हिंदी

Write the Distance of the Plane → R ⋅ ( 2 ^ I − ^ J + 2 ^ K ) = 12 from the Origin. - Mathematics

Advertisements
Advertisements

प्रश्न

Write the distance of the plane  \[\vec{r} \cdot \left( 2 \hat{i} - \hat{j} + 2 \hat{k} \right) = 12\] from the origin.

  
Advertisements

उत्तर

\[\text{ The given equation of the plane is} \]

\[ \vec{r} . \left( 2 \hat{i}  - \hat{j} + 2 \hat{k} \right) = 12 \text{ or } \vec{r} . \vec{n} = - 6, \text{ where } \vec{n} =2 \hat{i}  - \hat{j}  + 2 \hat{k}  \]

\[\left| \vec{n} \right| = \sqrt{4 + 1 + 4} = 3\]

\[\text{ For reducing the given equation to normal form, we need to divide both sides by } \left| \vec{n} \right|. \text{ Then, we get } \]

\[ \vec{r} . \frac{\vec{n}}{\left| \vec{n} \right|} = \frac{12}{\left| \vec{n} \right|}\]

\[ \Rightarrow \vec{r} . \left( \frac{2 \hat{i}  - \hat{j}  + 2 \hat{k} }{3} \right) = \frac{12}{3}\]

\[ \Rightarrow \vec{r} . \left( \frac{2}{3} \hat{i}  - \frac{1}{3} \hat{j} + \frac{2}{3} \hat{k}  \right) = 4 . . . \left( 1 \right)\]

\[\text{ The equation of the plane in normal form is } \]

\[ \vec{r} . \hat{n}  = d . . . \left( 2 \right)\]

\[( \text{ where d is the distance of the plane from the origin } )\]

\[\text{ Comparing (1) and (2) } ,\]

\[\text{ length of the perpendicular from the origin to the plane =d= 4 units } \]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: The Plane - Very Short Answers [पृष्ठ ८३]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 29 The Plane
Very Short Answers | Q 10 | पृष्ठ ८३

वीडियो 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, 0), (1, 2, 1), (−2, 2, −1)


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 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 vector equation of each one of following planes. 

x + y = 3

 

\[\vec{n}\] 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 \[\vec{n}\] .

 


The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane.

 

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 


Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.

 

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 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 that contains the point (1, −1, 2) and is perpendicular to each of the planes 2x + 3y − 2z = 5 and x + 2y − 3z = 8.


Find the equation of the plane passing through (abc) and parallel to the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k}  \right) = 2 .\]

 

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

 

Find the equation of a plane passing through the points (0, 0, 0) and (3, −1, 2) and parallel to the line \[\frac{x - 4}{1} = \frac{y + 3}{- 4} = \frac{z + 1}{7} .\]

 

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 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 + y + z = 7.

 

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

 

Find the coordinates of the foot of the perpendicular and the perpendicular distance of the  point P (3, 2, 1) from the plane 2x − y + z + 1 = 0. Also, find the image of the point in the plane.


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


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

 

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

 

Write the ratio in which the plane 4x + 5y − 3z = 8 divides the line segment joining the points (−2, 1, 5) and (3, 3, 2).

 

Write the equation of the plane  \[\vec{r} = \vec{a} + \lambda \vec{b} + \mu \vec{c}\]   in scalar product form.

 

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

 

Find the co-ordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, –1, 3) and C(2, –3, –1).


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


Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0.


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.


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.


`vec"AB" = 3hat"i" - hat"j" + hat"k"` and `vec"CD" = -3hat"i" + 2hat"j" + 4hat"k"` are two vectors. The position vectors of the points A and C are `6hat"i" + 7hat"j" + 4hat"k"` and `-9hat"j" + 2hat"k"`, respectively. Find the position vector of a point P on the line AB and a point Q on the line Cd such that `vec"PQ"` is perpendicular to `vec"AB"` and `vec"CD"` both.


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.


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×