Advertisements
Advertisements
प्रश्न
XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio
पर्याय
3 : 7
2 : 7
–3 : 7
–2 : 7
Advertisements
उत्तर
−3:7
Let A\[\equiv\](2, 3, 1) and B\[\equiv\]Let the line joining A and B be divided by the xz-plane at point P in the ratio\[\lambda: 1\]
Then, we have,
P\[\equiv \left( \frac{6\lambda + 2}{\lambda + 1}, \frac{7\lambda + 3}{\lambda + 1}, \frac{\lambda + 1}{\lambda + 1} \right)\]
Since P lies on the xz-plane, the y-coordinate of P will be zero.
\[\therefore \frac{7\lambda + 3}{\lambda + 1} = 0\]
\[ \Rightarrow 7\lambda + 3 = 0\]
\[ \Rightarrow \lambda = \frac{- 3}{7}\]
Hence, the xz-plane divides AB in the ratio \[-\]3 : 7
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.
If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find the values of a, b and c.
Name the octants in which the following points lie: (5, 2, 3)
Name the octants in which the following points lie:
(–5, 4, 3)
Name the octants in which the following points lie:
(–5, –3, –2)
Name the octants in which the following points lie:
(2, –5, –7)
Name the octants in which the following points lie:
(–7, 2 – 5)
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.
Planes are drawn through the points (5, 0, 2) and (3, –2, 5) parallel to the coordinate planes. Find the lengths of the edges of the rectangular parallelepiped so formed.
Determine the points in zx-plane are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).
Find the points on z-axis which are at a distance \[\sqrt{21}\]from the point (1, 2, 3).
Find the locus of P if PA2 + PB2 = 2k2, where A and B are the points (3, 4, 5) and (–1, 3, –7).
Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.
Are the points A(3, 6, 9), B(10, 20, 30) and C(25, –41, 5), the vertices of a right-angled triangle?
Verify the following:
(5, –1, 1), (7, –4,7), (1, –6,10) and (–1, – 3,4) are the vertices of a rhombus.
Find the ratio in which the sphere x2 + y2 + z2 = 504 divides the line joining the points (12, –4, 8) and (27, –9, 18).
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.
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
What is the locus of a point for which y = 0, z = 0?
The coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz- plane are
The perpendicular distance of the point P(3, 3,4) from the x-axis is
If the direction ratios of a line are 1, 1, 2, find the direction cosines of the line.
If a line makes an angle of 30°, 60°, 90° with the positive direction of x, y, z-axes, respectively, then find its direction cosines.
Find the co-ordinates of the foot of perpendicular drawn from the point A(1, 8, 4) to the line joining the points B(0, –1, 3) and C(2, –3, –1).
The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by ______.
A line makes equal angles with co-ordinate axis. Direction cosines of this line are ______.
Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0.
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 equations of the line passing through the point (3,0,1) and parallel to the planes x + 2y = 0 and 3y – z = 0.
The direction cosines of the vector `(2hati + 2hatj - hatk)` are ______.
The vector equation of the line through the points (3, 4, –7) and (1, –1, 6) is ______.
The line `vecr = 2hati - 3hatj - hatk + lambda(hati - hatj + 2hatk)` lies in the plane `vecr.(3hati + hatj - hatk) + 2` = 0.
The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is `vecr = (5hati - 4hatj + 6hatk) + lambda(3hati + 7hatj - 2hatk)`.
If the foot of perpendicular drawn from the origin to a plane is (5, – 3, – 2), then the equation of plane is `vecr.(5hati - 3hatj - 2hatk)` = 38.
