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Questions Related to maths

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

  1. $\cfrac{15}{2}$
  2. $\cfrac{3}{2}$
  3. $\cfrac{2}{5}$
  4. $\cfrac{5}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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

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

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) $\displaystyle x-y+z-6=0$

  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
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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

  1. $\displaystyle -\frac{3}{5}$
  2. $\displaystyle \frac{5}{3}$
  3. $\displaystyle -\frac{4}{3}$
  4. $\displaystyle \frac{3}{4}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 }} $

$\therefore  \lambda =\dfrac{5}{3}$

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

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

  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. $\displaystyle \cos \theta =-\frac{11\sqrt{7}}{70}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $\sin\theta =\dfrac{2\sqrt{6}}{15} \Rightarrow \theta =\sin\dfrac{2\sqrt{6}}{15}$

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

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.

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

The projection of a line on a plane involves finding the plane that contains the original line and the projected line. By finding the normal to this plane and using the given point, the constant k can be determined.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

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 

  1. $2 sin\frac{\pi }{10}$
  2. $2 sin\frac{\pi }{12}$
  3. $2 sin\frac{\pi }{4}$
  4. $2 sin\frac{\pi }{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The problem involves geometric properties of an ellipse. Solving for the orthocenter condition leads to the relation involving the eccentricity squared.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

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

  1. $\sqrt { 2 } -1$
  2. $\sqrt { 3 } -1$
  3. $\cfrac { 1 }{ 2 } $
  4. $\cfrac { 1 }{ 3 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Given foci (3,1) and (1,1), the center is (2,1) and 2ae = 2, so ae = 1. Using the point (1,3), the sum of distances to the foci equals 2a.

Multiple choice maths ellipse special cases of an ellipse eccentricity equation of ellipse

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?

  1. $\dfrac{\sqrt{2}}{2}$
  2. $\dfrac{\sqrt{3}}{2}$
  3. $\dfrac{1}{2}$
  4. $\dfrac{3}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer