Advertisements
Advertisements
प्रश्न
Write the ratio in which the plane 4x + 5y − 3z = 8 divides the line segment joining the points (−2, 1, 5) and (3, 3, 2).
Advertisements
उत्तर
\[\text{ We know that the ratio in which the plane ax + by + cz + d = 0 divides the line segment joining} \left( x_1 , y_1 , z_1 \right)\text{ and } \left( x_2 , y_2 , z_2 \right)is\]
\[\frac{- \left( a x_1 + b y_1 + c z_1 + d \right)}{a x_2 + b y_2 + c z_2 + d}\]
\[\text{ Here } ,a = 4; b = 5; c = - 3; d = - 8; x_1 = - 2; y_1 = 1; z_1 = 5; x_2 = 3; y_2 = 3; z_2 = 2\]
\[\text{ So, the required ratio } \]
\[ = \frac{- \left( 4 \left( - 2 \right) + 5 \left( 1 \right) - 3 \left( 5 \right) - 8 \right)}{4 \left( 3 \right) + 5 \left( 3 \right) - 3 \left( 2 \right) - 8}\]
\[ = \frac{- \left( - 8 + 5 - 15 - 8 \right)}{12 + 15 - 6 - 8}\]
\[ = \frac{26}{13}\]
\[ = \frac{2}{1} \text{ or } 2 : 1\]
APPEARS IN
संबंधित प्रश्न
Find the equations of the planes that passes through three points.
(1, 1, −1), (6, 4, −5), (−4, −2, 3)
Find the equations of the planes that passes through three points.
(1, 1, 0), (1, 2, 1), (−2, 2, −1)
The coordinates of the foot of the perpendicular drawn from the origin to a plane are (12, −4, 3). Find the equation of the plane.
A plane passes through the point (1, −2, 5) and is perpendicular to the line joining the origin to the point
Show that the normals to the following pairs of planes are perpendicular to each other.
x − y + z − 2 = 0 and 3x + 2y − z + 4 = 0
Show that the normals to the following pairs of planes are perpendicular to each other.
Show that the normal vector to the plane 2x + 2y + 2z = 3 is equally inclined to the coordinate axes.
Find the vector equation of a plane which is at a distance of 3 units from the origin and has \[\hat{k}\] as the unit vector normal to it.
Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5.
Find the equation of the plane passing through the origin and perpendicular to each of the planes x + 2y − z = 1 and 3x − 4y + z = 5.
Find the equation of the plane passing through the points whose coordinates are (−1, 1, 1) and (1, −1, 1) and perpendicular to the plane x + 2y + 2z = 5.
Find the equation of the plane passing through (a, b, c) and parallel to the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 2 .\]
Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - 5 \hat{k} \right) + 9 = 0 .\]
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the zx - plane .
Find the image of the point (1, 3, 4) in the plane 2x − y + z + 3 = 0.
Find the length and the foot of the perpendicular from the point (1, 1, 2) to the plane \[\vec{r} \cdot \left( \hat{i} - 2 \hat{j} + 4 \hat{k} \right) + 5 = 0 .\]
Find the direction cosines of the unit vector perpendicular to the plane \[\vec{r} \cdot \left( 6 \hat{i} - 3 \hat{j} - 2 \hat{k} \right) + 1 = 0\] passing through the origin.
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x − 3y + 4z − 6 = 0.
Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to both the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Hence, find the distance of point P (–2, 5, 5) from the plane obtained
Write the general equation of a plane parallel to X-axis.
Write the equation of the plane passing through (2, −1, 1) and parallel to the plane 3x + 2y −z = 7.
Write the intercept cut off by the plane 2x + y − z = 5 on x-axis.
Find the length of the perpendicular drawn from the origin to the plane 2x − 3y + 6z + 21 = 0.
If the line drawn from (4, −1, 2) meets a plane at right angles at the point (−10, 5, 4), find the equation of the plane.
Find the equation of the plane which bisects the line segment joining the points (−1, 2, 3) and (3, −5, 6) at right angles.
Find the vector equation of the plane which contains the line of intersection of the planes `vec("r").(hat"i"+2hat"j"+3hat"k"),-4=0, vec("r").(2hat"i"+hat"j"-hat"k")+5=0`and which is perpendicular to the plane`vec("r").(5hat"i"+3hat"j"-6hat"k"),+8=0`
Find the image of the point (1, 6, 3) in the line `x/1 = (y - 1)/2 = (z - 2)/3`.
Find the length and the foot of perpendicular from the point `(1, 3/2, 2)` to the plane 2x – 2y + 4z + 5 = 0.
Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.
The point at which the normal to the curve y = `"x" + 1/"x", "x" > 0` is perpendicular to the line 3x – 4y – 7 = 0 is:
Let A be the foot of the perpendicular from focus P of hyperbola `x^2/a^2 - y^2/b^2 = 1` on the line bx – ay = 0 and let C be the centre of hyperbola. Then the area of the rectangle whose sides are equal to that of PA and CA is,
A unit vector perpendicular to the plane ABC, where A, B and C are respectively the points (3, –1, 2), (1, –1, –3) and (4, –3, 1), is
The coordinates of the foot of the perpendicular drawn from the point A(1, 0, 3) to the join of the points B(4, 7, 1) and C(3, 5, 3) are
