 ### Equation of a plane in intercept form - class-XII

 Description: equation of a plane in intercept form Number of Questions: 34 Created by: Garima Pandit Tags: applications of vector algebra the plane three dimensional geometry - ii three dimensional geometry maths
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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 $\Rightarrow 0 - 0 + 52 = 15 \Rightarrow z \, int . = 3$
$\therefore A = (15, -5 , 3)$

From the point $P(a, b, c)$, let perpendiculars $PL$ and $PM$ be drawn to $YOZ$ and $ZOX$ planes, respectively. Then the equation of the plane $OLM$ is-

1. $\displaystyle \dfrac {x}{a}+\dfrac {y}{b}+\dfrac {z}{c}=0$

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

3. $\displaystyle \dfrac {x}{a}-\dfrac {y}{b}-\dfrac {z}{c}=0$

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

Correct Option: B
Explanation:

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

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. $\left (1, \dfrac {1}{2}, 1\right)$

Correct Option: A
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 $\left(-\cfrac{9}{2}, 9, 9\right)$

A plane makes intercept $3$ and $4$ with $x$ and $z$ axes and parallel to y-axis is

1. $3x+4z=12$

2. $4x+3z=12$

3. $3y+4z=12$

4. $4y+3x=12$

Correct Option: B
Explanation:

Consider the intercept $3$unit and $4$ unit with $x$ and $z$axes,

Now, equation of a plane which cut intercept $a,b$ and $c$ from $x-$axis, $y-$axis and $z-$axis is,

$\dfrac{x}{a}+\dfrac{y}{b}+\dfrac{z}{c}=1$

But given that, it is parallel to $y-$axis ,

So, $b=0$

$\dfrac{x}{a}+\dfrac{z}{c}=1$

Given that, $a=3$ unit and $c=4$ unit.

So,

$\dfrac{x}{3}+\dfrac{z}{4}=1$

$4x+3y=12$

So,

$\dfrac{x}{3}+\dfrac{z}{4}=1$

$4x+3z=12$

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

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

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

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

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

Correct Option: A
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

$p =\dfrac fg$ and $q =\dfrac{-f}{h}$

Hence, equation required plane $OLM$ is,

$\dfrac{x}{f}+\dfrac{y}{g}-\dfrac{z}{h} = 0$

If the plane $x-3y+5z=d$, passes through the point $(1, 2, 4)$, then the intercept on x, y, z axes are?

1. $15, -5, 3$

2. $1, -5, 3$

3. $-15, 5, -3$

4. $1, -6, 20$

Correct Option: A

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

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

Correct Option: A
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
$p =f/g$ and $q =-f/h$
Hence, equation required plane $OLM$ is,
$\dfrac{x}{f}+\dfrac{y}{g}-\dfrac{z}{h}=0$

A plane meet the co-ordinates axes in $A,B,C$ such that the centroid of triangle $ABC$ is the point $\alpha,\beta,\gamma.$ If the equation of the plane be $\displaystyle \frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=k$ then,$k=?$

1. $1$

2. $3$

3. $2$

4. $\alpha^{2}+\beta^{2}+\gamma^{2}$

Correct Option: B
Explanation:

Equation of plane ABC in intercept form is $\displaystyle \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1 .....(i)$

So$A=(a,0,0),B=(0,b,0),C=(0,0,c)$
Also centroid of $\displaystyle \bigtriangleup ABC\ is \left(\frac{a}{3},\frac{b}{3},\frac{c}{3}\right)$
But co-ordinates of centroid is $(\alpha,\beta,\gamma)$
$\displaystyle \therefore \alpha=\frac{a}{3}$
$\therefore a=3\alpha,b=3\beta,c=3\gamma$
Putting in (i) we get
$\displaystyle \frac{x}{3\alpha}+\frac{y}{3\beta}+\frac{y}{3\gamma}=1$
$\displaystyle \frac{x}{\alpha}+\frac{y}{\beta}+\frac{z}{\gamma}=3$

If a plane meets the coordinate axes in A, B and C such that the centroid of $\Delta ABC$ is $(1, 2, 4)$, then the equation of the plane is?

1. $x+2y+4z=6$

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

3. $x+2y+4z=7$

4. $4x+2y+z=7$

Correct Option: B
Explanation:

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

Then, it meets the coordinate axes in $A(a, 0, 0), B(0, b, 0), C(0, 0, c)$.

$\therefore$ centroid of $\Delta ABC$ is $G\left(\dfrac{a+0+0}{3}, \dfrac{0+b+0}{3}, \dfrac{0+0+c}{3}\right)$, i.e., $G\left(\dfrac{a}{3}, \dfrac{b}{3}, \dfrac{c}{3}\right)$

$\therefore \left(\dfrac{a}{3}=1, \dfrac{b}{3}=2, \dfrac{c}{3}=4\right)$

$\Rightarrow a=3, b=6, c=12$

Hence, the required equation of the plane is

$\dfrac{x}{3}+\dfrac{y}{6}+\dfrac{z}{12}=1$

$\Rightarrow 4x+2y+z=12$.

The equation of a plane passing through the point $A(2, -3, 7)$ and making equal intercepts on the axes, is?

1. $x+y+z=3$

2. $x+y+z=6$

3. $x+y+z=9$

4. $x+y+z=4$

Correct Option: B
Explanation:

Let the required equation of the plane be $\dfrac{x}{a}+\dfrac{y}{a}+\dfrac{z}{a}=1$, i.e., $x+y+z=a$

Since, it passes through the point $A(2, -3, 7)$, we have $2+(-3)+7=a$

$\Rightarrow a=6$

Hence, the required equation of the plane is $x+y+z=6$.

A variable plane moves so that the sum of the reciprocals of its intercepts on the coordinate axes is $\dfrac{1}{2}$. Then, the plane passes through the point

1. $(0, 0, 0)$

2. $(1, 1, 1)$

3. $\left(\dfrac{1}{2}, \dfrac{1}{2}, \dfrac{1}{2}\right)$

4. $(2, 2, 2)$

Correct Option: D

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

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

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?

1. $2, -3, 4$

2. $2, -3, -6$

3. $-6, -4, 3$

4. $-6, 4, 3$

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

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

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

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

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

1. $-3$

2. $3$

3. $5$

4. $-5$

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 -

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

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

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

2. $56$

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

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

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

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

2. $x+y+z=6$

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

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

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

1. $5y+7z=35$

2. $7y+5z=1$

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

4. $7y+5z=35$

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

1. $2$

2. $4$

3. $6$

4. $12$

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 $2$.

lf a plane meets the coordinate axes at $A,B,C$ , then equation of plane is such that centroid of triangle $ABC$ is $\left (\displaystyle \dfrac{1}{3}\dfrac{2} {3},\dfrac{4}{3}\right)$

1. $4x+2y+z=4$

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

3. $x+y+z=3$

4. $x+y+z=9$

Correct Option: A
Explanation:

Assume equation of plane is, $ax+by+cz=d$
Now this plane intersect axes at $A,B$ and $C$,
$\Rightarrow A = \left (\dfrac{d}{a}, 0, 0\right), B = \left (0, \dfrac{d}{b}, 0\right)$ and $C =\left (0, 0, \dfrac{d}{c}\right)$
So the centroid of triangle $ABC$ is, $\left (\dfrac{d}{3a}, \dfrac{d}{3b}, \dfrac{d}{3c}\right)$
Comparing this with given value $a=d, b = 2d$ and $c = 4d$
Hence equation of the plane is

$4x+2y+z = 4$

If from a point $P(a,b,c)$ perpendicular $PA$ and $PB$ are drawn to $yz$ and $zx$ planes, find the equation of the plane $OAB$:

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

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

3. None of these

Correct Option: A
Explanation:

The coordinates of $A$ and $B$ are $(0,b,c)$ and $(a,0,c)$ respectively.

The equation of the plane passing through $O(0,0,0),A(0,b,0)$ and $B(a,0,c)$ is given by
$\displaystyle \begin{vmatrix} x-0 & y-0 & z-0 \ 0-0 & b-0 & c-0 \ a-0 & 0-0 & c-0 \end{vmatrix}=0\Rightarrow bcx+acy-abz=0$
$\displaystyle \Rightarrow \frac { x }{ a } +\frac { y }{ b } -\frac { z }{ c } =0$

The equation of the plane which is parallel to y-axis and cuts off intercepts of length 2 and 3 from x-axis and z-axis is :

1. $3x + 2z = 1$

2. $3x+ 2z = 6$

3. $2x+ 3z = 6$

4. $3x+ 2z = 0$

Correct Option: B
Explanation:

The equation of plane parallel to $y-axis$ is, $ax+bz+1=0$      ----- ( 1 )

Here,  $x=2$ and $z=3$

Substituting $x=2$ and $z=0$ in equation ( 1 ),

$\Rightarrow$  $2a+0+1=0$

$\Rightarrow$  $x=\dfrac{-1}{2}$

Substituting $x=0$ and $z=3$ in equation ( 1 ),

$\Rightarrow$  $0+3b+1=0$

$\Rightarrow$  $b=\dfrac{-1}{3}$

Substituting value of $a$ and $b$ equation (  1 ) we get,

$\Rightarrow$  $\dfrac{-1}{2}x-\dfrac{1}{3}z+1=0$

$\Rightarrow$  $3x+2z=6$

The expression of $x+y+z=1$ in form of $x\cos { \alpha } +y\cos { \beta } +z\cos { \gamma } =p$ is _______.

1. $x+y+z=1$

2. $\cfrac { x }{ 2\sqrt { 3 } } +\cfrac { y }{ 2\sqrt { 3 } } +\cfrac { z }{ 2\sqrt { 3 } } =\cfrac { 1 }{ \sqrt { 3 } }$

3. $\cfrac { x }{ \sqrt { 3 } } +\cfrac { y }{ \sqrt { 3 } } +\cfrac { z }{ \sqrt { 3 } } =1$

4. $\cfrac { x }{ \sqrt { 3 } } +\cfrac { y }{ \sqrt { 3 } } +\cfrac { z }{ \sqrt { 3 } } =\cfrac { 1 }{ \sqrt { 3 } }$

Correct Option: D
Explanation:

$\cfrac { x }{ \sqrt { 3 } } +\cfrac { y }{ \sqrt { 3 } } +\cfrac { z }{ \sqrt { 3 } } =1$
$\quad \rightarrow P=\cfrac { \left| -1 \right| }{ \sqrt { 1+1+1 } } =\cfrac { 1 }{ \sqrt { 3 } }$
$\therefore x+y+z=1$
$\therefore \cfrac { x }{ \sqrt { 3 } } +\cfrac { y }{ \sqrt { 3 } } +\cfrac { z }{ \sqrt { 3 } } =\cfrac { 1 }{ \sqrt { 3 } }$

The sum of Y and Z intercepts of the plane $3x+4y-6z=12$ is ___________.

1. $10$

2. $4$

3. $1$

4. $5$

Correct Option: C
Explanation:

$3x+4y-6z=12$
$\therefore\dfrac{3x}{12}+\dfrac{4y}{12}+\dfrac{(-6)z}{12}=1$
$\therefore \dfrac{x}{4}+\dfrac{y}{3}+\dfrac{z}{(-2)}=1$
$\therefore$ y intercept $b=3$ and z intercept $c=-2$
$\therefore$ y intercept $+$ z intercept $=3+(-2)=1$

The plane $ax+by+cz=1$ meets the coordinate axes in $A, B$ and $C$. The centroid of the triangle is:

1. $(3a, 3b, 3c)$

2. $\left( \dfrac { a }{ 3 } ,\dfrac { b }{ 3 } ,\dfrac { c }{ 3 } \right)$

3. $\left( \dfrac { 3 }{ a } ,\dfrac { 3 }{ b }, \dfrac { 3 }{ c } \right)$

4. $\left( \dfrac { 1 }{ 3a } ,\dfrac { 1 }{ 3b } ,\dfrac { 1 }{ 3c } \right)$

Correct Option: D
Explanation:

The plane $ax + by + cz = 1$ meets the coordinate axis in $A,B,C$, then the coordinates will be,

$A\left( {a,0,0} \right)$, $B\left( {0,b,0} \right)$ and $C\left( {0,0,c} \right)$

The equation of the plane in intercept form is,

$\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{c}{z} = 1$

The intercepts that the plane make on the axis is $\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}$

Let C denotes the centroid, then,

$C = \left( {\dfrac{{\dfrac{1}{a} + 0 + 0}}{3},\dfrac{{0 + \dfrac{1}{b} + 0}}{3},\dfrac{{0 + 0 + \dfrac{1}{c}}}{3}} \right)$

Therefore, the coordinates of the centroid will be $\left( {\dfrac{1}{{3a}},\dfrac{1}{{3b}},\dfrac{1}{{3c}}} \right)$.

If a plane passes through a fixed point $\left ( 2, 3, 4 \right )$ and meets the axes of reference in $A$, $B$ and $C$, the point of intersection of the planes through $A$, $B$, $C$ parallel to the coordinate planes can be

1. $\left ( 6, 9, 12 \right )$

2. $\left ( 4, 12, 16 \right )$

3. $\left ( 1, 1, -1 \right )$

4. $\left ( 2, 3, -4 \right )$

Correct Option: A,B,C,D
Explanation:

Let us say a plane P $ax+by+cz=k$ passes through $\left( 2,3,4 \right)$ so $2a+3b+4c=k \quad -(1)$

$A\left( \dfrac { k }{ a } ,0,0 \right) ,\quad B\left( 0,\dfrac { k }{ b } ,0 \right) ,\quad C\left( 0,0,\dfrac { k }{ c } \right)$

Points of intersection will be $\left< \dfrac { k }{ a } ,\dfrac { k }{ b } ,\dfrac { k }{ c } \right>$

Let $\dfrac { k }{ a } =x\quad \dfrac { k }{ b } =y\quad \dfrac { k }{ c } =z$ so in $(1)$

$\dfrac { 2k }{ x } +\dfrac { 3k }{ y } +\dfrac { 4k }{ z } =k$

$\dfrac { 2 }{ x } +\dfrac { 3 }{ y } +\dfrac { 4 }{ z } =1\quad -(1)$

$(a)$ if $(x,y,z) = (6,9,12)$

$\dfrac { 2 }{ 6 } +\dfrac { 3 }{ 9 } +\dfrac { 4 }{ 12 } =\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 3 } +\dfrac { 1 }{ 3 } =1$ Hence true.

$(b)$ $\left< 4,12,16 \right>$

$\dfrac { 2 }{ 4 } +\dfrac { 3 }{ 12 } +\dfrac { 4 }{ 16 } =\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 4 } +\dfrac { 1 }{ 4 } =1$ Hence correct

$(c)$ $\left< 1,1,-1 \right>$

$\dfrac { 2 }{ 1 } +\dfrac { 3 }{ 1 } +\dfrac { 4 }{ -1 } =1$ Hence this is also correct.

$(d)$ $\left< 2,3,-4 \right>$

$\dfrac { 2 }{ 2 } +\dfrac { 3 }{ 3 } +\dfrac { 4 }{ -4 } =2-1=1$ This is also correct.

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