5+3\lambda}{\sqrt{5+\lambda^2}.\sqrt{14}}\quad ---(1)$
given $\theta =\cos^{-1}\left(\sqrt{\dfrac{5}{14}}\right)$
$\cos \theta =\left(\sqrt{\dfrac{5}{14}}\right)\quad ----(2)$
By eqn $(1)$ & $(2)$
$\dfrac{5+3\lambda }{\sqrt{5+\lambda^2}\sqrt{14}}=\sqrt{\dfrac{5}{14}}$
$5+3\lambda=\sqrt{5}.\sqrt{5+\lambda^2}$
$(5+3\lambda)^2=5.(5+\lambda^2)$
$25+9\lambda^2+30\lambda =25+5\lambda^2$
$4\lambda^2+30\lambda =0$
$\lambda (2\lambda +15)=0\Rightarrow \lambda =0, \dfrac{15}{2}$
but $\lambda \neq 0$
So $\lambda =\dfrac{15}{2}$ Ans

How is the line $\displaystyle \frac{x-4}{4}=\frac{y-12}{12}=\frac{z-8}{8}$ related to the planes
(A) $\displaystyle x-y+z=0$
(B) $\displaystyle x-y+z-6=0$

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. parallel to plane A but not B

  2. parallel to plane A and also lies in plane A but not parallel to B

  3. parallel to plane A and also lies in plane A

  4. none of these

Show answer
Correct Option: C
Explanation:

A line is inclined at Φ to a plane. The vector equation of the line is given by 

$\vec r =\vec a + \lambda \vec b $
Let $\theta$ be the angle between the line and the normal to the plane. Its value can be given by the following equation
$cos\theta=|\dfrac{\vec b . \vec n }{|\vec b|.|\vec n |}|$
Finding the value of the $Φ$ between the line and the plane we know that 
parallel to plane A and also lies in plane A.

If the angle $\theta $ between the line $\displaystyle \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}$ and the plane $2x-y+\sqrt{\lambda} z+4=0$ is such that $\displaystyle \sin \theta =\frac{1}{3}$, then value of $\lambda $ is

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\displaystyle -\frac{3}{5}$

  2. $\displaystyle \frac{5}{3}$

  3. $\displaystyle -\frac{4}{3}$

  4. $\displaystyle \frac{3}{4}$

Show answer
Correct Option: B
Explanation:

Angle between the line and plane is same as the angle between the line and normal to the plane
$\displaystyle \therefore \cos \left ( 90^{0}-\theta  \right )=\frac{a _{1}a _{2}+b _{1}b _{2}+c _{1}c _{2}}{\sqrt{a _{1}^{2}+b _{1}^{2}+c _{1}^{2}}\sqrt{a _{2}^{2}+b _{2}^{2}+c _{2}^{2}}}$
$\displaystyle \therefore\frac{1}{3}=\frac{\left ( 1\times 2+2\times \left ( -1 \right )+2\sqrt{\lambda } \right )}{\sqrt{1^{2}+2^{2}+2^{2}}\sqrt{2^{2}+1^{2}+\lambda }} $

$\therefore  \lambda =\dfrac{5}{3}$

If $\displaystyle \theta$ is the angle between the line 
$\vec r=2i+j-k+\left ( i+j+k \right )t$ and the plane
$\displaystyle \vec r\cdot \left ( 3i-4j+5k \right )=q$, then

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $\displaystyle \cos \theta =\frac{2\sqrt{6}}{15}$

  2. $\displaystyle \sin \theta =\frac{2\sqrt{6}}{15}$

  3. $\displaystyle \sin \theta =-\frac{11\sqrt{7}}{70}$

  4. $\displaystyle \cos \theta =-\frac{11\sqrt{7}}{70}$

Show answer
Correct Option: B
Explanation:

 $\theta$ is angle b/w $\xrightarrow [\gamma]{} =2\hat {  i}+j+k+(i+j+k)t$ and $\rightarrow.(3\hat { i }-4\hat { j }+5k)=q$

Angle b/w line and plane is given by 

$\sin\theta =\dfrac{4 _1a _2+b _1b _2+c _1c _2}{\sqrt{a _1^2+b _1^2+c _1^2}\sqrt{a _2^2+b _2^2+c _2^2}}$   

Where $(a _1,b _1,c _1)$ and $(a _2,b _2,c _2)$ are direction ratios of line and plane Respectively so here 

$a _1,b _1,c _1)=(1,1,1)$ and $(a _2,b _2,c _2)=(3,-4,5)$

So $\sin \theta=\dfrac{3-4+5}{\sqrt{1+1+1}\sqrt{9+16+25}}$

$\dfrac{4}{\sqrt{3}\sqrt{50}}=\dfrac{4}{\sqrt{3}5\sqrt{2}}=\dfrac{4}{\sqrt{6.5}}\times \dfrac{\sqrt{6}}{\sqrt{6}}=\dfrac{2\sqrt{6}}{5.3}=\dfrac{2\sqrt{6}}{15}$

so here $\sin\theta =\dfrac{2\sqrt{6}}{15} \Rightarrow \theta =\sin\dfrac{2\sqrt{6}}{15}$

The projection of line $\displaystyle\frac{x}{2}=\frac{y-1}{2}=\frac{z-1}{1}$ on a plane 'P' is $\displaystyle\frac{x}{1}=\frac{y-1}{1}=\frac{z-1}{-1}$. If the plane P passes through $(k, -2, 0)$, then k is greater than.

angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

  1. $2$

  2. $3$

  3. $5$

  4. $4$

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Correct Option: A

Eccentricity of the conic $3x^{2}+2xy-3y^{2}+x+y-2=0$

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. 2

  2. $\sqrt{2}$

  3. 3

  4. none

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Correct Option: A

If circle whose diameter is major axis of ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1(a>b>0)$ meets minor axis at point P and orthocentre of $\Delta PF _{1}F _{2}$ lies on ellipse where $F _{1}$  and $F _{2}$ are foci of ellipse, then square of eccentricity of ellipse, is 

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. $2 sin\frac{\pi }{10}$

  2. $2 sin\frac{\pi }{12}$

  3. $2 sin\frac{\pi }{4}$

  4. $2 sin\frac{\pi }{2}$

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Correct Option: A

If (3,4), (5,12) are two foci of the ellipse passing thrpsough the origin. Then the eccentricity of the ellipse is 

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. $\frac{\sqrt{17}}{3}$

  2. $\frac{\sqrt[2]{17}}{3}$

  3. $\frac{\sqrt{34}}{3}$

  4. $\frac{\sqrt{34}}{9}$

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Correct Option: B

An ellipse has foci (3, 1), (1, 1) and it passes through point (1, 3). Its eccentricity is equal to 

maths ellipse special cases of an ellipse eccentricity equation of ellipse

  1. $\sqrt { 2 } -1$

  2. $\sqrt { 3 } -1$

  3. $\cfrac { 1 }{ 2 } $

  4. $\cfrac { 1 }{ 3 } $

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Correct Option: A

The ellipse $E _1:\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$ is inscribed in a rectangle R whose sides are parallel to the coordinates axis. Another ellipse $E _2$ passing through the point $(0, 4)$ circumscribes the rectangle R. The eccentricity of the ellipse $E _2$ is?

maths ellipse special cases of an ellipse eccentricity equation of ellipse

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

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

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

  4. $\dfrac{3}{4}$

Show answer
Correct Option: A