English

In the Following Cases, Find the Coordinates of the Foot of the Perpendicular Drawn from the Origin.2x + 3y + 4z – 12 = 0 - Mathematics

Advertisements
Advertisements

Question

In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

2x + 3y + 4z – 12 = 0

Solution

Let the coordinates of the foot of perpendicular P from the origin to the plane be (x1y1z1).

2x + 3y + 4z − 12 = 0

⇒ 2x + 3y + 4z = 12 … (1)

The direction ratios of normal are 2, 3, and 4

`:. sqrt((2)^2 + (3)^2 + (4)^2) =  sqrt29`

Dividing both sides of equation (1) by sqrt29, we obtain

This equation is of the form lx + my + nz = d, where lmn are the direction cosines of normal to the plane and d is the distance of normal from the origin.

The coordinates of the foot of the perpendicular are given by

(ldmdnd).

Therefore, the coordinates of the foot of the perpendicular are

shaalaa.com
  Is there an error in this question or solution?
Chapter 11: Three Dimensional Geometry - Exercise 11.3 [Page 493]

APPEARS IN

NCERT Mathematics [English] Class 12
Chapter 11 Three Dimensional Geometry
Exercise 11.3 | Q 4.1 | Page 493

RELATED QUESTIONS

Find the vector equation of the plane passing through the points `hati +hatj-2hatk, hati+2hatj+hatk,2hati-hatj+hatk`. Hence find the cartesian equation of the plane.


Find the vector equation of the plane which is at a distance of 5 units from the origin and which is normal to the vector  `2hati + hatj + 2hatk.`


Parametric form of the equation of the plane is `bar r=(2hati+hatk)+lambdahati+mu(hat i+2hatj+hatk)` λ and μ are parameters. Find normal to the plane and hence equation of the plane in normal form. Write its Cartesian form.


Write the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane `vec r.(hati+hatj+hatk)=2`


Find the vector equation of the plane passing through three points with position vectors ` hati+hatj-2hatk , 2hati-hatj+hatk and hati+2hatj+hatk` . Also find the coordinates of the point of intersection of this plane and the line `vecr=3hati-hatj-hatk lambda +(2hati-2hatj+hatk)`

 


Find the vector equation of the plane with intercepts 3, –4 and 2 on x, y and z-axis respectively.


Find the equation of the plane which contains the line of intersection of the planes

`vecr.(hati-2hatj+3hatk)-4=0" and"`

`vecr.(-2hati+hatj+hatk)+5=0`

and whose intercept on x-axis is equal to that of on y-axis.


Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is `2hati-3hatj+6hatk`


Find the vector equation of a line passing through the points A(3, 4, –7) and B(6, –1, 1).


Find the Cartesian equation of the following planes:

`vecr.(hati + hatj-hatk) = 2`


Find the Cartesian equation of the following planes:

`vecr.(2hati + 3hatj-4hatk) = 1`


In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

3y + 4z – 6 = 0


In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

x + y + z = 1


In the following cases, find the coordinates of the foot of the perpendicular drawn from the origin.

5y + 8 = 0


Find the cartesian form of the equation of the plane `bar r=(hati+hatj)+s(hati-hatj+2hatk)+t(hati+2hatj+hatj)`


Find the equation of the plane through the line of intersection of `vecr*(2hati-3hatj + 4hatk) = 1`and `vecr*(veci - hatj) + 4 =0`and perpendicular to the plane `vecr*(2hati - hatj + hatk) + 8 = 0`. Hence find whether the plane thus obtained contains the line x − 1 = 2y − 4 = 3z − 12.


The Cartesian equation of the line is 2x - 3 = 3y + 1 = 5 - 6z. Find the vector equation of a line passing through (7, –5, 0) and parallel to the given line.


Find the image of a point having the position vector: `3hati - 2hatj + hat k` in the plane `vec r.(3hati - hat j + 4hatk) = 2`


Find the vector and Cartesian equations of the line passing through (1, 2, 3) and parallel to the planes \[\vec{r} \cdot \left( \hat{i}  - \hat{j} + 2 \hat{k}  \right) = 5 \text{ and } \vec{r} \cdot \left( 3 \hat{i} + \hat{j}  + 2 \hat{k} \right) = 6\]

 

Find the vector and Cartesian forms of the equation of the plane passing through the point (1, 2, −4) and parallel to the lines \[\vec{r} = \left( \hat{i} + 2 \hat{j}  - 4 \hat{k}  \right) + \lambda\left( 2 \hat{i}  + 3 \hat{j}  + 6 \hat{k}  \right)\] and \[\vec{r} = \left( \hat{i}  - 3 \hat{j}  + 5 \hat{k}  \right) + \mu\left( \hat{i}  + \hat{j}  - \hat{k} \right)\] Also, find the distance of the point (9, −8, −10) from the plane thus obtained.  

 


Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes  \[\vec{r} \cdot \left( \hat{i}  - \hat{j}  + 2 \hat{k}  \right) = 5 \text{ and } \vec{r} \cdot \left( 3 \hat{i}  + \hat{j}  + \hat{k}  \right) = 6 .\]

 


Find the equation of the plane passing through the intersection of the planes `vec(r) .(hat(i) + hat(j) + hat(k)) = 1"and" vec(r) . (2 hat(i) + 3hat(j) - hat(k)) +4 = 0 `and parallel to x-axis. Hence, find the distance of the plane from x-axis.


Find the Cartesian equation of the plane, passing through the line of intersection of the planes `vecr. (2hati + 3hatj - 4hatk) + 5 = 0`and `vecr. (hati - 5hatj + 7hatk) + 2 = 0`  intersecting the y-axis at (0, 3).


Find the vector and cartesian equation of the plane passing through the point (2, 5, - 3), (-2, -3, 5) and (5, 3, -3). Also, find the point of intersection of this plane with the line passing through points (3, 1, 5) and (-1, -3, -1).


Find the vector and Cartesian equations of the plane passing through the points (2, 2 –1), (3, 4, 2) and (7, 0, 6). Also find the vector equation of a plane passing through (4, 3, 1) and parallel to the plane obtained above.


Find the vector equation of the plane that contains the lines `vecr = (hat"i" + hat"j") + λ (hat"i" + 2hat"j" - hat"k")` and the point (–1, 3, –4). Also, find the length of the perpendicular drawn from the point (2, 1, 4) to the plane thus obtained.


The vector equation of the line through the points (3, 4, –7) and (1, –1, 6) is ______.


The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is `vec"r" = 5hat"i" - 4hat"j" + 6hat"k" + lambda(3hat"i" + 7hat"j" + 2hat"k")`.


Find the vector and the cartesian equations of the plane containing the point `hati + 2hatj - hatk` and parallel to the lines `vecr = (hati + 2hatj + 2hatk) + s(2hati - 3hatj + 2hatk)` and `vecr = (3hati + hatj - 2hatk) + t(hati - 3hatj + hatk)`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×