Advertisements
Advertisements
प्रश्न
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.
पर्याय
True
False
Advertisements
उत्तर
This statement is True.
Explanation:

From the figure normal to the plane is `vecn = vec(OP) = 5hati - 3hatj - 2hatk`
Plane passing through the point P(5, – 3, – 2).
∴ Equation of the plane is `5(x - 5) - 3(y + 3) - 2(z + 2)` = 0
or `5x - 3y - 2z` = 38
APPEARS IN
संबंधित प्रश्न
Coordinate planes divide the space into ______ octants.
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:
(4, –3, 5)
Name the octants in which the following points lie:
(–5, –4, 7)
Name the octants in which the following points lie:
(–7, 2 – 5)
Find the image of:
(–5, 4, –3) in the xz-plane.
Planes are drawn parallel to the coordinate planes through the points (3, 0, –1) and (–2, 5, 4). Find the lengths of the edges of the parallelepiped so formed.
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.
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 coordinates of the point which is equidistant from the four points O(0, 0, 0), A(2, 0, 0), B(0, 3, 0) and C(0, 0, 8).
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:
(0, 7, 10), (–1, 6, 6) and (–4, 9, –6) are vertices of a right-angled 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.
What is the locus of a point for which y = 0, z = 0?
The ratio in which the line joining the points (a, b, c) and (–a, –c, –b) is divided by the xy-plane is
The coordinates of the foot of the perpendicular from a point P(6,7, 8) on x - axis are
The length of the perpendicular drawn from the point P (3, 4, 5) on y-axis 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.
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).
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 ______.
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.
If a variable line in two adjacent positions has direction cosines l, m, n and l + δl, m + δm, n + δn, show that the small angle δθ between the two positions is given by δθ2 = δl2 + δm2 + δn2
O is the origin and A is (a, b, c). Find the direction cosines of the line OA and the equation of plane through A at right angle to OA.
Find the length and the foot of perpendicular from the point `(1, 3/2, 2)` to the plane 2x – 2y + 4z + 5 = 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
Show that the points `(hati - hatj + 3hatk)` and `3(hati + hatj + hatk)` are equidistant from the plane `vecr * (5hati + 2hatj - 7hatk) + 9` = 0 and lies on opposite side of it.
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 vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is ______.
The vector equation of the line through the points (3, 4, –7) and (1, –1, 6) is ______.
The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is `vecr = (5hati - 4hatj + 6hatk) + lambda(3hati + 7hatj - 2hatk)`.
