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

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

  1. $4x-y-2z=6$
  2. $4x-y-2z=-6$
  3. $4x-y+2z=6$
  4. $4x+y-2z=6$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $(4,-1,-2)$.
$4x-y-2z=6$

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

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:

  1. $9x+2y-5z+8 =0$
  2. $9x +2y -5z +10=0$
  3. $9x-2y +5z +6=0$
  4. $9x -2y -5z+4=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $(3, -2, 7)$
$\therefore 9(x - 3) - 2(y + 2) - 5(z - 7)=0$
$\Rightarrow 9x - 27 - 2y - 4 - 5z + 35 = 0$
$\Rightarrow 9x - 2y - 5z + 4 = 0$

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

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

  1. $x + 2y - 2z = 0$
  2. $3x + 2y - 2z = 0$
  3. $x - 2y + z = 0$
  4. $5x + 2y - 4z = 0$
Reveal answer Fill a bubble to check yourself
C Correct answer
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 $26x-52y+26z=0$ $i.e.$ $x-2y+z=0$

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

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

  1. 2

  2. -1

  3. 0

  4. 1
Reveal answer Fill a bubble to check yourself
D Correct answer
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
Multiple choice maths vectors:planes in three dimensions cartesian equation of plane general form of the equation of a plane lines in space

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

  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 } } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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

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-

  1. $2x+y=5$
  2. $2x-y=5$
  3. $2x+z=5$
  4. $2x-z=5$
Reveal answer Fill a bubble to check yourself
C Correct answer
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.

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

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-

  1. $4x+3y+5z=25$
  2. $4x+3y+5z=50$
  3. $3x+4y+5z=49$
  4. $x+7y-5z=2$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 $4x+3y+5z=50$

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

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

  1. $2x-3y+z-2\sqrt {14}=0$
  2. $2x-3y+z-\sqrt {14}=0$
  3. $2x-3y+z+2=0$
  4. $2x-3y+z-2=0$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $2x-3y+z=2\sqrt{14}$.

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

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:

  1. $x-y+z=1$
  2. $x+y+z=5$
  3. $x+2y-z=1$
  4. $2x-y+z=5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The plane passes through (3, 2, 0) and contains the line with direction (1, 5, 4). The normal vector is the cross product of the vector from (3, 2, 0) to a point on the line (3, 6, 4) and the line's direction vector.