हिंदी

Find the Coordinates of the Point Where the Line Through (5, 1, 6) and (3, 4, 1) Crosses the Zx − Plane. - Mathematics

Advertisements
Advertisements

प्रश्न

Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX − plane.

उत्तर

It is known that the equation of the line passing through the points, (x1y1z1) and (x2y2z2), is

Any point on the line is of the form (5 − 2k, 3k + 1, 6 −5k).

Since the line passes through ZX-plane,

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 11: Three Dimensional Geometry - Exercise 11.4 [पृष्ठ ४९८]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 12
अध्याय 11 Three Dimensional Geometry
Exercise 11.4 | Q 11 | पृष्ठ ४९८

संबंधित प्रश्न

In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

z = 2


In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.

5y + 8 = 0


Find the equation of the plane with intercept 3 on the y-axis and parallel to ZOX plane.


If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (­−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.


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 point where the line through (3, ­−4, −5) and (2, − 3, 1) crosses the plane 2x + z = 7).


Find the equation of the plane passing through the point (2, 3, 1), given that the direction ratios of the normal to the plane are proportional to 5, 3, 2.

 

Reduce the equation 2x − 3y − 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane. 


Reduce the equation \[\vec{r} \cdot \left( \hat{i}  - 2 \hat{j}  + 2 \hat{k}  \right) + 6 = 0\] to normal form and, hence, find the length of the perpendicular from the origin to the plane.

 


Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0.

 

The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane. 


Find a unit normal vector to the plane x + 2y + 3z − 6 = 0.

 

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

 

Find the vector equation of the plane which is at a distance of \[\frac{6}{\sqrt{29}}\] from the origin and its normal vector from the origin is  \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} .\] Also, find its Cartesian form. 

 

Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.

 

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

 

Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line \[\vec{r} = \hat{i} + 3 \hat{j}  - 2 \hat{k}  + \lambda\left( \hat{i}  - \hat{j}  + \hat{k}  \right) .\]

 

Write a vector normal to the plane  \[\vec{r} = l \vec{b} + m \vec{c} .\]

 

Write the value of k for which the line \[\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{k}\]  is perpendicular to the normal to the plane  \[\vec{r} \cdot \left( 2 \hat{i}  + 3 \hat{j}  + 4 \hat{k}  \right) = 4 .\]


Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is \[2 \hat{i} - 3 \hat{j} + 6 \hat{k} \] .


The equation of the plane \[\vec{r} = \hat{i} - \hat{j}  + \lambda\left( \hat{i}  + \hat{j} + \hat{k}  \right) + \mu\left( \hat{i}  - 2 \hat{j}  + 3 \hat{k}  \right)\]  in scalar product form is

 

 

 

 

 
 
 

Find the image of the point having position vector `hat"i" + 3hat"j" + 4hat"k"` in the plane `hat"r" * (2hat"i" - hat"j" + hat"k") + 3` = 0.


The equations of x-axis in space are ______.


Find the equation of a plane which is at a distance `3sqrt(3)` units from origin and the normal to which is equally inclined to coordinate axis.


The plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(α) with x-axis. The value of α is equal to ______.


The unit vector normal to the plane x + 2y +3z – 6 = 0 is `1/sqrt(14)hat"i" + 2/sqrt(14)hat"j" + 3/sqrt(14)hat"k"`.


What will be the cartesian equation of the following plane. `vecr * (hati + hatj - hatk)` = 2


Find the vector and cartesian equations of the planes that passes through (1, 0, – 2) and the normal to the plane is `hati + hatj - hatk`


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×