Tag: lines in space

Quizzes (0) Questions (31)

Questions Related to lines in space

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.

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 passing through the points $(2, 2, 1)$ and $(9, 3, 6)$, and perpendicular to the plane $2x+6y+6z-1=0$ is-

  1. $3x+4y+5z=9$
  2. $3x+4y-5z=9$
  3. $3x-4y+5z=9$

  4. None of the above.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Equation of plane passes through $(2,2,1)$ is given by,
$a(x-2)+b(y-2)+c(z-1) = 0.......(A)$
Given it also passes through $(9,3,6)$
$\Rightarrow 7a+b+5c=0 ......(1)$
and this plane is perpendicular to plane $2x+6y+6z-1=0$
$\Rightarrow 2a+6a+6c=0 .....(2)$
Solving (1) and (2), we get $ a= \dfrac{-3a}{5}, b = \dfrac{-4a}{5}$
Putting these values in (A) our required plane is,
$3x+4y-5z=9$

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 $\overrightarrow { r } =\left( 1+\lambda -\mu  \right) i+\left( 2-\lambda  \right) j+\left( 3-2\lambda +2\mu  \right) k$ 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

We have $\overrightarrow { r } =\left( 1+\lambda -\mu  \right) i+\left( 2-\lambda  \right) j+\left( 3-2\lambda +2\mu  \right) k$

$\Rightarrow \overrightarrow { r } =\left( i+2j+k \right) +\lambda \left( i=j-2k \right) +\mu \left( -i+2k \right) $
which is a plane through $\overrightarrow { a } =i+2j+3k$ and parallel to the vectors
$\overrightarrow { b } =i-j-2k$ and $\overrightarrow { c } =-i+2k$
Therefore, it is perpendicular to the vector
$\overrightarrow { n } =\overrightarrow { b } \times \overrightarrow { c } =-2i-k$
Hence, its vector equation is 
$\left( \overrightarrow { r } -\overrightarrow { a }  \right) .\overrightarrow { n } =0\Rightarrow \overrightarrow { r } .\overrightarrow { n } =\overrightarrow { a } .\overrightarrow { n } \ \Rightarrow \overrightarrow { r } .\left( -2i-k \right) =-2-3\Rightarrow \overrightarrow { r } \left( 2i+k \right) =5$
So, the cartesian equation is
$\left( xi+yj+zk \right) .\left( 2i+k \right) =5\Rightarrow 2x+z=5$

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

If $lx+my+nz=p$ is equation of plane in normal form, then :

  1. $l^2+m^2+n^2=1$

  2. l, m , n are d.c's of a normal to the plane

  3. p > 0

  4. All of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In the normal form lx + my + nz = p, l, m, and n are the direction cosines of the normal to the plane, which satisfy the condition l^2 + m^2 + n^2 = 1.

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 through the points $(2,3,1)$ and $(4,-5,3)$ and parallel to $x$-axis is

  1. $x-z-1=0$
  2. $4x+y-11=0$
  3. $y+4z-7=0$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
The line segment passing through $(2,3,1)$ and $(4,-5,3)$ is given by $2i-8j+2k$. 
Hence, the normal to the plane is given by $(2i-8j+2k) \times i=2j+8k$. 
Hence, the equation of plane is given by $2y+8z+d=0$. 
Since it passes through $(2,3,1)$ we get $6+8+d=0 \Rightarrow d=-14$. 
Thus, the equation of plane is given by $y+4z-7=0$.  
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 passing through the point $(1, 1, 1)$ and perpendicular to each of the planes $x+ 2y+ 3z= 7$ and $2x- 3y +4z= 0$, is

  1. $17x- 2y +7z= 12$
  2. $17x+ 2y -7z= 12$
  3. $17x+ 2y +7z= 12$
  4. $17x- 2y -7z= 12$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let $ax+by+cz=1$ be the desired plane.
Since, it is perpendicular to $x+2y+3z=7$ & $2x-3y+4z=0$
Therefore, $a+2b+3c=0$        .... (1)
and $2a-3b+4c=0$       ...(2)
and it passes through $(1,1,1)$
Therefore, $a+b+c=1$      ...(3)
Solving $(1),(2)$ and $(3)$ simultaneously, we get

$a=\dfrac {17}{12}$, $b=\dfrac {1}{6}$, $c=-\dfrac {7}{12}$
Therefore, desired plane is $17x+2y-7z=12$

Ans: B

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 form of the plane 
$ { r } =(s-2t)\hat { i+(3-t)\hat { j+(2s+t)\hat { k }  }  } $ is 

  1. $ 2 x-5 y-z-15=0$
  2. $2 x-5 y+z-15=0$
  3. $2 x-5 y-z+15+0$
  4. $2 x+5 y-z+15=0$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since $\overrightarrow{r}=x\hat{i}+y\hat{j}+z\hat{k}$ is the proof any vector $(x,y,z)$on the plane. The given equation can be written as.

$x\hat{i}+y\hat{j}++z\hat{k}=(s—2t)\hat{i}+(3-t)\hat{j}+(2s+t)\hat{k}\\x=(s-2t)\quad y=(3-t)\quad z=2s+t$

Similarly We get $x-2y=s-6$ and $y+z=3+2s$

Now eliminating $s$ we get $2(x-2y)-(y+z)=-15$

$2x-5y-z+15=0$ is the required form of the equation.