English

Find the Vector Equation of the Plane Passing Through Three Points with Position Vectors ^ I + ^ J − 2 ^ K , 2 ^ I − ^ J + ^ K and ^ I + 2 ^ J + ^ K . - Mathematics

Advertisements
Advertisements

Question

Find the vector equation of the plane passing through three points with position vectors  \[\hat{i}  + \hat{j}  - 2 \hat{k}  , 2 \hat{i}  - \hat{j}  + \hat{k}  \text{ and }  \hat{i}  + 2 \hat{j}  + \hat{k}  .\]  Also, find the coordinates of the point of intersection of this plane and the line  \[\vec{r} = 3 \hat{i}  - \hat{j}  - \hat{k}  + \lambda\left( 2 \hat{i}  - 2 \hat{j} + \hat{k} \right) .\]

 
Sum

Solution

\[\text{ Let A(1, 1 , -2),B(2, -1, 1) and C(1, 2, 1) be the points represented by the given position vectors } .\]
\[\text{ The required plane passes through the point A (1, 1, -1) whose position vector is }  \vec{a} = \hat{i} + \hat{j}  - 2 \hat{k}  \text{ and is normal to the vector } \vec{n} \text{ given by } \]
\[ \vec{n} = \vec{AB} \times \vec{AC .} \]
\[\text{ Clearly } , \vec{AB} = \vec{OB} - \vec{OA} = \left( 2 \hat{i}  - \hat{j}  + \hat{k}  \right) - \left( \hat{i}  + \hat{j}  - 2 \hat{k}  \right) = \hat{i}  - 2 \hat{j}  + 3 \hat{k}  \]
\[ \vec{AC} = \vec{OC} - \vec{OA} = \left( \hat{i}  + 2 \hat{j} + \hat{k}  \right) - \left( \hat{i}  + \hat{j}  -2 \hat{k}  \right) = 0 \hat{i}  + \hat{j}  + 3 \hat{k}  \]
\[ \vec{n} = \vec{AB} \times \vec{AC} = \begin{vmatrix}\hat{i}  & \hat{j}  & \hat{k}  \\ 1 & - 2 & 3 \\ 0 & 1 & 3\end{vmatrix} = - 9 \hat{i}  - 3 \hat{j} + \hat{k}  \]
\[\text{ The vector equation of the required plane is } \]
\[ \vec{r} . \vec{n} = \vec{a} . \vec{n} \]
\[ \Rightarrow \vec{r} . \left( - 9 \hat{i}  - 3 \hat{j}  + \hat{k}  \right) = \left( \hat{i}  + \hat{j} -2 \hat{k}  \right) . \left( - 9 \hat{i} - 3 \hat{j}  + \hat{k}  \right)\]
\[ \Rightarrow \vec{r} . \left( - 9 \hat{i}  - 3 \hat{j} + \hat{k}  \right) = - 9 - 3 - 2\]
\[ \Rightarrow \vec{r} . \left[ - \left( 9 \hat{i} + 3 \hat{j}  - \hat{k} \right) \right] = - 14\]
\[ \Rightarrow \vec{r} . \left( 9 \hat{i}  + 3 \hat{j}  - \hat{k}  \right) = 14\]
\[\text{ To find the point of intersection of this plane } \]
\[\text{ The given equation of the line is } \]
\[ \vec{r} = \left( 3 \hat{i}  - \hat{j} - \hat{k}  \right) + \lambda \left( 2 \hat{i}  - 2 \hat{j}  + \hat{k}  \right)\]
\[ \Rightarrow \vec{r} = \left( 3 + 2\lambda \right) \hat{i}  + \left( - 1 - 2\lambda \right) \hat{j}  + \left( - 1 + \lambda \right) \hat{k}  \]
\[\text{ The coordinates of any point on this line are in the form of }  \left( 3 + 2\lambda \right) \hat{i}  + \left( - 1 - 2\lambda \right) \hat{j}  + \left( - 1 + \lambda \right) \hat{k}  \text{ or } \left( 3 + 2\lambda, - 1 - 2\lambda, - 1 + \lambda \right)\]
\[\text{ Since this point lies on the plane }  \vec{r} .\left( 9 \hat{i}  + 3 \hat{j}  - \hat{k}  \right) = 14, \]
\[\left[ \left( 3 + 2\lambda \right) \hat{i}  + \left( - 1 - 2\lambda \right) \hat{j}  + \left( - 1 + \lambda \right) \hat{k}  \right] . \left( 9 \hat{i}  + 3 \hat{j}  - \hat{k}  \right) = 14\]
\[ \Rightarrow 27 + 18\lambda - 3 - 6\lambda + 1 - \lambda = 14\]
\[ \Rightarrow 11\lambda = - 11\]
\[ \Rightarrow \lambda = - 1\]
\[\text{ So, the coordinates of the point are } \]
\[\left( 3 + 2\lambda, - 1 - 2\lambda, - 1 + \lambda \right)\]
\[ = \left( 3 - 2, - 1 + 2, - 1 - 1 \right)\]
\[ = \left( 1, 1, - 2 \right)\]

shaalaa.com
  Is there an error in this question or solution?
Chapter 29: The Plane - Exercise 29.13 [Page 74]

APPEARS IN

RD Sharma Mathematics [English] Class 12
Chapter 29 The Plane
Exercise 29.13 | Q 11 | Page 74

RELATED QUESTIONS

Find the equation of the plane through the intersection of the planes 3x – y + 2z – 4 = 0 and x + y + z – 2 = 0 and the point (2, 2, 1).


Find the vector equation of the plane passing through the intersection of the planes `vecr.(2hati + 2hatj - 3hatk) = 7, vecr.(2hati + 5hatj + 3hatk) = 9` and through the point (2, 1, 3)


Find the equation of the plane passing through the line of intersection of the planes `vecr.(hati + hatj + hatk) = 1` and `vecr.(2hati + 3hatj -hatk) + 4 = 0` and parallel to x-axis.


Find the equation of the plane which contains the line of intersection of the planes `vecrr.(hati + 2hatj + 3hatk) - 4 = 0, vecr.(2hati + htj - hatk) + 5 = 0`,  and which is perpendicular to the plane `vecr.(5hati + 3hatj - 6hatk) + 8 = 0`.


Find the equation of the plane through the intersection of the planes 3x − 4y + 5z = 10 and 2x + 2y − 3z = 4 and parallel to the line x = 2y = 3z.

 

Find the distance of the point (−1, −5, −10) from the point of intersection of the line \[\vec{r} = \left( 2 \hat{i}  - \hat{j} + 2 \hat{k}  \right) + \lambda\left( 3 \hat{i}+ 4 \hat{j} + 2 \hat{k}  \right)\] and the plane  \[\vec{r} . \left( \hat{i}  - \hat{j}  + \hat{k} \right) = 5 .\]

 

Find the distance of the point (2, 12, 5) from the point of intersection of the line \[\vec{r} = 2 \hat{i}  - 4 \hat{j}+ 2 \hat{k}  + \lambda\left( 3 \hat{i}  + 4 \hat{j}  + 2 \hat{k} \right)\] and \[\vec{r} . \left( \hat{i}  - 2 \hat{j}  + \hat{k}  \right) = 0\]

  

Find the distance of the point P(−1, −5, −10) from the point of intersection of the line joining the points A(2, −1, 2) and B(5, 3, 4) with the plane  \[x - y + z = 5\] . 

 


Find the distance of the point (1, -5, 9) from the plane

\[x - y + z =\] 5  measured along the line \[x = y = z\]  . 
 

Show that the lines  \[\frac{x + 4}{3} = \frac{y + 6}{5} = \frac{z - 1}{- 2}\] and 3x − 2y + z + 5 = 0 = 2x + 3y + 4z − 4 intersect. Find the equation of the plane in which they lie and also their point of intersection.

  

Show that the plane whose vector equation is \[\vec{r} \cdot \left( \hat{i}  + 2 \hat{j} - \hat{k}  \right) = 3\] contains the line whose vector equation is \[\vec{r} = \hat{i} + \hat{j}  + \lambda\left( 2 \hat{i}  + \hat{j} + 4 \hat{k}  \right) .\]

 

Find the equation of the plane determined by the intersection of the lines \[\frac{x + 3}{3} = \frac{y}{- 2} = \frac{z - 7}{6} \text{ and  }\frac{x + 6}{1} = \frac{y + 5}{- 3} = \frac{z - 1}{2}\]

 

Find the coordinates of the point where the line \[\frac{x - 2}{3} = \frac{y + 1}{4} = \frac{z - 2}{2}\] intersect the plane x − y + z − 5 = 0. Also, find the angle between the line and the plane.

  

Find the distance of the point with position vector

\[- \hat{i}  - 5 \hat{j}  - 10 \hat{k} \]  from the point of intersection of the line \[\vec{r} = \left( 2 \hat{i}  - \hat{j}  + 2 \hat{k}  \right) + \lambda\left( 3 \hat{i}  + 4 \hat{j}  + 12 \hat{k}  \right)\]  with the plane \[\vec{r} \cdot \left( \hat{i} - \hat{j}+ \hat{k}  \right) = 5 .\]
 

The plane 2x − (1 + λ) y + 3λz = 0 passes through the intersection of the planes


The equation of the plane through the line x + y + z + 3 = 0 = 2x − y + 3z + 1 and parallel to the line \[\frac{x}{1} = \frac{y}{2} = \frac{z}{3}\] is 

 

 


The distance of the point (−1, −5, −10) from the point of intersection of the line \[\vec{r} = 2 \hat{i}- \hat{j} + 2 \hat{k}  + \lambda\left( 3 \hat{i}  + 4 \hat{j}+ 12 \hat{k}  \right)\]   and the plane \[\vec{r} \cdot \left( \hat{i} - \hat{j} + \hat{k}  \right) = 5\] is 

 
 

The equation of the plane which cuts equal intercepts of unit length on the coordinate axes is


Show that the lines `("x"-1)/(3) = ("y"-1)/(-1) = ("z"+1)/(0) = λ and  ("x"-4)/(2) = ("y")/(0) = ("z"+1)/(3)` intersect. Find their point of intersection. 


Find the distance of the point (–1, –5, – 10) from the point of intersection of the line `vec"r" = 2hat"i" - hat"j" + 2hat"k" + lambda(3hat"i" + 4hat"j" + 2hat"k")` and the plane `vec"r" * (hat"i" - hat"j" + hat"k")` = 5


Show that the lines `(x - 1)/2 = (y - 2)/3 = (z - 3)/4` and `(x - 4)/5 = (y - 1)/2` = z intersect. Also, find their point of intersection.


Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.


The plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is `"a"x + "b"y +- (sqrt("a"^2 + "b"^2) tan alpha)`z = 0.


Find the equation of the plane through the intersection of the planes `vec"r" * (hat"i" + 3hat"j") - 6` = 0 and `vec"r" * (3hat"i" - hat"j" - 4hat"k")` = 0, whose perpendicular distance from origin is unity.


Find the equation of line parallel to the y-axis and drawn through the point of intersection of x – 4y + 1 = 0 and 2x + y – 7 = 0.


ABCD be a parallelogram and M be the point of intersection of the diagonals, if O is any point, then OA + OB + OC + OD is equal to


The equation of the curve passing through the point `(0, pi/4)` whose differential equation is sin x cos y dx + cos x sin y dy = 0, is


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×