Advertisements
Advertisements
प्रश्न
Name the octants in which the following points lie:
(2, –5, –7)
Advertisements
उत्तर
The x-coordinate, the y-coordinate and the z-coordinate of the point (2, −5, −7) are positive, negative and negative, respectively.
Therefore, this point lies in XOY'Z' octant.
APPEARS IN
संबंधित प्रश्न
Name the octants in which the following points lie:
(1, 2, 3), (4, –2, 3), (4, –2, –5), (4, 2, –5), (–4, 2, –5), (–4, 2, 5),
(–3, –1, 6), (2, –4, –7).
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:
(7, 4, –3)
Name the octants in which the following points lie:
(–7, 2 – 5)
Find the image of:
(–5, 4, –3) in the xz-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.
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).
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.
Verify the following:
(0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles triangle.
Find the locus of the point, the sum of whose distances from the points A(4, 0, 0) and B(–4, 0, 0) is equal to 10.
Write the length of the perpendicular drawn from the point P(3, 5, 12) on x-axis.
The ratio in which the line joining (2, 4, 5) and (3, 5, –9) is divided by the yz-plane is
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 x-coordinate of a point on the line joining the points Q(2, 2, 1) and R(5, 1, –2) is 4. Find its z-coordinate.
Find the coordinates of the point where the line through (3, – 4, – 5) and (2, –3, 1) crosses the plane passing through three points (2, 2, 1), (3, 0, 1) and (4, –1, 0)
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).
If a line makes angles `pi/2, 3/4 pi` and `pi/4` with x, y, z axis, respectively, then its direction cosines 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 equations of the two lines through the origin which intersect the line `(x - 3)/2 = (y - 3)/1 = z/1` at angles of `pi/3` each.
Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 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 length and the foot of perpendicular from the point `(1, 3/2, 2)` to the plane 2x – 2y + 4z + 5 = 0.
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 plane ax + by = 0 is rotated about its line of intersection with the plane z = 0 through an angle α. Prove that the equation of the plane in its new position is ax + by `+- (sqrt(a^2 + b^2) tan alpha)z ` = 0
If l1, m1, n1 ; l2, m2, n2 ; l3, m3, n3 are the direction cosines of three mutually perpendicular lines, prove that the line whose direction cosines are proportional to l1 + l2 + l3, m1 + m2 + m3, n1 + n2 + n3 makes equal angles with them.
If the directions cosines of a line are k, k, k, then ______.
The sine of the angle between the straight line `(x - 2)/3 = (y - 3)/4 = (z - 4)/5` and the plane 2x – 2y + z = 5 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 planes `vecr.(2hati - 3hatj + hatk)` = 1 and `vecr.(hati - hatj)` = 4 is `cos^-1((-5)/sqrt(58))`.
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.
