Advertisements
Advertisements
प्रश्न
If the direction ratios of a line are 1, 1, 2, find the direction cosines of the line.
Advertisements
उत्तर
The direction cosines are given by
`l = a/sqrt(a^2 + b^2 + c^2)`
`m = b/sqrt(a^2 + b^2 + c^2)`
`n = c/sqrt(a^2 + b^2 + c^2)`
Here a, b, c are 1, 1, 2, respectively
Therefore, `l = 1/sqrt(1^2 + 1^2 + 2)`
`m = 1/sqrt(1^2 + 1^2 + 2^2)`
`n = 2/sqrt(1^2 + 1^2 + 2^2)`
i.e., `l = 1/sqrt(6)`
`m = 1/sqrt(6)`
`n = 2/sqrt(6)`
i.e. `+-(1/sqrt(6), 1/sqrt(6), 2/sqrt(6))` are D.C’s of the line.
APPEARS IN
संबंधित प्रश्न
The x-axis and y-axis taken together determine a plane known as_______.
Coordinate planes divide the space into ______ octants.
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:
(7, 4, –3)
Name the octants in which the following points lie:
(2, –5, –7)
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.
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.
Find the locus of P if PA2 + PB2 = 2k2, where A and B are the points (3, 4, 5) and (–1, 3, –7).
Verify the following:
(0, 7, 10), (–1, 6, 6) and (–4, 9, –6) are vertices of a right-angled triangle.
Verify the following:
(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.
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.
Show that the plane ax + by + cz + d = 0 divides the line joining the points (x1, y1, z1) and (x2, y2, z2) in the ratio \[- \frac{a x_1 + b y_1 + c z_1 + d}{a x_2 + b y_2 + c z_2 + d}\]
Write the coordinates of the foot of the perpendicular from the point (1, 2, 3) on y-axis.
The ratio in which the line joining (2, 4, 5) and (3, 5, –9) is divided by the yz-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
What is the locus of a point for which y = 0, z = 0?
Find the direction cosines of the line passing through the points P(2, 3, 5) and Q(–1, 2, 4).
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)
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 ______.
If a line makes angles α, β, γ with the positive directions of the coordinate axes, then the value of sin2α + sin2β + sin2γ is ______.
Find the equation of the plane through the points (2, 1, 0), (3, –2, –2) and (3, 1, 7).
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.
Two systems of rectangular axis have the same origin. If a plane cuts them at distances a, b, c and a′, b′, c′, respectively, from the origin, prove that
`1/a^2 + 1/b^2 + 1/c^2 = 1/(a"'"^2) + 1/(b"'"^2) + 1/(c"'"^2)`
Find the length and the foot of perpendicular from the point `(1, 3/2, 2)` to the plane 2x – 2y + 4z + 5 = 0.
The locus represented by xy + yz = 0 is ______.
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 cartesian equation of the plane `vecr * (hati + hatj - hatk)` is ______.
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)))`.
The angle between the planes `vecr.(2hati - 3hatj + hatk)` = 1 and `vecr.(hati - hatj)` = 4 is `cos^-1((-5)/sqrt(58))`.
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.
