मराठी

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.

बेरीज

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
  या प्रश्नात किंवा उत्तरात काही त्रुटी आहे का?
पाठ 11: Three Dimensional Geometry - Exercise [पृष्ठ २३७]

APPEARS IN

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

व्हिडिओ ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्‍न

If O be the origin and the coordinates of P be (1, 2, −3), then find the equation of the plane passing through P and perpendicular to OP.


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 \[\vec{n}\] .

 


find the equation of the plane passing through the point (1, 2, 1) and perpendicular to the line joining the points (1, 4, 2) and (2, 3, 5). Find also the perpendicular distance of the origin from this plane


Find the vector equation of the plane passing through the points P (2, 5, −3), Q (−2, −3, 5) and R (5, 3, −3).


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 \[\frac{1}{p^2} = \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} .\]

 


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

\[\vec{r} \cdot \left( \hat{i}  + 2 \hat{j} + 3 \hat{k} \right) = 7 \text{ and }  \vec{r} \cdot \left( \lambda \hat{i} + 2 \hat{j}  - 7 \hat{k}  \right) = 26\]

 


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

 3x − 6y − 2z = 7 and 2x + y − λz = 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 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 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 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 equation of a plane which passes through the point (3, 2, 0) and contains the line  \[\frac{x - 3}{1} = \frac{y - 6}{5} = \frac{z - 4}{4}\] .

 


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

 

Find the coordinates of the foot of the perpendicular from the point (1, 1, 2) to the plane 2x − 2y + 4z + 5 = 0. Also, find the length of the perpendicular.

 

Find the length and the foot of perpendicular from the point \[\left( 1, \frac{3}{2}, 2 \right)\]  to the plane \[2x - 2y + 4z + 5 = 0\] .

 

Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to both the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Hence, find the distance of point P (–2, 5, 5) from the plane obtained


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 equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2y −z = 7.


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

 

If O be the origin and the coordinates of P be (1, 2,−3), then find the equation of the plane passing through P and perpendicular to OP.


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`


`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.


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×