Tag: maths

Questions Related to maths

The intercepts of the plane $2x-3y+5z-30=0$ are 

  1. $15,-10,6$

  2. $5,10,6$

  3. $1/8,-1/6,1/4$

  4. $3,-4,6$


Correct Option: A
Explanation:

We have,

$\begin{array}{l} 2x-3y+5z=30 \ \frac { x }{ { 15 } } +\frac { y }{ { -10 } } +\frac { z }{ 6 } =1 \end{array}$
Intercept we $(15,-10,6)$
Then,
OPtion $A$ is correct answer.

If $A=(3,1,-2) , B=(-1,0,1)$ and $l, m$ are the projections of AB on the y-axis, zx plane respectively then $3l^2-m+1$=

  1. 0

  2. 1

  3. 11

  4. 27


Correct Option: A

A plane $
\pi
 $ makes intercept 3 and 4 respectively on z-axis and x-axis. If $
\pi
 $ is parallel to y-axis, then its equation is


  1. 3x+4z=12

  2. 3z+4x=12

  3. 3y+4z=12

  4. 3z+4y=12


Correct Option: A
Explanation:
$X-$intercept $a=4$
and $Z-$intercept $c=3$
Required equation $\dfrac{x}{4}+\dfrac{z}{3}=1$
or $3x+4z=12$

The lengths of the intercepts on the co-ordinate axes made by the plane $5x+2y+z-13=0$ are

  1. $5, 2, 1$ unit

  2. $\dfrac{13}{5}, \dfrac{13}{2}, 13$ unit

  3. $\dfrac{5}{13}, \dfrac{2}{13}, \dfrac{1}{13}$ unit

  4. $1, 2, 5$ unit


Correct Option: B
Explanation:

Solution:

Given that:
$5x+2y+z-13=0$ 
or, $\cfrac{x}{\frac{13}{5}}+\cfrac{y}{\frac{13}{2}}+\cfrac{z}{13}=1$
$\therefore$ Lengths of intercepts are $\cfrac{13}5,\cfrac{13}2$ and $13.$
Hence, B is the correct option.

Equation of a plane making X-intercept $4$, Y-intercept ($-6$), Z-intercept $3$ is _______.

  1. $3x-4y+6z=12$

  2. $3x-2y+4z=12$

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

  4. $4x-3y+2z=12$


Correct Option: B
Explanation:

Equation of a plane which cuts intercepts $4,-6,3$ on axes is
$\cfrac { x }{ 4 } +\cfrac { y }{ (-6) } +\cfrac { z }{ 3 } =1$
$\therefore$ $3x-2y+4z=12$

Two system of rectangular axes have the same origin. If a plane cuts them at distances, $a$, $b$, $c$ and ${a} _{1}$,${b} _{1}$ , ${c} _{1}$ from the origin, then

  1. $\dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } } =\dfrac { 1 }{ { a } _{ 1 }^{ 2 } } +\dfrac { 1 }{ { b } _{ 1 }^{ 2 } } +\dfrac { 1 }{ { c } _{ 1 }^{ 2 } }$

  2. $\dfrac { 1 }{ { a }^{ 2 } } -\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } } =\dfrac { 1 }{ { a } _{ 1 }^{ 2 } } -\dfrac { 1 }{ { b } _{ 1 }^{ 2 } } +\dfrac { 1 }{ { c } _{ 1 }^{ 2 } }$

  3. ${ a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }={ a } _{ 1 }^{ 2 }+{ b } _{ 1 }^{ 2 }+{ c } _{ 1 }^{ 2 }$

  4. ${ a }^{ 2 }-{ b }^{ 2 }+{ c }^{ 2 }={ a } _{ 1 }^{ 2 }-{ b } _{ 1 }^{ 2 }+{ c } _{ 1 }^{ 2 }$


Correct Option: A
Explanation:
Let the equation of the plane be
$\dfrac { x }{ a } +\dfrac { y }{ b } +\dfrac { z }{ c } =1$  and $\dfrac { x }{ a _1 } +\dfrac { y }{ b _1 } +\dfrac { z }{ c _1 } =1$
$ax+by+cz+d=0\quad perpendicular\quad distance\quad from\quad origin\quad is\quad \dfrac { \left| d \right|  }{ \sqrt { { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 } }  } $
as they have the same origin their perpendicular distance is constant.
$\dfrac { 1 }{ \sqrt { \dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } }  }  } =\dfrac { 1 }{ \sqrt { \dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } }  }  }$
$\dfrac { 1 }{ { a _1}^{ 2 } } +\dfrac { 1 }{ { b _1 }^{ 2 } } +\dfrac { 1 }{ { c _1 }^{ 2 } } =\dfrac { 1 }{ { a }^{ 2 } } +\dfrac { 1 }{ { b }^{ 2 } } +\dfrac { 1 }{ { c }^{ 2 } } $


A plane $x-3y+5z=d$ passes through the point $(1,2,4)$. Intercepts on the axes are

  1. $15,-5,3$

  2. $1,-5,3$

  3. $-15,5,-3$

  4. $1,-6,20$


Correct Option: A
Explanation:

$(1, 2, 4)$ must satisfy this plane

$(1)(1) - (3)(2) + (5)(4) = d$ 
$\Rightarrow $ plane $\Rightarrow x - 3y + 5z = 15$
$x$ intercept $\Rightarrow x - 0 + 0 = 15 \Rightarrow 15 = x \, ml$
$y$ intercept $\Rightarrow 0 - 3y + 0 = 15 \Rightarrow y \, int = -5$
$z$ intercept $\Rightarrow 0 - 0 + 52 = 15 \Rightarrow z \, int . = 3$
$\therefore A = (15, -5 , 3)$