Advertisements
Advertisements
प्रश्न
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.
उत्तर
\[\text{ We know that the lines } \]
\[\frac{x - x_1}{l_1} = \frac{y - y_1}{m_1} = \frac{z - z_1}{n_1} \text{ and } \frac{x - x_2}{l_2} = \frac{y - y_2}{m_2} = \frac{z - z_2}{n_2} \text{ are coplanar if } \]
\[\begin{vmatrix}x_2 - x_1 & y_2 - y_1 & z_2 - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{vmatrix} = 0\]
\[\text{ and the equation of the plane containing these lines is } \]
\[\begin{vmatrix}x - x_1 & y - y_1 & z - z_1 \\ l_1 & m_1 & n_1 \\ l_2 & m_2 & n_2\end{vmatrix} = 0\]
\[\text{ Here } ,\]
\[ x_1 = - 1; y_1 = 3; z_1 = - 2; x_2 = 0; y_2 = 7; z_2 = - 7; l_1 = - 3; m_1 = 2; n_1 = 1; l_2 = 1; m_2 = - 3; n_2 = 2\]
\[\text{ Now } ,\]
\[\begin{vmatrix}0 + 1 & 7 - 3 & - 7 + 2 \\ - 3 & 2 & 1 \\ 1 & - 3 & 2\end{vmatrix}\]
\[ = \begin{vmatrix}1 & 4 & - 5 \\ - 3 & 2 & 1 \\ 1 & - 3 & 2\end{vmatrix}\]
\[ = 1 \left( 7 \right) - 4 \left( - 7 \right) - 5 \left( 7 \right)\]
\[ = 7 + 28 - 35\]
\[ = 0\]
\[\text{ So, the given lines are coplanar } .\]
\[\text{ The equation of the plane containing the given lines is } \]
\[\begin{vmatrix}x + 1 & y - 3 & z + 2 \\ - 3 & 2 & 1 \\ 1 & - 3 & 2\end{vmatrix} = 0\]
\[ \Rightarrow \left( x + 1 \right) \left( 7 \right) - \left( y - 3 \right) \left( - 7 \right) + \left( z + 2 \right) \left( 7 \right) = 0\]
\[ \Rightarrow 7x + 7y + 7z = 0\]
\[ \Rightarrow x + y + z = 0\]
APPEARS IN
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 points.
(2, 3, 4), (−3, 5, 1) and (4, −1, 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 following points are coplanar.
(0, −1, 0), (2, 1, −1), (1, 1, 1) and (3, 3, 0)
Find the coordinates of the point P where the line through A (3, -4 , -5 ) and B (2, -3 , 1) crosses the plane passing through three points L(2,2,1), M(3,0,1) and N(4, -1,0 ) . Also, find the ratio in which P diveides the line segment AB.
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 \[\left( \vec{r} \cdot \vec{n} = d \right):\]\[\vec{r} = \hat{i} - \hat{j} + \lambda\left( \hat{i} + \hat{j} + \hat{k} \right) + \mu\left( 4 \hat{i} - 2 \hat{j} + 3 \hat{k} \right)\]
Find the vector equation of the following planes in non-parametric form. \[\vec{r} = \left( 2 \hat{i} + 2 \hat{j} - \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) + \mu\left( 5 \hat{i} - 2 \hat{j} + 7 \hat{k} \right)\]
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
Show that the lines \[\vec{r} = \left( 2 \hat{j} - 3 \hat{k} \right) + \lambda\left( \hat{i} + 2 \hat{j} + 3 \hat{k} \right) \text{ and } \vec{r} = \left( 2 \hat{i} + 6 \hat{j} + 3 \hat{k} \right) + \mu\left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right)\] are coplanar. Also, find the equation of the plane containing them.
Find the values of \[\lambda\] for which the lines
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 \[\alpha\].
If the straight lines \[\frac{x - 1}{2} = \frac{y + 1}{k} = \frac{z}{2}\] and \[\frac{x + 1}{2} = \frac{y + 1}{2} = \frac{z}{k}\] are coplanar, find the equations of the planes containing them.
Find the equation of the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7).
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)
