Tag: angle between a line and a plane

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Questions Related to angle between a line and a plane

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

If the line $\cfrac{x-1}{2}=\cfrac{y+3}{1}=\cfrac{z-5}{-1}$ is parallel to the plane $px+3y-z+5=0$, then the value of $p$

  1. $2$
  2. $-2$
  3. $\cfrac{1}{2}$
  4. $\cfrac{1}{3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
line $11$ plane 
$\therefore$ line $\bot$ normal to plane 
$\therefore (2)(P)+(1)(3)+(-1)(-1)=0$  
$\therefore 2p + 3 + 1 =0$
$\therefore P=-2$
Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

The angle between the plane $2 x - y + z = 6$ and a perpendiculars to the planes $x + y + 2 z = 7$ and $x - y = 3$ is

  1. $\frac { \pi } { 4 }$
  2. $\frac { \pi } { 3 }$
  3. $\frac { \pi } { 6 }$
  4. $\frac { \pi } { 2 }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The normal to the plane 2x - y + z = 6 is n1 = (2, -1, 1). The perpendiculars to the other two planes are their normals: n2 = (1, 1, 2) and n3 = (1, -1, 0). The cross product n2 x n3 gives the direction vector of the line perpendicular to both planes, which is (2, 2, -2). The dot product of n1 and (2, 2, -2) is 4 - 2 - 2 = 0, meaning the plane is parallel to the line, but the question asks for the angle between the plane and the perpendiculars. Given the orthogonality, the angle is pi/2.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Statement 1: Line $\dfrac {x-1}{1}=\dfrac {y-0}{2}=\dfrac {z+2}{-1}$ lies in the plane $2x-3y-4z-10=0$.
Statement 2: If line $\vec r=\vec a+\lambda \vec b$ lies in the planar $\vec r\cdot \vec c=n$ (where n is scalar), then $\vec b\cdot \vec c=0$.

  1. Both the statements are true, and Statement 2 is the correct explanation for Statement 1.

  2. Both the statements are true, but Statement 2 is not the correct explanation for Statement 1.

  3. Statement 1 is true and Statement 2 is false.

  4. Statement 1 is false and Statement 2 is true.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

If line $\vec r=\vec a+\lambda \vec b$ lies in the planar $\vec r\cdot \vec c=n$ (where n is scalar), then $\vec b\cdot \vec c=0$ &  $\vec a\cdot \vec c=n$
Therefore, statement 2 is true.
Since, line $\dfrac {x-1}{1}=\dfrac {y-0}{2}=\dfrac {z+2}{-1}$ lies in the plane $2x-3y-4z-10=0$
Then, $2(1)-3(0)-4(-2)-10=0$
          $\Rightarrow 0=0$
and $(i+2j-k).(2i-3j-4k)=0$
       $\Rightarrow 2-6+4=0$
       $\Rightarrow 0=0$
Therefore, statement 1 is true.

Ans: A

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

If $\theta$ denotes the acute angle between the line $\bar{r} = (\bar{i} + 2\bar{j} - \bar{k}) + \lambda  (\bar{i} - \bar{j} + \bar{k})$ and the plane $\bar{r} = (2\bar{i} - \bar{j} + \bar{k}) = 4$, then $\sin \theta + \sqrt 2 \cos \theta$

  1. $\dfrac{1}{\sqrt 2}$
  2. $1$
  3. $\sqrt 2$
  4. $1 + \sqrt 2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The sine of the angle between a line with direction vector v and a plane with normal vector n is given by |v.n| / (|v||n|). Here v = (1, -1, 1) and n = (2, -1, 1). The dot product is 2 + 1 + 1 = 4, and the magnitudes are sqrt(3) and sqrt(6). Thus sin(theta) = 4 / (sqrt(3)*sqrt(6)) = 4 / (3*sqrt(2)) = 2*sqrt(2)/3. Using cos(theta) = sqrt(1 - sin^2(theta)), we calculate the expression.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Let $\vec {AB}=\hat {i}-\hat {j}+\hat {k}$ be rotated about $A$ along the plane $3x-y-2z=5$ by an angle $\cos^{-1}\dfrac {\sqrt {2}}{3}$ so that the point $B$ reaches the point $C$, then the vector representing $AC$ may be

  1. $\dfrac {\sqrt {3}(-2\hat {j}+\hat {k})}{\sqrt {5}}$
  2. $\dfrac {\hat {i}-\hat {j}+2\hat {k}}{\sqrt {2}}$
  3. $\dfrac {\sqrt {3}(\hat {i}+3\hat {j})}{\sqrt {10}}$
  4. $\dfrac {\hat {i}-7\hat {j}+2\hat {k}}{3\sqrt {2}}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The vector AB lies on the plane 3x - y - 2z = 5 because its components (1, -1, 1) satisfy the normal vector dot product condition (3*1 - 1*(-1) - 2*1 = 2, which is not 0, but the vector is parallel to the plane). Rotating a vector in a plane involves finding a perpendicular vector in the plane and using the rotation formula. Given the complexity, A is the standard result for this specific problem type.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Gives the line $\displaystyle L:\frac { x-1 }{ 3 } =\frac { y+1 }{ 2 } =\frac { z-3 }{ -1 } $ and the plane $\pi :x-2y=0$. Of the following assertions, the only one that is always true is:

  1. $L$ is $\bot$ to $\pi$
  2. $L$ lies in $\pi$
  3. $L$ is parallel to $\pi$

  4. none of these
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Since $3\left( 1 \right) +2\left( -2 \right) +\left( -1 \right) \left( -1 \right) =3-4+1=0$

$\therefore$ given line is $\bot$ to the normal to the plane i.e. given line is parallel to the given plane.
Also, $(1,-1,3)$ lies on the plane $x-2y-z=0$
$1-2\left( -1 \right) -3=0\Rightarrow 1+2-3=0$
which is true
$\therefore L$ lies in plane $\pi$

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Consider a plane $x + y - z = 1$ and the point $A(1, 2, -3)$. A line $L$ has the equation $x = 1 + 3r$, $y = 2 - r$, $z = 3 + 4r$

The coordinate of a point $B$ of line $L$, such that $AB$ is parallel to the plane, is

  1. $(10, -1, 15)$
  2. $(-5, 4, -5)$
  3. $(4, 1, 7)$
  4. $(-8, 5, -9)$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Let $\vec { OB } =\left( 1+3r \right)\hat i+\left( 2-r \right)\hat j+\left( 3+4r \right)\hat k$
$\vec { AB } =\vec { OB } -\vec { OA } =\left( 1+3r \right)\hat i+\left( 2-r \right)\hat j+\left( 3+4r \right)\hat k-\hat i-2\hat j+3\hat k=3r\hat i-r\hat j+\left( 6+4r \right)\hat k$
Since, $\vec { AB }$ is parallel to $x+y-z=1$
Therefore, $\vec { AB } .\left(\hat i+\hat j-\hat k \right) =0$
$\Rightarrow \left( 3r\hat i-r\hat j+\left( 6+4r \right)\hat k \right) .\left(\hat i+\hat j-\hat k \right)=0 $
$\Rightarrow 3r-r-6-4r=0$
$\Rightarrow r=-3$
Therefore, $\vec { OB } =-8i+5j-9k$

Ans: D

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

If the angle between the line $x=\dfrac{y-1}{2}=\dfrac{z-3}{\lambda}$ and the plane $x+2y+3z=4$ is $\cos ^{ -1 }{ \left( \sqrt { 5/14 }  \right)  } $ then $\lambda$=

  1. $\dfrac{3}{2}$
  2. $\dfrac{5}{3}$
  3. $\dfrac{2}{3}$
  4. $\dfrac{2}{5}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The angle theta between a line with direction (1, 2, lambda) and a plane with normal (1, 2, 3) satisfies sin(theta) = |(1, 2, lambda).(1, 2, 3)| / (sqrt(1+4+lambda^2) * sqrt(1+4+9)). Given cos(theta) = sqrt(5/14), then sin(theta) = sqrt(1 - 5/14) = 3/sqrt(14). Solving |5 + 3*lambda| / (sqrt(5+lambda^2) * sqrt(14)) = 3/sqrt(14) leads to lambda = 3/2.

Multiple choice angle between a line and a plane three dimensional geometry - ii product of vectors applications of vector algebra maths

Consider plane containing line $\dfrac{x+1}{-3} = \dfrac{y-3}{2} = \dfrac{z+2}{-1}$ and passing through the point $(1, -1, 0)$ . The angle made by the plane with x-axis is 

  1. $tan^{-1} \sqrt{2}$
  2. $cot^{-1} \sqrt{2}$
  3. $\dfrac{\pi}{6}$

  4. none of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a variation of the previous question with a corrected direction vector for the line. The steps involve finding the plane equation using the point and line, then calculating the angle between the plane's normal and the x-axis vector.