Tag: equation of a plane in intercept form

Questions Related to equation of a plane in intercept form

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

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} + \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 $

Show answer
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?

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. $15, -5, 3$

  2. $1, -5, 3$

  3. $-15, 5, -3$

  4. $1, -6, 20$

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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

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: 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,$k=?$

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. $1$

  2. $3$

  3. $2$

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

Show answer
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?

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

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

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

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

Show answer
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?

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. $x+y+z=3$

  2. $x+y+z=6$

  3. $x+y+z=9$

  4. $x+y+z=4$

Show answer
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

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. $(0, 0, 0)$

  2. $(1, 1, 1)$

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

  4. $(2, 2, 2)$

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Correct Option: D

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,
$\dfrac{x}{a}-\dfrac{y}{b}-\dfrac{z}{c}=0$