Tag: maths

Questions Related to maths

The equation of the plane which makes with the coordinate axes, a triangle with centroid $(\alpha, \beta, \gamma)$ is given by?

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

  1. $\alpha x+\beta y+\gamma z=1$

  2. $\alpha x+\beta y+\gamma z=3$

  3. $\dfrac{x}{\alpha}+\dfrac{y}{\beta}+\dfrac{z}{\gamma}=1$

  4. $\dfrac{x}{\alpha}+\dfrac{y}{\beta}+\dfrac{z}{\gamma}=3$

Show answer
Correct Option: D
Explanation:

Let the equation of the plane be $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

Then, it meets the axes at $A(a, 0, 0), B(0, b, 0)$ adn $C(0, 0, c)$.
$\therefore$ centroid of $\Delta ABC$ is $G\left(\dfrac{a}{3}, \dfrac{b}{3}, \dfrac{c}{3}\right)$.
$\therefore$ $\left(\dfrac{a}{3}=\alpha, \dfrac{b}{3}=\beta and \dfrac{c}{3}=\gamma\right)\Rightarrow a=3\alpha, b=3\beta$ and $c=3\gamma$.
$\therefore$ the required equation of the plane is
$\dfrac{x}{3\alpha}+\dfrac{y}{3\beta}+\dfrac{z}{3\gamma}=1$
$\Rightarrow \dfrac{x}{\alpha}+\dfrac{y}{\beta}+\dfrac{z}{\gamma}=3$.

The intercepts made by the plane $\vec{r}\cdot (2\hat{i}-3\hat{j}+4\hat{k})=12$ are?

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

  1. $2, -3, 4$

  2. $2, -3, -6$

  3. $-6, -4, 3$

  4. $-6, 4, 3$

Show answer
Correct Option: C
Explanation:

The given plane is $2x-3y+4z=12$

$\Rightarrow \dfrac{x}{6}+\dfrac{y}{-4}+\dfrac{z}{3}=1$
$\therefore$ required intercepts are $6, -4, 3$.

From a point $P\left ( a,\, b,\, c \right )$ perpendiculars $PM$ and $PN$ are drawn to $zx$ and $xy$-planes respectively, $O$ is the origin. An equation of the plane $OMN$ is

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

  1. $\displaystyle \frac{x}{a}\, -\, \frac{y}{b}\, -\, \frac{z}{c}= 0$

  2. $\displaystyle \frac{x}{a}\, -\, \frac{y}{b}\, +\, \frac{z}{c}= 0$

  3. $\displaystyle \frac{x}{a}\, +\, \frac{y}{b}\, +\, \frac{z}{c}= 0$

  4. $\displaystyle \frac{x}{a}\, +\, \frac{y}{b}\, -\, \frac{z}{c}= 0$

Show answer
Correct Option: A
Explanation:

If perpendicular $PM$ and $PN$ drawn from the point $P(a,b,c)$ to the plane $zx$ and $xy$, then coordinates of $M$ and $N$ are,
$M = (a,0,c)$ and $N = (a,b,0)$
Now general equation of plane passes through origin is given by,
$x+p y+q z = 0$
Also this plane passes through $M$ and $N$
$\Rightarrow a +qc=0$ ...(1)
and $ a+p b = 0$ ...(2)
Solving (1) and (2), we get
$p =-\dfrac {a}{b}$ and $q =-\dfrac {a}{c}$
Hence, equation required plane $OMN$ is,
$\dfrac{x}{a}-\dfrac{y}{b}-\dfrac{z}{c}=0$

A variable plane moves so that the sum of reciprocals of its intercepts on the three coordinate axes is constant $\lambda$. It passes through a fixed point, which has coordinates

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

  1. $\left( \lambda ,\lambda ,\lambda  \right) $

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

  3. $\left( -\lambda ,-\lambda ,-\lambda  \right) $

  4. $\displaystyle \left( -\frac { 1 }{ \lambda  } ,-\frac { 1 }{ \lambda  } ,-\frac { 1 }{ \lambda  }  \right) $

Show answer
Correct Option: B
Explanation:

Let the equation of the variable plane be

$\displaystyle \frac { x }{ a } +\frac { y }{ b } +\frac { z }{ c } =1$    ...(1)
The intercepts on the coordinate axes are $a,b,c$.
The sum of reciprocals of intercepts in constant $\lambda$, therefore
$\displaystyle \frac { 1 }{ a } +\frac { 1 }{ b } +\frac { 1 }{ c } =\lambda \Rightarrow \frac { \left( 1/\lambda  \right)  }{ a } +\frac { \left( 1/\lambda  \right)  }{ b } +\frac { \left( 1/\lambda  \right)  }{ c } =\lambda $
$\displaystyle \therefore \left( \frac { 1 }{ \lambda  } ,\frac { 1 }{ \lambda  } ,\frac { 1 }{ \lambda  }  \right) $ lies on the plane (1)
Hence, the variable plane (1) always passes through the fixed point $\displaystyle \left( \frac { 1 }{ \lambda  } ,\frac { 1 }{ \lambda  } ,\frac { 1 }{ \lambda  }  \right) $

A plane meets the coordinate axes in $A, B, C$ such that the centroid of the triangle $ABC$ is the point $(1,\, r,\, r^2)$. The plane passes through the point $(4, 8, 15)$, if $r$ is equal to

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

  1. $-3$

  2. $3$

  3. $5$

  4. $-5$

Show answer
Correct Option: A,D
Explanation:

Here $A(a,0,0)$, $B(0,b,0)$ and $C(0,0,c)$
Now, centroid $G(1,r,r^2)=\left( \dfrac { a }{ 3 } ,\dfrac { b }{ 3 } ,\dfrac { c }{ 3 }  \right) $
Therefore, $a=3$, $b=3r$ and $c=3r^2$
Now, $\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$ passes through $(4,8,15)$
Therefore, $\dfrac{4}{3}+\dfrac{8}{3r}+\dfrac{15}{3r^2}=1$

$\Rightarrow {4}r^{2}+{8}r +15=3r^{2}$
$\Rightarrow r^{2}+8r+15=0$
$\Rightarrow (r+3)(r+5)=0$
$\Rightarrow r=-3,-5$

If from the point $P(f, g, h)$ perpendiculars $PL, PM$ be drawn to $yz$ and $zx$ planes then the equation to the plane $OLM$ is -

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

  1. $\displaystyle \frac{x}{f}\, +\, \displaystyle \frac{y}{g}\, +\, \displaystyle \frac{z}{h}\, =\, 0$

  2. $\displaystyle \frac{x}{f}\, +\, \displaystyle \frac{y}{g}\, -\, \displaystyle \frac{z}{h}\, =\, 0$

  3. $\displaystyle \frac{x}{f}\, -\, \displaystyle \frac{y}{g}\, +\, \displaystyle \frac{z}{h}\, =\, 0$

  4. $ - \displaystyle \frac{x}{f}\, +\, \displaystyle \frac{y}{g}\, +\, \displaystyle \frac{z}{h}\, =\, 0$

Show answer
Correct Option: B
Explanation:

If perpendicular $PL$ and $PM$ drawn from the point $P(f,g,h)$ to the plane $yz$ and $zx$ then coordinates of $L$ and $M$ are,
$L = (0,g,h)$ and $M = (f,0,h)$
Now general equation of plane passes through origin is given by,
$x+p y+q z = 0$
Also this plane passes through $L$ and $M$
$\Rightarrow pg +qh=0 ...(1)$
and $ f+q h = 0 ...(2)$
Solving (1) and (2), we get
$p =\dfrac {f}{g}$ and $q =-\dfrac {f}{h}$
Hence, equation required plane $OLM$ is,
$\dfrac{x}{f}+\dfrac{y}{g}-\dfrac{z}{h} = 0$

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

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

  1. $\displaystyle \dfrac{8}{21}$

  2. $56$

  3. $\displaystyle \dfrac{56}{5}$

  4. $\displaystyle \dfrac{217}{10}$

Show answer
Correct Option: C
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} $.

Equation of the plane whose intercepts are $1,2,3$ is

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

  1. $6x+2y+3z=1$

  2. $x+y+z=6$

  3. $6x+3y+2z=6$

  4. $6x-3y-2z=1$

Show answer
Correct Option: C
Explanation:

The equation of the plane with intercepts $a,b,c$ on the axes is,
$ \dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} $ $= 1$
Since, $a = 1 b = 2, c = 3$
The equation of the plane is,
$ \dfrac{x}{1} + \dfrac{y}{2} + \dfrac{z}{3} $ $= 1$
$\therefore 6x + 3y + 2z = 6$

$5, 7$ are the intercepts of a plane on the $y$ - axis, $z$ - axis respectively. If the plane is parallel to the $x$-axis, then the equation of that plane is

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

  1. $5y+7z=35$

  2. $7y+5z=1$

  3. $\displaystyle \dfrac{y}{5}+\dfrac{Z}{7}=35$

  4. $7y+5z=35$

Show answer
Correct Option: D
Explanation:

The equation of the plane with intercepts $a,b,c$ on the axes is $ \displaystyle \dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} $ $= 1$
$a =$ infinite (plane is parallel to the $x$-axis, $b = 5$, $c = 7$
$ \therefore  \dfrac{y}{5} + \dfrac{z}{7} $ $= 1$
$\therefore 7y + 5z = 35$

The sum of the intercepts of the plane which bisects the line segment joining $(0,1,2)$ and $(2,3,0)$ perpendicularly is

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

  1. $2$

  2. $4$

  3. $6$

  4. $12$

Show answer
Correct Option: A
Explanation:

The dr's of the normal to the plane will be $(2,2,-2)$.
Hence, the plane is , $2x + 2y - 2z = d$
The midpoint of the line segment is $(1,2,1)$.
Hence, $d = 4$.
Hence, the intercepts are $\left (  \dfrac{1}{2} , 1 ,  \dfrac{1}{2} \right )$
The sum of intercepts is $2$.