हिंदी

Determine the Value of λ for Which the Following Planes Are Perpendicular to Each Other. → R ⋅ ( ^ I + 2 ^ J + 3 ^ K ) = 7 and → R ⋅ ( λ ^ I + 2 ^ J − 7 ^ K ) = 26 - Mathematics

Advertisements
Advertisements

प्रश्न

Determine the value of λ for which the following planes are perpendicular to each other.

\[\vec{r} \cdot \left( \hat{i}  + 2 \hat{j} + 3 \hat{k} \right) = 7 \text{ and }  \vec{r} \cdot \left( \lambda \hat{i} + 2 \hat{j}  - 7 \hat{k}  \right) = 26\]

 

योग
Advertisements

उत्तर

` \text{ We know that the planes } \vec{r} . \vec{n_1} = d_1 , \vec{r} . \vec{n_2} = d_2 \text{ are perpendicular to each other only if } \vec{n_1} . \vec{n_2} =0.`

\[\text{ Here} , \vec{n_1} = \hat{i}| + 2 \hat{j} + 3 \hat{k} ; \vec{n_2} = \lambda \hat{i} + 2 \hat{j} - 7 \hat{k} \]

\[\text{ The given planes are perpendicular.} \]

\[ \Rightarrow \vec{n_1} . \vec{n_2} = 0\]

\[ \Rightarrow \left( \hat{i} + 2 \hat{j} + 3 \hat{k}  \right) . \left( \lambda \hat{i}  + 2 \hat{j} - 7 \hat{k}  \right) = 0\]

\[ \Rightarrow \lambda + 4 - 21 = 0\]

\[ \Rightarrow \lambda - 17 = 0\]

\[ \Rightarrow \lambda = 17\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: The Plane - Exercise 29.06 [पृष्ठ २९]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 29 The Plane
Exercise 29.06 | Q 4.1 | पृष्ठ २९

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Find the equations of the planes that passes through three points.

(1, 1, −1), (6, 4, −5), (−4, −2, 3)


Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes `vecr = (hati - hatj + 2hatk)  = 5`and `vecr.(3hati + hatj + hatk) = 6`


Find the vector equation of the line passing through the point (1, 2, − 4) and perpendicular to the two lines: 

`(x -8)/3 = (y+19)/(-16) = (z - 10)/7 and (x - 15)/3 = (y - 29)/8 = (z- 5)/(-5)`


Find the Cartesian form of the equation of a plane whose vector equation is 

  \[\vec{r} \cdot \left( - \hat{i} + \hat{j}  + 2 \hat{k} \right) = 9\]

 


Find the vector equations of the coordinate planes.

 

Show that the normals to the following pairs of planes are perpendicular to each other. 

x − y + z − 2 = 0 and 3x + 2y − z + 4 = 0 


Find the vector equation of a plane which is at a distance of 3 units from the origin and has \[\hat{k}\] as the unit vector normal to it.


find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane


Find the vector equation of the plane passing through the points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3).


Find the vector equation of the plane passing through points A (a, 0, 0), B (0, b, 0) and C(0, 0, c). Reduce it to normal form. If plane ABC is at a distance p from the origin, prove that \[\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} .\]

 


Find the vector equation of the plane passing through the points (1, 1, −1), (6, 4, −5) and (−4, −2, 3).


Determine the value of λ for which the following planes are perpendicular to each other. 

 3x − 6y − 2z = 7 and 2x + y − λz = 5

 

Obtain the equation of the plane passing through the point (1, −3, −2) and perpendicular to the planes x + 2y + 2z = 5 and 3x + 3y + 2z = 8.

 

Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y − z = 1 and 3x − 4y + z = 5.

 

Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.

 

Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis.

 

Find the equation of the plane passing through the intersection of the planes x − 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios proportional to 1, 2, 1. Also, find the perpendicular distance of (1, 1, 1) from this plane


Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \[\vec{r} \cdot \left( \hat{i}  + 2 \hat{j}  - 5 \hat{k}  \right) + 9 = 0 .\]

 

Find the image of the point with position vector \[3 \hat{i} + \hat{j}  + 2 \hat{k} \]  in the plane  \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j}  + \hat{k}  \right) = 4 .\]  Also, find the position vectors of the foot of the perpendicular and the equation of the perpendicular line through \[3 \hat{i}  + \hat{j}  + 2 \hat{k} .\]

 
 

Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x − 2y + 4z + 5 = 0. Also, find the length of the perpendicular.

 

Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane \[\vec{r} \cdot \left( \hat{i}  - 2 \hat{j}  + 4 \hat{k}  \right) + 5 = 0 .\]

 

Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0.


Find the length and the foot of perpendicular from the point \[\left( 1, \frac{3}{2}, 2 \right)\]  to the plane \[2x - 2y + 4z + 5 = 0\] .

 

Write the equation of the plane parallel to XOY- plane and passing through the point (2, −3, 5).

 

Write the equation of the plane parallel to the YOZ- plane and passing through (−4, 1, 0).

 

Write the intercepts made by the plane 2x − 3y + 4z = 12 on the coordinate axes.

 

Write the distance between the parallel planes 2x − y + 3z = 4 and 2x − y + 3z = 18.  


If the line drawn from (4, −1, 2) meets a plane at right angles at the point (−10, 5, 4), find the equation of the plane.


Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles.


Find the image of the point (1, 6, 3) in the line `x/1 = (y - 1)/2 = (z - 2)/3`.


The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by ______.


Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.


Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.


Show that the points `(hat"i" - hat"j" + 3hat"k")` and `3(hat"i" + hat"j" + hat"k")` are equidistant from the plane `vec"r" * (5hat"i" + 2hat"j" - 7hat"k") + 9` = 0 and lies on opposite side of it.


If the foot of perpendicular drawn from the origin to a plane is (5, – 3, – 2), then the equation of plane is `vec"r".(5hat"i" - 3hat"j" - 2hat"k")` = 38.


A unit vector perpendicular to the plane ABC, where A, B and C are respectively the points (3, –1, 2), (1, –1, –3) and (4, –3, 1), is


The coordinates of the foot of the perpendicular drawn from the point A(1, 0, 3) to the join of the points B(4, 7, 1) and C(3, 5, 3) are


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×