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.
Name the octants in which the following points lie:
(4, –3, 5)
Name the octants in which the following points lie:
(–5, –4, 7)
Find the image of:
(–4, 0, 0) in the xy-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.
Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, –4).
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.
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).
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.
Verify the following:
(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.
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.
Write the length of the perpendicular drawn from the point P(3, 5, 12) on x-axis.
What is the locus of a point for which y = 0, z = 0?
Find the point on y-axis which is at a distance of \[\sqrt{10}\] units from the point (1, 2, 3).
Find the point on x-axis which is equidistant from the points A (3, 2, 2) and B (5, 5, 4).
What is the locus of a point for which y = 0, z = 0?
The perpendicular distance of the point P (6, 7, 8) from xy - plane is
The length of the perpendicular drawn from the point P (3, 4, 5) on y-axis is
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)
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 α, β, γ are the angles that a line makes with the positive direction of x, y, z axis, respectively, then the direction cosines of the 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 ______.
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 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 equation of the plane which is perpendicular to the plane 5x + 3y + 6z + 8 = 0 and which contains the line of intersection of the planes x + 2y + 3z – 4 = 0 and 2x + y – z + 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
The plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(α) with x-axis. The value of α is equal to ______.
The direction cosines of the vector `(2hati + 2hatj - hatk)` are ______.
The intercepts made by the plane 2x – 3y + 5z +4 = 0 on the co-ordinate axis are `-2, 4/3, - 4/5`.
The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is `vecr = (5hati - 4hatj + 6hatk) + lambda(3hati + 7hatj - 2hatk)`.
