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Angle between Line and Plane

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An angle is formed when two lines intersect each other at a point. The very famous tool known as a protractor is used to measure the angle between two lines in degrees.

Angles not only found in geometry problems also we are influenced in our day to day life. For example, if you say Sen has been on 360 degrees of the city, you would be saying Sen has been all over the city. Also We cannot be created our rectangular class model without using angles and its measures. In this section we will get detailed step-by-step solution to problems based on this topic, which will definitely help you to find angle between line and plane by your own pace.

How to find Angle between Line & Plane

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If we are asked to  find the angle between plane, $5x - 2y + z + 1$ = $0$ and line $-2i - j + 3k$, follow below steps to get angle between a line and a plane.
Step 1: Write the normal vector of a plane. 
            Normal vector of a plane = $<M, N, Q>$ = $(5, -2, 1)$

Step 2: Write direction vector of given line.
            Direction vector of line = $<m, n, q>$ = $(-2, -1,  3)$

Step 3: Substitute values among the formula.

            $Sin\ (\phi)$ = $(\frac{|Mm + Nn + Qq|)}{(\sqrt{M^{2} + N^{2} + Q^{2}}\ \sqrt{m^{2} + n^{2} + q^{2}})})$

Step 4: Calculate and get your desired result.

             $Sin(\phi)$

        = $\frac{|5 . (-2) + (-2) . (-1) + 1 . 3}{\sqrt{5^{2} + (-2)^{2} + (1)^{2}} . \sqrt{(-2)^{2} + (-1)^{2} + (3)^{2}}}$ 

        = $\frac{|-10 + 2 + 3 |}{(\sqrt{(25 + 4 + 1)}\ •\ \sqrt{(4 + 1 + 9)})}$

        = $\frac{5}{(\sqrt{30}\ •\ \sqrt{14})}$

        = $\frac{5}{\sqrt{(420)}}$

        = $0.24398$

     $\phi$ = $14.1215$

Formula

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The angle between line $l$ and a plane, say $P$ is the angle between $l$ and a line passing through $P$.

Let $\phi$ be the angle between line $l$ and plane $P$, then it is defined by the below equation.

$Sin\ (\phi)$ = $\frac{(|Mm + Nn + Qq|)}{(\sqrt{M^{2} + N^{2} + Q^{2}}\ \sqrt{m^{2} + n^{2} + q^{2}})}$

Where $M, N, Q$ = components of normal vector of plane equation, $Mx + Ny + Qz + S$ = $0$

$a, b, c$ = components of the direction vector of line.

Proof

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Suppose equation of line $s$ = $mx + ny + qz$ = $0$ and equation of plane, $u$ = $Mx + Ny + Qz + Z$ = $0$

From the equation of plane Normal vector, $u$ = $\{ M, N, Q \}$

From the equation of line directional vector, $s$ = $\{ m, n, q \}$

A very famous relationship between normal and directional vectors is defined by scalar  product. 

If we have two vectors u and s, Scalar product of vectors defines as

$vec \{u \}\ .\ vec \{s \}$ = $|vec \{u \}|\ .\ |vec\{s \}|\ Cos\ (\theta)$

or $Cos\ (\theta)$ = $\frac{(vec \{u \}\ .\ vec \{s \})}{(|vec \{u \}|\ .\ |vec \{s \}|)}$

use module of vectors in coordinate form, $(\phi$ = $90^{\circ}\ -\ \theta)$ property and above formula to get the angle between plane and a line.

$Sin\ (\phi)$ = $\frac{(|Mm + Nn + Qq|)}{(\sqrt{M^{2} + N^{2} + Q^{2}} \sqrt{m^{2} + n^{2} + q^{2}})}$

Examples

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Example 1: 

Find the angle between $x - y + 3z + 12$ = $0$ and $\frac{(x - 2)}{3}$ = $\frac{(y - 2)}{5}$ = $\frac{-(2z - 14)}{8}$.

Solution: 

We are given with equation of plane and line  $x - y + 3z + 12$ = $0$ and

$\frac{(x - 2)}{3}$
= $\frac{(y - 2)}{5}$ = $\frac{-(2z - 14)}{8}$ or $\frac{(x - 2)}{3}$ = $\frac{(y - 2)}{5}$ = $\frac{-(z - 7)}{4}$.

First step is to find the direction vectors of both the equations:

Normal Vector of Plane: $<1, -1, 3>$

Direction Vector of Line : $<3, 5, -4>$

using above formula, we get

Let $\theta$ be the angle between line and plane projection

$sin\ (\theta)$ =  $\frac{| 1\ •\ 3 + (-1)\ •\ 5 + (3)\ •\ (-4) |}{\sqrt{(1^{2} + (-1)^{2} + (3)^{2})}\ •\ \sqrt{(3^{2} + 5^{2} + (-4)^{2}})}$

          = $\frac{| 3 - 5 - 12 |}{(\sqrt{(1 + 1 + 9)}\ •\ \sqrt{(9 + 25 + 16)})}$

          = $\frac{14}{(\sqrt{11}\ •\ \sqrt{50})}$

          = $\frac{14}{\sqrt{(550)}}$

          = $0.5969$ (approx)

       $\theta$ = $36.6481$
Example 2: 

Find angle between line $\frac{(x + 3)}{7}$ = $\frac{(y - 5)}{5}$ = $\frac{(z - 1)}{3}$ and plane $2x + y - 3z + 5$ = $0$

Direction vector of the lines = $(7, 5, 3)$

Normal vector of the plane = $(2, 1, -3)$

Solution:

Let $\phi$ be the Angle between line and plane:

$sin\ \phi$ =  $\frac{| 2\ •\ 7 + 1\ •\ 5 + (-3)\ •\ 3 |}{\sqrt{(2^{2} + 1^{2} + (-3)^{2})}\ •\ \sqrt{(7^{2} + 5^{2} + 3^{2})}}$

         = $\frac{| 14 + 5 - 9 |}{(\sqrt{(4 + 1 + 9)}\ •\ \sqrt{(49 + 25 + 9)}}$

         = $\frac{10}{(\sqrt{14}\ •\ \sqrt{83})}$

         = $\frac{10}{\sqrt{(1162)}}$

         = $0.293357$ (approx)

        This implies $\phi$ = $17.059$
Example 3:

Determine the angle between the lines: $3x + 5y + z + 10$ = $0, 2x - 4y + 5z + 13$ = $0$ and the plane $5x + y - 2z$ = $0$

Solution: 

Here we are given with two lines.

First calculate the perpendicular line passing through given lines using cross product

Let $A$ = $3x + 5y + z + 10$ = $0$, and $B$ = $2x - 4y + 5z + 13$ = $0$

=> $A \times B$ = $29i - 13j - 22k$ new equation of line.

Direction vector of line: $(29, -13, -22)$

Normal vector of plane: $(5, 1, -2)$

$sin\ \phi$ = $\frac{| 5\ •\ 29 + 1\ •\ (-13) + (-2)\ .\ (-22)|}{\sqrt{(5^{2} + (1)^{2} + (-2)^{2})}\ •\ (\sqrt{(29)^{2} + (-13)^{2} + (-22)^{2})}}$

         = $\frac{| 145 - 13 + 44 |}{(\sqrt{(25 + 1 + 4)}\ •\ \sqrt{(841 + 169 + 484)})}$

         = $\frac{176}{(\sqrt{30}\ •\ \sqrt{1494})}$

         = $\frac{176}{\sqrt{(44820)}}$

         = $0.83133$

      $\phi$ = $56.235$, which is angle between line and given plane.