Advertisements
Advertisements
Question
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
Options
2
3
6
7
Advertisements
Solution
3
Suppose d is the diameter of the sphere. Then
\[d^2 = \left( - 1 - 3 \right)^2 + \left( 2 - 4 \right)^2 + \left( 3 + 1 \right)^2 \]
\[ \Rightarrow d^2 = \left( - 4 \right)^2 + \left( - 2 \right)^2 + \left( 4 \right)^2 \]
\[ \Rightarrow d^2 = 16 + 4 + 16\]
\[ \Rightarrow d^2 = 36\]
\[ \Rightarrow d = 6\]
Hence, radius of the sphere is 3 units.
APPEARS IN
RELATED QUESTIONS
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).
Coordinate planes divide the space into ______ octants.
Name the octants in which the following points lie: (5, 2, 3)
Name the octants in which the following points lie:
(–7, 2 – 5)
Find the image of:
(–5, 0, 3) in the xz-plane.
Find the distances of the point P(–4, 3, 5) from the coordinate axes.
Determine the points in zx-plane are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).
Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, –4).
Find the point on y-axis which is equidistant from the points (3, 1, 2) and (5, 5, 2).
Prove that the point A(1, 3, 0), B(–5, 5, 2), C(–9, –1, 2) and D(–3, –3, 0) taken in order are the vertices of a parallelogram. Also, show that ABCD is not a rectangle.
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 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.
Verify the following:
(5, –1, 1), (7, –4,7), (1, –6,10) and (–1, – 3,4) are the vertices of a rhombus.
Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Show that the points A(1, 2, 3), B(–1, –2, –1), C(2, 3, 2) and D(4, 7, 6) are the vertices of a parallelogram ABCD, but not a rectangle.
The ratio in which the line joining the points (a, b, c) and (–a, –c, –b) is divided by the xy-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 perpendicular distance of the point P (6, 7, 8) from xy - plane is
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.
A plane meets the co-ordinates axis in A, B, C such that the centroid of the ∆ABC is the point (α, β, γ). Show that the equation of the plane is `x/alpha + y/beta + z/γ` = 3
If a line makes angles `pi/2, 3/4 pi` and `pi/4` with x, y, z axis, respectively, then its direction cosines are ______.
Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.
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 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 vector equation of the line through the points (3, 4, –7) and (1, –1, 6) is ______.
The cartesian equation of the plane `vecr * (hati + hatj - hatk)` is ______.
The angle between the planes `vecr.(2hati - 3hatj + hatk)` = 1 and `vecr.(hati - hatj)` = 4 is `cos^-1((-5)/sqrt(58))`.
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.
