मराठी
CUET (UG)
MCQ ऑनलाइन सराव परीक्षा१४८
संकल्पना स्पष्टीकरण६२०
अभ्यासक्रम

Equation of a Plane - Plane Passing Through the Intersection of Two Given Planes

Advertisements

Topics

Notes

Let `π_1` and `π_2` be two planes with equations `vec r  . hat n _1 = d_1` and `vec r . hat n _2 = d_2` respectively.  The position vector of any point on the line of intersection must satisfy both the equations fig.

If `vec t ` is the position vector of a point on the line , then
`vec t . hat n_1 = d_1` and `vec t . hat n _2 = d_2` 
Therefore , for all real values of  λ, we have
`vec t . (hat n _1 + lambda hat n_2) = d_1 + lambda d_2`
Since `vec t` is arbitrary, it satisfies for any point on the line.
Hence , the equation `vec r . (vec n_1 + lambda vec n_2) = d_1 + lambda d_2`   represents a plane `π_3` which is such  that if any vector ` vec r` satisfies both the equations `π_1` and `π_2`, it also satisfies the equation `π_3` i.e., any plane passing through the intersection of the planes 
`vec r . vec n_1 = d_1` and `vec r . vec n_2 = d_2`
has the equation  `vec r . (vec n_1 + lambda vec n_2) = d_1 + lambda d_2`    ...(1)

Cartesian form:
In Cartesian system, let  `vec n_1 = A_1 hat i + B_2 hat j + C_1 hat k`
`vec n_2 = A_2 hat i + B_2 hat j + C _2 hat k` 
and `vec r = x hat i + y hat j + z hat k`
Then (1) becomes 
`x (A_1 + lambda A_2) + y(B_1 + lambda B_2) + z(C_1 +lambda C_2) = d_1 + lambda d_2`
or `(A_1x +B_1y + C_1z -d_1) + lambda (A_2x + B_2y + C_2z -d_2) = 0`     ..(2)
which is the required Cartesian form of the equation of the plane passing through the intersection of the given planes for each value of λ.

Related QuestionsVIEW ALL [36]

Find the equation of the plane containing the line \[\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1}\]  and the point (0, 7, −7) and show that the line  \[\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}\] also lies in the same plane.

 

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

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 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.

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

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

 

Related concepts

Advertisements
Share
Notifications

Englishहिंदीमराठी
Login
Create free account


      Forgot password?
Course
Use app×