Advertisements
Advertisements
प्रश्न
Find the equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 5 = 0.
Advertisements
उत्तर
The given planes are
P1: 5x + 3y + 6z + 8 = 0
P2: x + 2y + 3z – 4 = 0
P3: 2x + y – z + 5 = 0
Now, the equation of the plane passing through the line of intersection of P1 and P3 is
(x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0
(1 + 2λ)x + (2 + λ)y + (3 – λ)z – 4 + 5λ = 0 ......(i)
From the question its understood that plane (i) is perpendicular to P1, then
5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0
5 + 10λ + 6 + 3λ + 18 – 6λ = 0
7λ + 29 = 0
λ = `(-29)/7`
Putting the value of the equation (i), we get
`[1 + 2((-29)/7)]x + [2 - 29/7]y + [3 + 29/7]z - 4 + 5((-29)/7)` = 0
⇒ `(-15)/7x - 15/7y + 50/7z - 4 - 145/7` = 0
–15x – 15y + 50z – 28 – 145 = 0
–15x – 15y + 50z – 173 = 0
⇒ 51x + 15y – 50z + 173 = 0
Thus, the required equation of plane is 51x + 15y – 50z + 173 = 0.
APPEARS IN
संबंधित प्रश्न
Three vertices of a parallelogram ABCD are A (3, –1, 2), B (1, 2, –4) and C (–1, 1, 2). Find the coordinates of the fourth vertex.
Name the octants in which the following points lie: (5, 2, 3)
Name the octants in which the following points lie:
(7, 4, –3)
Name the octants in which the following points lie:
(–5, –4, 7)
Name the octants in which the following points lie:
(–5, –3, –2)
Name the octants in which the following points lie:
(–7, 2 – 5)
Find the image of:
(–2, 3, 4) in the yz-plane.
Find the image of:
(–4, 0, 0) in the xy-plane.
A cube of side 5 has one vertex at the point (1, 0, –1), and the three edges from this vertex are, respectively, parallel to the negative x and y axes and positive z-axis. Find the coordinates of the other vertices of the cube.
Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).
Prove that the triangle formed by joining the three points whose coordinates are (1, 2, 3), (2, 3, 1) and (3, 1, 2) is an equilateral triangle.
Show that the points A(3, 3, 3), B(0, 6, 3), C(1, 7, 7) and D(4, 4, 7) are the vertices of a square.
If A(–2, 2, 3) and B(13, –3, 13) are two points.
Find the locus of a point P which moves in such a way the 3PA = 2PB.
Find the equation of the set of the points P such that its distances from the points A(3, 4, –5) and B(–2, 1, 4) are equal.
Write the distance of the point P(3, 4, 5) from z-axis.
What is the locus of a point for which y = 0, z = 0?
Find the ratio in which the line segment joining the points (2, 4,5) and (3, −5, 4) is divided by the yz-plane.
The ratio in which the line joining the points (a, b, c) and (–a, –c, –b) is divided by the xy-plane is
Let (3, 4, –1) and (–1, 2, 3) be the end points of a diameter of a sphere. Then, the radius of the sphere is equal to
XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio
The coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz- plane are
The coordinates of the foot of the perpendicular from a point P(6,7, 8) on x - axis are
The perpendicular distance of the point P(3, 3,4) from the x-axis is
The length of the perpendicular drawn from the point P(a, b, c) from z-axis is
If the direction ratios of a line are 1, 1, 2, find the direction cosines of the line.
The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by ______.
If a line makes angles `pi/2, 3/4 pi` and `pi/4` with x, y, z axis, respectively, then its direction cosines 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
Find the equation of the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7).
Find the foot of perpendicular from the point (2,3,–8) to the line `(4 - x)/2 = y/6 = (1 - z)/3`. Also, find the perpendicular distance from the given point to the line.
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 locus represented by xy + yz = 0 is ______.
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)hati + 2/sqrt(14)hatj + 3/sqrt(14)hatk`.
The angle between the line `vecr = (5hati - hatj - 4hatk) + lambda(2hati - hatj + hatk)` and the plane `vec.(3hati - 4hatj - hatk)` + 5 = 0 is `sin^-1(5/(2sqrt(91)))`.
