Advertisements
Advertisements
प्रश्न
What is the locus of a point for which y = 0, z = 0?
Advertisements
उत्तर
We know that on x - axis both y = 0, z = 0.
Hence, the locus of a point for which y = 0, z = 0 is x - axis.
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).
Name the octants in which the following points lie:
(–5, 4, 3)
Name the octants in which the following points lie:
(4, –3, 5)
Name the octants in which the following points lie:
(7, 4, –3)
Name the octants in which the following points lie:
(–5, –3, –2)
The coordinates of a point are (3, –2, 5). Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.
Find the points on z-axis which are at a distance \[\sqrt{21}\]from the point (1, 2, 3).
Prove that the triangle formed by joining the three points whose coordinates are (1, 2, 3), (2, 3, 1) and (3, 1, 2) is an equilateral triangle.
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).
Are the points A(3, 6, 9), B(10, 20, 30) and C(25, –41, 5), the vertices of a right-angled triangle?
Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
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).
XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio
What is the locus of a point for which y = 0, z = 0?
The length of the perpendicular drawn from the point P (3, 4, 5) on y-axis is
If the direction ratios of a line are 1, 1, 2, find the direction cosines of the line.
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 α, β, γ with the positive directions of the coordinate axes, then the value of sin2α + sin2β + sin2γ is ______.
If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane
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.
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 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.
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.
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 area of the quadrilateral ABCD, where A(0, 4, 1), B(2, 3, –1), C(4, 5, 0) and D(2, 6, 2), is equal to ______.
The locus represented by xy + yz = 0 is ______.
The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` is ______.
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 line `vecr = (5hati - hatj - 4hatk) + lambda(2hati - hatj + hatk)` and the plane `vec.(3hati - 4hatj - hatk)` + 5 = 0 is `sin^-1(5/(2sqrt(91)))`.
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.
