Tag: maths

Questions Related to maths

Vector form of plane $2x-z+1=0$ is _________

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $F.(2,-1,0)=1$

  2. $F.(2,-1,0)+1=0$

  3. $F.(2,0,-1)+1=0$

  4. $F.(2,0,-1)=1$

Show answer
Correct Option: C
Explanation:


$2x-z+1=0$
$\therefore$ $2x+0y-1z=-1$
$\therefore$ $(x,y,z).(2,0,-1)=-1$
$\therefore$ $F.(2,0,-1)+1=0$

Find the equation of the plane through the points $(1, 0, -1), (3, 2, 2)$ and parallel to the line $\dfrac{x-1}{1}=\dfrac{y-1}{-2}=\dfrac{z-2}{3}$.

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

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

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

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

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

Show answer
Correct Option: A
Explanation:

$L:\cfrac { x-1 }{ 1 } =\cfrac { y-1 }{ -2 } =\cfrac { z-2 }{ 3 } \ P(1,0,-1),Q(3,2,2)$

$ \therefore$ Direction of  $\overrightarrow { PQ } =2\hat { i } +2\hat { j } +3\hat { k } $
Plane is parallel to line $L$ & contains $\overrightarrow { PQ } $. (normal to plane $\bot$ to $PQ$ & $L$)
$\therefore \overrightarrow { n } =\left| \begin{matrix} \hat { i }  & \hat { j }  & \hat { k }  \ 2 & 2 & 3 \ 1 & -2 & 3 \end{matrix} \right| =(12)\hat { i } -3\hat { j } +(-6)\hat { k } $
direction ratios of normal are $(4,-1,-2)$
$\therefore$ Equation plane passing through $(1,0,-1)$ and having directions of normal $(4,-1,-2)$.
$4x-y-2z=6$

The equation of the plane passing through the straight line $\dfrac{x-1}{2}=\dfrac{y+1}{-1}=\dfrac{z-3}{4}$ and perpendicular to plane $x+2y +z=12$ is:

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $9x+2y-5z+8 =0$

  2. $9x +2y -5z +10=0$

  3. $9x-2y +5z +6=0$

  4. $9x -2y -5z+4=0$

Show answer
Correct Option: D
Explanation:
Let the DR of the normal of the plane be $<a, b, c>$
Since it passes through a line.
$\therefore$ Normal of the plane must be perpendicular to the line
$\therefore a + 2b + c = 0$ ...... $(1)$ ($\perp$ to another plane)
$2a - b + 4c = 0$ ...... $(2)$
On solving :
$\dfrac{a}{8 + 1} = \dfrac{-b}{4 - 2} = \dfrac{c}{-1 - 4}$
$\Rightarrow \dfrac{a}{9} = \dfrac{b}{-2} = \dfrac{c}{-5}$
$a = 9, b = -2, c = -5$
Any point on the straight line
$(2\alpha + 1, -\alpha - 1, 4\alpha + 3)$
putting $\alpha = 1$
$(3, -2, 7)$
$\therefore$ Equation of the line plane with normal DR $(9, -2, -5)$ and passing through $(3, -2, 7)$
$\therefore 9(x - 3) - 2(y + 2) - 5(z - 7)=0$
$\Rightarrow 9x - 27 - 2y - 4 - 5z + 35 = 0$
$\Rightarrow 9x - 2y - 5z + 4 = 0$

Equation of the plane containing the straight lines $\dfrac{x}{2} = \dfrac{y}{3} = \dfrac{z}{4}$ and perpendicular to the plane containing the straight lines  $\dfrac{x}{3} = \dfrac{y}{4} = \dfrac{z}{2}$ and $\dfrac{x}{4} = \dfrac{y}{2} = \dfrac{z}{3}$

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $x + 2y - 2z = 0$

  2. $3x + 2y - 2z = 0$

  3. $x - 2y + z = 0$

  4. $5x + 2y - 4z = 0$

Show answer
Correct Option: C
Explanation:

Vector normal to plane $P _1$ is $\overrightarrow{n _1}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\3&4&2\4&2&3\end{vmatrix}$


                                                       $=\hat{i}(12-4)-\hat{j}(9-8)+\hat{k}(6-8)$
                                                       $=8\hat{i}-\hat{j}-10\hat{k}$
Plane $P _2$ is perpendicular to this plane and it leaking the line $\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{2}{4}$

$\overrightarrow{n _2}$ is normal to

$\overrightarrow{n _2}=\begin{vmatrix}\hat{i}&\hat{j}&\hat{k}\8&-1&-10\2&3&4\end{vmatrix}$

       $=\hat{i}(-4+30)-\hat{j}(32+20)+\hat{k}(24+2)$

       $=26\hat{i}-52\hat{j}+26\hat{k}$

This plane $T _2$ having normal $n _2$ passing through $(0,0,0)$ is $26x-52y+26z=0$ $i.e.$ $x-2y+z=0$

Let a,b,c be any real numbers.Suppose that there are real numbers x,y,z not all zero such that $x=cy+bz , y=az+cx$ and $z=bx+ay$, then ${a^2} + {b^2} + {c^2} + 2abc $ is equal to

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. 2

  2. -1

  3. 0

  4. 1

Show answer
Correct Option: D
Explanation:
$a, b, c$ real numbers

$x, y, z$ real numbers not all zero

$x=cy+bz\rightarrow x-cy-bz=0 --- (1)$

$y=az+cx\rightarrow cx+y-az=0---(2)$

$z=bx+cy\rightarrow -bx-ay+z=0 --- (3)$

The system of equation have trivial solution then

$\begin{vmatrix} 1 & -c & -b \\ c & -1 & a \\ b & a & -1 \end{vmatrix}=0$

$\Rightarrow 1(1-a^{2})+c(-c-ab)-b(a+b)=0$

$\Rightarrow a^{2}+b^{2}+c^{2}+2abc=1$

$D$ is correct

The direction cosines of the normal to the plane $x+2y-3z+4=0$ are

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $\cfrac { -1 }{ \sqrt { 14 } } ,\cfrac { -2 }{ \sqrt { 14 } } ,\cfrac { 3 }{ \sqrt { 14 } } $

  2. $\cfrac { 1 }{ \sqrt { 14 } } ,\cfrac { 2 }{ \sqrt { 14 } } ,\cfrac { 3 }{ \sqrt { 14 } } $

  3. $\cfrac { -1 }{ \sqrt { 14 } } ,\cfrac { 2 }{ \sqrt { 14 } } ,\cfrac { 3 }{ \sqrt { 14 } } $

  4. $\cfrac { 1 }{ \sqrt { 14 } } ,\cfrac { -2 }{ \sqrt { 14 } } ,\cfrac { -3 }{ \sqrt { 14 } } $

Show answer
Correct Option: A
Explanation:

Clearly, the normal to the plane has DR's $\equiv(1, 2, -3)$

$\therefore $ DCs  are $\equiv \pm \left(\dfrac{1}{\sqrt{1^2 + 2^2 + (-3)^2}} , \dfrac{2}{\sqrt{14}} , \dfrac{-3}{\sqrt{14}}\right)$
$= \pm \left(\dfrac{1}{\sqrt{14}} , \dfrac{2}{\sqrt{14}} , \dfrac{-3}{\sqrt{14}} \right)$
$\therefore A$

The Cartesian equation of the plane $\vec r=(1+\lambda-\mu)\hat i+(2-\lambda)\hat j+(3-2\lambda+2\mu)\hat k$ is-

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $2x+y=5$

  2. $2x-y=5$

  3. $2x+z=5$

  4. $2x-z=5$

Show answer
Correct Option: C
Explanation:

Given, $\vec{r} = (1+\lambda-\mu)\hat{i}+(2-\lambda)\hat{j}+(3-2\lambda+2\mu)\hat{k}$
$\Rightarrow x\hat{i}+y\hat{j}+z\hat{k} = (1+\lambda-\mu)\hat{i}+(2-\lambda)\hat{j}+(3-2\lambda+2\mu)\hat{k}$
Comparing coefficient, we get
$ 1+\lambda-\mu = x, 2-\lambda=y, 3-2\lambda+2\mu=z$
$\Rightarrow\lambda = 2-y, \mu=1+\lambda - x = 3-y-x$
Eliminating $\mu$ and $\lambda$, we get
$2x+z=5$ which is required equation of plane in cartesian form.

The equation of a plane which passes through the point of intersection of lines $\dfrac {x-1}{3}=\dfrac {y-2}{1}=\dfrac {z-3}{2}$, and $\dfrac {x-3}{1}=\dfrac {y-1}{2}=\dfrac {z-2}{3}$ and at greatest distance from point $(0, 0, 0)$ is-

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $4x+3y+5z=25$

  2. $4x+3y+5z=50$

  3. $3x+4y+5z=49$

  4. $x+7y-5z=2$

Show answer
Correct Option: B
Explanation:

Any point on the first line is $P\left( 3\lambda +1,\lambda +2,2\lambda +3 \right) $
and on the second line is $Q\left( u+3,2u+1,3u+2 \right) $
$P$ and $Q$ represent the same point if $\lambda =u=1$
And the point of intersection of the given line is $P\left( 4,3,5 \right) $
The plane given in (a),(b),(c) and (d) all pass through $P$.
The plane at greatest distance is one which is at a distance equalt to $OP$ from the origin.
So, the distance of the plane from origin is $\sqrt { { 4 }^{ 2 }+{ 3 }^{ 2 }+{ 5 }^{ 2 } } =\sqrt { 50 } $
The equation of plane is $4x+3y+5z=50$

Let $A (1, 1, 1), B(2, 3, 5)$ and $C(-1, 0, 2)$ be three points, then equation of a plane parallel to the plane $ABC$ and at the distance $2$ is

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $2x-3y+z-2\sqrt {14}=0$

  2. $2x-3y+z-\sqrt {14}=0$

  3. $2x-3y+z+2=0$

  4. $2x-3y+z-2=0$

Show answer
Correct Option: A
Explanation:

$\vec{AB}=i+2j+4k$
$\vec{BC}=3i+3j+3k$

Hence, $\vec{AB}\times\vec{BC}=3(2i-3j+k)$
Now the unit normal of the plane of $ABC$ will be 
$=\dfrac{2i-3j+k}{\sqrt{14}}$
The required plane is parallel to the plane $ABC$.
Hence, its unit normal will be parallel to the normal of $ABC$.
Therefore, the equation of the required plane is 
$r.(\dfrac{2i-3j+k}{\sqrt{14}})=d$
$2x-3y+z=d\sqrt{14}$
Now $d$ is $2$.
Hence, the equation is $2x-3y+z=2\sqrt{14}$.

The plane which passes through the point $(3, 2, 0)$ and the line $\dfrac {x-3}{1}=\dfrac {y-6}{5}=\dfrac {z-4}{4}$ is:

maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

  1. $x-y+z=1$

  2. $x+y+z=5$

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

  4. $2x-y+z=5$

Show answer
Correct Option: A