मराठी

Find the Vector Equation of Each One of Following Planes. (Ii) X + Y − Z = 5 - Mathematics

Advertisements
Advertisements

प्रश्न

Find the vector equation of each one of following planes. 

x + y − z = 5

 

बेरीज

उत्तर

 `\text{ The given equation of plane is } `

\[x + y - z = 8\]

\[ \Rightarrow \left( \text{ x }\hat{i} + \text{  y} \hat{j}  + \text{ z} \hat{k}  \right) . \left( \hat{i}  + \hat{j}  - \hat{k} \right) = 8\]

\[ \Rightarrow \vec{r} . \left( \hat{i} + \hat{j} - \hat{k} \right) = 8, \text{ which is the vector equation of the plane }.\]

\[(\text{ Because the vector equation of the plane is }  \vec{r} . \vec{n} = \vec{a} . \vec{n} )\]

shaalaa.com
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 29: The Plane - Exercise 29.03 [पृष्ठ १३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
पाठ 29 The Plane
Exercise 29.03 | Q 4.2 | पृष्ठ १३

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector.`3hati + 5hatj - 6hatk`


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

(1, 1, −1), (6, 4, −5), (−4, −2, 3)


Find the equation of the plane passing through (abc) and parallel to the plane `vecr.(hati + hatj + hatk) = 2`


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 vector equation of a plane passing through a point with position vector \[2 \hat{i} - \hat{j} + \hat{k} \] and perpendicular to the vector  \[4 \hat{i} + 2 \hat{j} - 3 \hat{k} .\] 


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 equation of each one of following planes. 

2x − y + 2z = 8


\[\vec{n}\] is a vector of magnitude \[\sqrt{3}\] and is equally inclined to an acute angle with the coordinate axes. Find the vector and Cartesian forms of the equation of a plane which passes through (2, 1, −1) and is normal to \[\vec{n}\] .

 


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

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

Find the vector equation of the plane passing through the points (1, 1, 1), (1, −1, 1) and (−7, −3, −5).


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 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 points (1, −1, 2) and (2, −2, 2) and which is perpendicular to the plane 6x − 2y + 2z = 9.

 

Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.

 

Find the vector equation of the line through the origin which is perpendicular to the plane  \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j}  + 3 \hat{k}  \right) = 3 .\]

 

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

 

Find the equation of the plane through the points (2, 2, −1) and (3, 4, 2) and parallel to the line whose direction ratios are 7, 0, 6.

 

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 image of the point (0, 0, 0) in the plane 3x + 4y − 6z + 1 = 0.

 

Find the distance of the point (1, −2, 3) from the plane x − y + z = 5 measured along a line parallel to  \[\frac{x}{2} = \frac{y}{3} = \frac{z}{- 6} .\]

 


Find the image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0.

 

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


Write the general equation of a plane parallel to X-axis.

 

Write the value of k for which the planes x − 2y + kz = 4 and 2x + 5y − z = 9 are perpendicular.

 

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

 

Write the intercept cut off by the plane 2x + y − z = 5 on x-axis.

 

Find the length of the perpendicular drawn from the origin to the plane 2x − 3y + 6z + 21 = 0.

 

Find a vector of magnitude 26 units normal to the plane 12x − 3y + 4z = 1.


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 vector equation of the plane which contains the line of intersection of the planes `vec("r").(hat"i"+2hat"j"+3hat"k"),-4=0, vec("r").(2hat"i"+hat"j"-hat"k")+5=0`and which is perpendicular to the plane`vec("r").(5hat"i"+3hat"j"-6hat"k"),+8=0`


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


Find the foot of perpendicular from the point (2, 3, –8) to the line `(4 - x)/2 = y/6 = (1 - z)/3`. Also, find the perpendicular distance from the given point to the line.


Find the equations of the line passing through the point (3, 0, 1) and parallel to the planes x + 2y = 0 and 3y – z = 0.


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.


The equation of a line, which is parallel to `2hat"i" + hat"j" + 3hat"k"` and which passes through the point (5, –2, 4), is `(x - 5)/2 = (y + 2)/(-1) = (z - 4)/3`.


The point at which the normal to the curve y = `"x" + 1/"x", "x" > 0` is perpendicular to the line 3x – 4y – 7 = 0 is:


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×