मराठी

Find-vector-equation-plane-passing-through-points-3-4-2-7-0-6-perpendicular-plane-2x-5y-15-0 - Mathematics

Advertisements
Advertisements

प्रश्न

Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line \[\vec{r} = \hat{i} + 3 \hat{j}  - 2 \hat{k}  + \lambda\left( \hat{i}  - \hat{j}  + \hat{k}  \right) .\]

 
Advertisements

उत्तर

\[\text{ The equation of any plane passing through (3, 4, 2) is } \]

\[a \left( x - 3 \right) + b \left( y - 4 \right) + c \left( z - 2 \right) = 0 . . . \left( 1 \right)\]

\[\text{ It is given that (1) is passing through (7, 0, 6). So, } \]

\[a \left( 7 - 3 \right) + b \left( 0 - 4 \right) + c \left( 6 - 2 \right) = 0 \]

\[ \Rightarrow 4a - 4b + 4c = 0\]

\[ \Rightarrow a - b + c = 0 . . . \left( 2 \right)\]

\[\text{ It is given that (1) is perpendicular to the plane 2x - 5y + 0z + 15z = 0 . So, } \]

\[2a - 5b + 0c = 0 . . . \left( 3 \right)\]

\[\text{ Solving (1), (2) and (3), we get } \]

\[\begin{vmatrix}x - 3 & y - 4 & z - 2 \\ 1 & - 1 & 1 \\ 2 & - 5 & 0\end{vmatrix} = 0\]

\[ \Rightarrow 5 \left( x - 3 \right) + 2 \left( y - 4 \right) - 3 \left( z - 2 \right) = 0\]

\[ \Rightarrow 5x + 2y - 3z = 17\]

\[\text{ Or }  \vec{r} .\left( 5 \hat{i}  + 2 \hat{j}  - 3 \hat{k}  \right)= 17\]

\[\text{ Showing that the plane contains the line } \]

\[\text{ The line }  \vec{r} = \left( \hat{i} + 3 \hat{j} - 2 \hat{k}  \right) + \lambda \left( \hat{i} - \hat{j}  + \hat{k}  \right) \text{ passes through a point whose positon vector is } \vec{a} = \hat{i}  + 3 \hat{j} - 2 \hat{k}  \text{ and is parallel to the vector } \vec{b} = \hat{i} - \hat{j}  + \hat{k} . \]

\[\text{ If the plane }  \vec{r} .\left( 5 \hat{i}  + 2 \hat{j}  - 3 \hat{k}  \right)=17 \text{ contains the given line, then } \]

\[(1) \text{ it should pass through the point } \hat{i}  + 3 \hat{j} - 2 \hat{k}  \]

\[(2) \text{ it should be parallel to the line } \]

\[\text{ Now } ,\left( \hat{i} + 3 \hat{j}  - 2 \hat{k}  \right).\left( 5 \hat{i}  + 2 \hat{j}  - 3 \hat{k}  \right)= 5 + 6 + 6 = 17\]

\[\text{ So, the plane passes through the point } \hat{i}+ 3 \hat{j} - 2 \hat{k}  . \]

\[\text{ The normal vector to the given plane is } \vec{n} = \hat{i}  - \hat{j}  + \hat{k .} \]

\[\text{ We observe that } \]

\[ \vec{b} . \vec{n} = \left( \hat{i}  - \hat{j} + \hat{k}  \right) . \left( 5 \hat{i} + 2 \hat{j}  - 3 \hat{k}  \right) = 5 - 2 - 3 = 0\]

\[\text{ Therefore, the plane is parallel to the line. } \]

\[\text{ Hence, the given plane contains the given line } .\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: The Plane - Exercise 29.13 [पृष्ठ ७४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 29 The Plane
Exercise 29.13 | Q 8 | पृष्ठ ७४

संबंधित प्रश्‍न

In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

z = 2


In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

x + y + z = 1


Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the YZ-plane


Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX − plane.


Find the coordinates of the point where the line through (3, ­−4, −5) and (2, − 3, 1) crosses the plane 2x + z = 7).


The planes: 2− y + 4z = 5 and 5x − 2.5y + 10z = 6 are

(A) Perpendicular

(B) Parallel

(C) intersect y-axis

(C) passes through `(0,0,5/4)`


If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP.

 

Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane.


Reduce the equation \[\vec{r} \cdot \left( \hat{i}  - 2 \hat{j}  + 2 \hat{k}  \right) + 6 = 0\] to normal form and, hence, find the length of the perpendicular from the origin to the plane.

 


The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane. 


Find a unit normal vector to the plane x + 2y + 3z − 6 = 0.

 

Find the equation of a plane which is at a distance of \[3\sqrt{3}\]  units from the origin and the normal to which is equally inclined to the coordinate axes.

 

Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.

 

Find the equation of the plane which contains the line of intersection of the planes \[x + 2y + 3z - 4 = 0 \text { and } 2x + y - z + 5 = 0\] and whose x-intercept is twice its z-intercept.


Prove that the line of section of the planes 5x + 2y − 4z + 2 = 0 and 2x + 8y + 2z − 1 = 0 is parallel to the plane 4x − 2y − 5z − 2 = 0.

 

Find the value of λ such that the line \[\frac{x - 2}{6} = \frac{y - 1}{\lambda} = \frac{z + 5}{- 4}\]  is perpendicular to the plane 3x − y − 2z = 7.

 
 

Find the equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line \[\frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{- 1}\]

 

Write the plane  \[\vec{r} \cdot \left( 2 \hat{i}  + 3 \hat{j}  - 6 \hat{k}  \right) = 14\]  in normal form.

 
 

Write a vector normal to the plane  \[\vec{r} = l \vec{b} + m \vec{c} .\]

 

Write the value of k for which the line \[\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{k}\]  is perpendicular to the normal to the plane  \[\vec{r} \cdot \left( 2 \hat{i}  + 3 \hat{j}  + 4 \hat{k}  \right) = 4 .\]


Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is \[2 \hat{i} - 3 \hat{j} + 6 \hat{k} \] .


Find the image of the point having position vector `hat"i" + 3hat"j" + 4hat"k"` in the plane `hat"r" * (2hat"i" - hat"j" + hat"k") + 3` = 0.


The equations of x-axis in space are ______.


If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane.


The plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(α) with x-axis. The value of α is equal to ______.


The unit vector normal to the plane x + 2y +3z – 6 = 0 is `1/sqrt(14)hat"i" + 2/sqrt(14)hat"j" + 3/sqrt(14)hat"k"`.


What will be the cartesian equation of the following plane. `vecr * (hati + hatj - hatk)` = 2


In the following cases find the c9ordinates of foot of perpendicular from the origin `2x + 3y + 4z - 12` = 0


Find the vector and cartesian equations of the planes that passes through (1, 0, – 2) and the normal to the plane is `hati + hatj - hatk`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×