Tag: lines in space

Questions Related to lines in space

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

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

  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. $by+cz+d=0, b\neq 0,c\neq 0$

Show answer
Correct Option: D
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,

$by +cz+d =0,b≠0, c≠0.$

Equation of plane through $(2, 1,4)$ and having $\mathrm{d}.\mathrm{c}$'s of its normal $\alpha,\ \beta,\ \gamma$ is

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

  1. $\alpha x+\beta y+\gamma z =2\alpha+\beta+4\gamma$

  2. $\displaystyle \dfrac{x-2}{\alpha}+\dfrac{y-1}{\beta}+\dfrac{z-4}{\gamma}=0$

  3. $\alpha x+\beta y+\gamma z =1$

  4. $\displaystyle \dfrac{\alpha x}{2}+\dfrac{\beta y}{1}+\dfrac{\gamma z}{4}=0$

Show answer
Correct Option: A
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  $.

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

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

  1. $a=0$ and $c=0$

  2. $a+d=0$

  3. $c+d-5=0$

  4. $a+c+d-4=0$

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Correct Option: D

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

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

  1. $29x+31y+3z=63$

  2. $23x+29y-29z=23$

  3. $23x+29y+3z=55$

  4. $31x+27y+3z=71$

Show answer
Correct Option: C
Explanation:
Sol. By verification $2001=23 \times29 \times3$
$\therefore 23x+29y+3z=55$

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:

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

  3. $x+5y-3z=0$

  4. $4x-5y-7z=0$

Show answer
Correct Option: D
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
$4x-5y-7z=0$

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

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

  1. $r.(20i+23j+26k) = 69$

  2. $r.(2i+23j+26k) = 69$

  3. $r.(2i+2j+3k) = 69$

  4. $r.(20i+3j+26k) = 69$

Show answer
Correct Option: A
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 $\vec r.(20\hat i + 23\hat j + 26\hat k) = 69$

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

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

  1. $3y - 2x -4z + 5 =0$

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

  3. $2x - 3y + 4z +5 =0$

  4. $\displaystyle \frac{x - 1}{2} = \frac{y-1}{-3} = \frac{z-1}{4}$

Show answer
Correct Option: A
Explanation:

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

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

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

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

  1. $14x-5y-7z=0$

  2. $2x+7y-z=0$

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

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

Show answer
Correct Option: A,B
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. 

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

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

  1. $\displaystyle x + y + z = 0$

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

  3. $\displaystyle 4x + y - 2z = 3$

  4. $\displaystyle 3x + 2y + z = 0$

Show answer
Correct Option: A,C
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 $4x+y-2z=3$

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

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

  1. $4y :-:3z:+:1 =:0$

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

  3. $3x:-:2y:+:1 =:0$

  4. $2x:+:3y:+:1=:0$

Show answer
Correct Option: A,B,C
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.