हिंदी

The Equation of Plane Containing the Two Lines X−1 2 = Y + 1 − 1 = Z − 0 3 a Nd X − 2 = Y − 2 − 3 = Z + 1 − 1 (A) 8x + Y − 5z − 7 = 0 (B) 8x + Y + 5z − 7 = 0 (C) 8x − Y − 5z − 7 = 0 (D) None of These - Mathematics

Advertisements
Advertisements

प्रश्न

The equation of the plane containing the two lines

\[\frac{x - 1}{2} = \frac{y + 1}{- 1} = \frac{z - 0}{3} \text{ and }\frac{x}{- 2} = \frac{y - 2}{- 3} = \frac{z + 1}{- 1}\]
 
 

विकल्प

  •  8x + y − 5z − 7 = 0

  •  8x + y + 5z − 7 = 0

  • 8x − y − 5z − 7 = 0

  •  None of these

     
MCQ
Advertisements

उत्तर

 None of these

\[\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z- 0}{3}\] and 

\[\frac{x}{- 2} = \frac{y - 2}{- 3} = \frac{z + 1}{- 1}\]

Now, if these two lines lie on a plane, so the direction ratio of lines will be perpendicular to the plane's normal vector.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: The Plane - MCQ [पृष्ठ ८५]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 29 The Plane
MCQ | Q 6 | पृष्ठ ८५

संबंधित प्रश्न

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

2x + 3y – z = 5


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.

 

If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP.

 

Find the intercepts made on the coordinate axes by the plane 2x + y − 2z = 3 and also find the direction cosines of the normal to the plane.


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 value of λ such that the line \[\frac{x - 2}{6} = \frac{y - 1}{\lambda} = \frac{z + 5}{- 4}\]  is perpendicular to the plane 3x − y − 2z = 7.

 
 

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

 

Write the plane  \[\vec{r} \cdot \left( 2 \hat{i}  + 3 \hat{j}  - 6 \hat{k}  \right) = 14\]  in normal form.

 
 

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

 

 

 

 

 
 
 

The equations of x-axis in space are ______.


If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane.


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


Find the vector equation of a plane which is at a distance of 7 units from the origin and which is normal to the vector `3hati + 5hatj - 6hatk`


In the following cases find the c9ordinates of foot of perpendicular from the origin `2x + 3y + 4z - 12` = 0


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×