हिंदी

Show that the points ijk(i^-j^+3k^) and ijk3(i^+j^+k^) are equidistant from the plane rijkr→⋅(5i^+2j^-7k^)+9 = 0 and lies on opposite side of it. - Mathematics

Advertisements
Advertisements

प्रश्न

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.

योग
Advertisements

उत्तर

Given points are `"P"(hat"i" - hat"j" + 3hat"k")` and `"Q"3(hat"i" + hat"j" + hat"k")` and the plane `vec"r" * (5hat"i" + 2hat"j" - 7hat"k") + 9` = 0

Perpendicular distance of `"P"(hat"i" - hat"j" + 3hat"k")` from the plane

`vec"r" * (5hat"i" + 2hat"j" - 7hat"k") + 9 = |((hat"i" - hat"j" + 3hat"k")*(5hat"i" + 2hat"j" - 7hat"k") + 9)/sqrt((5)^2 + (2)^2 + (-7)^2)|`

= `|(5 - 2 - 21 + 9)/sqrt(25 + 4 + 49)|`

= `|(-9)/sqrt(78)|`

And perpendicular distance of `"Q"(3hat"i" + 3hat"j" + 3hat"k")` from the plane

= `|((3hat"i" + 3hat"j" + 3hat"k")*(5hat"i" + 2hat"j" - 7hat"k") + 9)/sqrt(25 + 4 + 29)|`

= `|(15 + 6 - 21 + 9)/sqrt(78)|`

= `|9/sqrt(78)|`

Hence, the two points are equidistant from the given plane.

Opposite sign shows that they lie on either side of the plane.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Three Dimensional Geometry - Exercise [पृष्ठ २३७]

APPEARS IN

एनसीईआरटी एक्झांप्लर Mathematics [English] Class 12
अध्याय 11 Three Dimensional Geometry
Exercise | Q 25 | पृष्ठ २३७

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

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

(1, 1, 0), (1, 2, 1), (−2, 2, −1)


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 (1, 2, 3) and parallel to the planes `vecr = (hati - hatj + 2hatk)  = 5`and `vecr.(3hati + hatj + hatk) = 6`


Find the vector equation of each one of following planes. 

2x − y + 2z = 8


Find the vector equation of each one of following planes. 

x + y − z = 5

 


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

 


Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.

 

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

 


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 \[3 \hat{i}  + 4 \hat{j}  + 2 \hat{k} , 2 \hat{i} - 2 \hat{j} - \hat{k}  \text{ and }  7 \hat{i}  + 6 \hat{k}  .\]

 

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

 2x − 4y + 3z = 5 and x + 2y + λz = 5


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


Find the coordinates of the foot of the perpendicular drawn from the point (5, 4, 2) to the line \[\frac{x + 1}{2} = \frac{y - 3}{3} = \frac{z - 1}{- 1} .\]

 Hence, or otherwise, deduce the length of the perpendicular.

 
 

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

 

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

 

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


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

  

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

 

Write the equation of a plane which is at a distance of \[5\sqrt{3}\] units from origin and the normal to which is equally inclined to coordinate axes.

 

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


`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.


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×