Tag: vectors:planes in three dimensions

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Questions Related to vectors:planes in three dimensions

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

The general equation of plane which is parallel to x-axis is

  1. $ax+by+cz+d=0, a\neq 0,b\neq 0,c\neq 0$
  2. $by+ax+d=0, a\neq 0,b\neq 0$
  3. $ax+cz+d=0, a\neq 0.c\neq 0$
  4. $by+cz+d=0, b\neq 0,c\neq 0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Generally a plane in 3-space has the equation

$ax + by + cz +d = 0,$

here, it is parallel to $x$ axis

hence the equation becomes,
$by + cz +d = 0,$

where at least one of the numbers a, b, c and d must be nonzero
finally the equation becomes,

$by +cz+d =0,b≠0, c≠0.$
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 plane through $(2, 1,4)$ and having $\mathrm{d}.\mathrm{c}$'s of its normal $\alpha,\ \beta,\ \gamma$ is

  1. $\alpha x+\beta y+\gamma z =2\alpha+\beta+4\gamma$
  2. <p class="MsoNormal">$\displaystyle \dfrac{x-2}{\alpha}+\dfrac{y-1}{\beta}+\dfrac{z-4}{\gamma}=0$</p>
  3. $\alpha x+\beta y+\gamma z =1$
  4. <p class="MsoNormal">$\displaystyle \dfrac{\alpha x}{2}+\dfrac{\beta y}{1}+\dfrac{\gamma z}{4}=0$</p>
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Since, the direction ratios of the normal are $ \alpha , \beta , \gamma $
The equation of the plane will be of the form, $ \alpha x + \beta y + \gamma z  = d$
And the plane passes through the point $(2,1,4)$.
Hence, equation of plane is $ \alpha x + \beta y + \gamma z $ = $ 2 \alpha + \beta  + 4 \gamma  $.

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

If the equation of the plane passing through the points $(1,2,3)$, $(-1,2,0)$ and perpendicular to the $zx$ - plane is $ax + by + cz + d$ $=$ $ 0$ $(a>0)$, then

  1. $a=0$ and $c=0$
  2. $a+d=0$
  3. $c+d-5=0$
  4. $a+c+d-4=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Since the plane is perpendicular to the zx-plane (y=0), its normal vector must be parallel to the y-axis (0, 1, 0). Thus, the equation is of the form by + d = 0. Given the points, we solve for the coefficients.

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

A plane $\Pi$ passes through the point $(1,1,1)$. If $b,c, a$ are the direction ratios of a normal to the plane, where $a, b, c (a<b<c)$ are the prime factors of $2001$, then the equation of the plane $\pi$ is

  1. $29x+31y+3z=63$
  2. $23x+29y-29z=23$
  3. $23x+29y+3z=55$
  4. $31x+27y+3z=71$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Sol. By verification $2001=23 \times29 \times3$
$\therefore 23x+29y+3z=55$
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 origin and containing the lines whose d.cs are proportional to $1,-2,2$ and $2,3,-1$ is:

  1. $x-2y+2z=0$
  2. $2x+3y-z=0$
  3. $x+5y-3z=0$
  4. $4x-5y-7z=0$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Equation of plane passing through origin is given by,
$ax+by+cz=0$
Also this line containing line whose d.cs are $(1,-2,2)$ and $(2,3,-1)$
$\Rightarrow a-2b+2c=0$ and $2a+3b-c=0$
Solving these, $ b= \dfrac{5c}{7}, a = -\dfrac{4c}{7}$
Hence, plane equation is
$4x-5y-7z=0$

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

The vector equation of the plane passing through the planes $r.(i+j+k)=6$ and $r.(2i+3j+4k)=-5$ and the point $(1,1,1)$ is

  1. $r.(20i+23j+26k) = 69$
  2. $r.(2i+23j+26k) = 69$
  3. $r.(2i+2j+3k) = 69$
  4. $r.(20i+3j+26k) = 69$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The vector equation of plane passing through the intersection of planes $\vec r.\vec {n _1}=d _1$ and $\vec r.\vec {n _2}$ and also through the point $x _1,y _1,z _1$ is 


$\vec r.(\vec {n _1}+\lambda \vec {n _2})=d _1+\lambda d _2$

According to question plane passes through 

$\vec r.(\hat i+\hat j+\hat k)=6$

comparing it with $\vec r.\vec {n _1}=d _1$

$\vec {n _1}=\hat i+\hat j+\hat k$
And 
$d _1=6$
Now, other plane by it also passes 

$\vec r.(2\hat i+3\hat j+4\hat k)=-5$

$=\vec r.(-2\hat i-3\hat j - 4\hat k)=5$
Comparing it with $\vec r.\vec {n _2}=d _2$

$\vec {n _2}=-2\hat i-3\hat j-4\hat k$
And 
$d _2=5$
Now, equation of the required plane 

$\vec r.[(\hat i+\hat j+\hat k)-\lambda (2\hat i+3\hat j+4\hat k)]=5\lambda +6$   ----- (i)
Now, 
Putting $\vec r=x\hat i+y\hat j+z\hat k$

$(x\hat i+y\hat j+z\hat k).[(\hat i+\hat j+\hat k)-\lambda (2\hat i+3\hat j+4\hat k)]=5\lambda +6$

$(1-2\lambda)x+(1-3\lambda)y+(1-4\lambda)z=5\lambda +6$   ----   (ii)

Since it passes through $(1,1,1)$ 
Therefore,

$(1-2\lambda)1+(1-3\lambda)1+(1-4\lambda)1=5\lambda +6$
Hence 
$\lambda =\dfrac{-3}{14}$

Put the value of $\lambda $ in (i)

$\vec r.[(\hat i+\hat j+\hat k)-(\dfrac{-3}{14}) (2\hat i+3\hat j+4\hat k)]=5(\dfrac{-3}{14}) +6$

$\vec r.[(1+\dfrac{6}{14})\hat i+(1+\dfrac{9}{14})\hat j+(1+\dfrac{12}{14}\hat k)]=\dfrac{69}{14}$

$\vec r.(20\hat i + 23\hat j + 26\hat k) = 69$

So, the required equation of plane is $\vec r.(20\hat i + 23\hat j + 26\hat k) = 69$

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 plane $\bar{r}.(2, -3, 4) = 5$ is _____

  1. $3y - 2x -4z + 5 =0$
  2. $2x - 3y + 4z =0$
  3. $2x - 3y + 4z +5 =0$
  4. $\displaystyle \frac{x - 1}{2} = \frac{y-1}{-3} = \frac{z-1}{4}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\vec{r}=x\hat{i}+y\hat{j}+z\hat{k}$

$\vec{r}.(2\hat{i}-3\hat{j}+4\hat{k})=5$
$\implies 2x-3y+4z=5$
$\implies 3y-2x-4z+5=0$

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(s) of the plane,  which is/are equally inclined to the lines $\dfrac {x-1}{2}=\dfrac {y}{-2}=\dfrac {z+2}{-1}$ and $\dfrac {x+3}{8}=\dfrac {y-4}{1}=\dfrac {z}{-4}$ and passing through the origin is/are

  1. $14x-5y-7z=0$
  2. $2x+7y-z=0$
  3. $3x-4y-z=0$
  4. $x+2y-5z=0$
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

The planes equally inclined to both the lines will have their normals along the angle bisectors of the two lines.  
The unit vector along the first line is $ \dfrac{2 \hat{i} -2 \hat{j} -\hat{k} } { 3 }$.
The unit vector along the second line is  $ \dfrac{ 8\hat{i}+ \hat{j} -4\hat{k} }{ 9} $.

The vectors along the angle bisectors can be written as : 
$ (\dfrac{2}{3} +\dfrac{8}{9}) \hat{i}  + ( \dfrac{-2}{3} + \dfrac{1}{9} )\hat{j} + (\dfrac{-1}{3} + \dfrac{-4}{9} )\hat{k}  = \dfrac{1}{9} (14\hat{i} -5\hat{j} -7\hat{k}) =0 $ 
and 
$ (\dfrac{2}{3} -\dfrac{8}{9}) \hat{i}  + ( \dfrac{-2}{3} - \dfrac{1}{9} )\hat{j} + (\dfrac{-1}{3} - \dfrac{-4}{9} )\hat{k} = \dfrac{1}{9} (-2\hat{i} -7\hat{j} +\hat{k} ) $.

Hence, options A and B represent equations of planes which have normals along the angle bisector and pass through the origin. 

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

A plane through the line $\displaystyle \frac{x - 1}{1} = \frac{y + 1}{-2} = \frac{z}{1}$ has the equation

  1. $\displaystyle x + y + z = 0$
  2. $\displaystyle 3x + 2y - z = 1$
  3. $\displaystyle 4x + y - 2z = 3$
  4. $\displaystyle 3x + 2y + z = 0$
Reveal answer Fill a bubble to check yourself
A,C Correct answer
Explanation
A plane eqn through the line $\dfrac{x-1}{1}=\dfrac{y+1}{-2}=\dfrac{z}{1}$      ...(i)
is given by $A(x-1) +B(y+1)+Cz=0$       ...(ii)
where $A,B,C$  are direction ratio which will perpence to line 
hence 
$A-2B+C=0$
so possible volves of $(A,B,C) $ are $(1,1,1)$ and $(4,1,-2)$
so eq of plane from (ii)
$\Rightarrow  1(x-1)+1(y+1)+1(z)=0$
$\Rightarrow x-1+y+1+z=0$
$\Rightarrow x+y+z=0$
          $ or $
$4(x-1)+(y+1)-2z=0$
$\Rightarrow 4x-4+y+1-2z=0$
$\Rightarrow 4x+y-2z-3=0$
$4x+y-2z=3$
So, here eq of plane is $x+y+z=0$ 
or $4x+y-2z=3$

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 a plane through the line $\displaystyle \frac{x\, -\, 1}{2}= \frac{y\, -\, 2}{3}= \frac{z\, -\, 3}{4}$ and parallel to a coordinate axis is

  1. $4y \:-\:3z\:+\:1 =\:0$
  2. $2x\:-\:z\:+\:1 =\:0$
  3. $3x\:-\:2y\:+\:1 =\:0$
  4. $2x\:+\:3y\:+\:1=\:0$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation
General equation of a plane is
$ax+by+cz+d=0$.........where a,b,c,d are constants taking any value

Let us consider the plane parallel to x-axis. 

Hence, the equation of the plane can be written as $ay+bz =1 $. 

The plane passes through $(1,2,3)$ and the normal to the plane is perpendicular to the vector along the line. 

Hence, 
$ 2a+3b =1$ and $ 3a+4b =0 $

On solving we get, $a= -4$ and $b=3$. 

Hence, $-4y+3z =1$ or $4y-3z +1 =0 $.

Let us consider the plane parallel to the y-axis.

Hence, the equation of the plane can be written as: 

$ cx+dz = 1 $

$c+3d =1 $ and $ 2c+4d =0 $. 

$\Rightarrow c = -2 $ and $d=1$ 

Hence, the equation of the plane is $ 2x-z+1 =0 $.

Similarly, we can find the equation for the plane parallel to the z-axis. The equation of the plane can be written as $ex+fy = 1 $.

After substituting the value of the point $(1,2,3)$ and using the information of the normal to the plane being perpendicular to the line, we get the equation of the plane as $ 3x-2y+1 =0$.

Hence, all three options are correct.