हिंदी

Find the Equation of the Plane Passing Through the Following Points. (Iv) (2, 3, 4), (−3, 5, 1) and (4, −1, 2) - Mathematics

Advertisements
Advertisements

प्रश्न

Find the equation of the plane passing through the following points. 

(2, 3, 4), (−3, 5, 1) and (4, −1, 2) 

 

योग
Advertisements

उत्तर

 The equation of the plane passing through points (2, 3, 4), (−3, 5, 1) and (4, −1, 2) is given by 

\[\begin{vmatrix}x - 2 & y - 3 & z - 4 \\ - 3 - 2 & 5 - 3 & 1 - 4 \\ 4 - 2 & - 1 - 3 & 2 - 4\end{vmatrix} = 0\]

\[\]

\[ \Rightarrow \begin{vmatrix}x - 2 & y - 3 & z - 4 \\ - 5 & 2 & - 3 \\ 2 & - 4 & - 2\end{vmatrix} = 0\]

\[\]

\[ \Rightarrow - 16 \left( x - 2 \right) - 16 \left( y - 3 \right) + 16 \left( z - 4 \right) = 0\]

\[ \Rightarrow \left( x - 2 \right) + \left( y - 3 \right) - \left( z - 4 \right) = 0\]

\[ \Rightarrow x + y - z = 1\]

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 29: The Plane - Exercise 29.01 [पृष्ठ ४]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 12
अध्याय 29 The Plane
Exercise 29.01 | Q 1.4 | पृष्ठ ४

संबंधित प्रश्न

Find the equations of the planes parallel to the plane x-2y + 2z-4 = 0, which are at a unit distance from the point (1,2, 3).


Find the equation of the plane passing through the following points.

 (2, 1, 0), (3, −2, −2) and (3, 1, 7)


Find the equation of the plane passing through the following points.

 (−5, 0, −6), (−3, 10, −9) and (−2, 6, −6)


Find the equation of the plane passing through the following point

 (1, 1, 1), (1, −1, 2) and (−2, −2, 2)


Find the equation of the plane passing through the following point

(0, −1, 0), (3, 3, 0) and (1, 1, 1)

 

 


Show that the four points (0, −1, −1), (4, 5, 1), (3, 9, 4) and (−4, 4, 4) are coplanar and find the equation of the common plane.


Show that the following points are coplanar. 

 (0, 4, 3), (−1, −5, −3), (−2, −2, 1) and (1, 1, −1)

 

Find the vector equations of the following planes in scalar product form  \[\left( \vec{r} \cdot \vec{n} = d \right):\] \[\vec{r} = \left( 2 \hat{i} - \hat{k} \right) + \lambda \hat{i} + \mu\left( \hat{i} - 2 \hat{j} - \hat{k}
\right)\]

 

Find the vector equations of the following planes in scalar product form  \[\left( \vec{r} \cdot \vec{n} = d \right):\]\[\vec{r} = \left( \hat{i}  + \hat{j}  \right) + \lambda\left( \hat{i}  + 2 \hat{j}  - \hat{k}  \right) + \mu\left( - \hat{i}  + \hat{j} - 2 \hat{k} \right)\]


Find the vector equations of the following planes in scalar product form  \[\left( \vec{r} \cdot \vec{n} = d \right):\]\[\vec{r} = \hat{i} - \hat{j} + \lambda\left( \hat{i}  + \hat{j}  + \hat{k}  \right) + \mu\left( 4 \hat{i}  - 2 \hat{j}  + 3 \hat{k} \right)\]

 


Find the Cartesian forms of the equations of the following planes.

\[\vec{r} = \left( 1 + s + t \right) \hat{i}  + \left( 2 - s + t \right) \hat{i}  + \left( 3 - 2s + 2t \right) \hat{k}\]

 


Find the vector equation of the following planes in non-parametric form. \[\vec{r} = \left( \lambda - 2\mu \right) \hat{i} + \left( 3 - \mu \right) \hat{j}  + \left( 2\lambda + \mu \right) \hat{k} \]


Find the equation of the plane through (3, 4, −1) which is parallel to the plane \[\vec{r} \cdot \left( 2 \hat{i} - 3 \hat{j}  + 5 \hat{k} \right) + 2 = 0 .\]

 

Find the equation of the plane passing through the line of intersection of the planes 2x − 7y + 4z − 3 = 0, 3x − 5y + 4z + 11 = 0 and the point (−2, 1, 3).


Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line 

\[\frac{x + 3}{2} = \frac{y - 3}{7} = \frac{z - 2}{5} .\]
  

Show that the lines \[\frac{x + 1}{- 3} = \frac{y - 3}{2} = \frac{z + 2}{1} \text{ and }\frac{x}{1} = \frac{y - 7}{- 3} = \frac{z + 7}{2}\]  are coplanar. Also, find the equation of the plane containing them. 

 

Show that the lines  \[\frac{5 - x}{- 4} = \frac{y - 7}{4} = \frac{z + 3}{- 5}\] and  \[\frac{x - 8}{7} = \frac{2y - 8}{2} = \frac{z - 5}{3}\] are coplanar.

 

Show that the lines  \[\frac{x + 3}{- 3} = \frac{y - 1}{1} = \frac{z - 5}{5}\] and  \[\frac{x + 1}{- 1} = \frac{y - 2}{2} = \frac{z - 5}{5}\]  are coplanar. Hence, find the equation of the plane containing these lines.

 

Find the values of  \[\lambda\] for which the lines

\[\frac{x - 1}{1} = \frac{y - 2}{2} = \frac{z + 3}{\lambda^2}\]and  \[\frac{x - 3}{1} = \frac{y - 2}{\lambda^2} = \frac{z - 1}{2}\]  are coplanar . 

If the lines  \[x =\]  5 ,  \[\frac{y}{3 - \alpha} = \frac{z}{- 2}\] and   \[x = \alpha\] \[\frac{y}{- 1} = \frac{z}{2 - \alpha}\] are coplanar, find the values of  \[\alpha\].

 


The points (1, 2, 3), (–2, 3, 4) and (7, 0, 1) are collinear.


The equation of the circle passing through the foci of the ellipse `x^2/16 + y^2/9` = 1 and having centre at (0, 3) is


Find the equations of the planes that passes through three points (1, 1, – 1), (6, 4, – 5),(– 4, – 2, 3)


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×