Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर

Clearly, PBEC and QDAF are the planes parallel to the yz-plane such that their distances from the yz-plane are 5 and 3, respectively.
\[\therefore\]PA = Distance between planes PBEC and QDAF
= 5\[-\]3
= 2
PB is the distance between planes PAFC and BDQE that are parallel to the zx-plane and are at distances 0 and\[-\]2,respectively, from the zx-plane.
\[\therefore\]PB = 0
\[-\](\[-\]2)
= 2
PC is the distance between parallel planes PBDA and CEQF that are at distances 2 and 5, respectively, from the xy-plane.
\[\therefore\]PC = 2\[-\]5
=\[-\]3
APPEARS IN
संबंधित प्रश्न
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:
(7, 4, –3)
Name the octants in which the following points lie:
(–5, –4, 7)
Name the octants in which the following points lie:
(–5, –3, –2)
Name the octants in which the following points lie:
(2, –5, –7)
Find the image of:
(5, 2, –7) in the xy-plane.
Find the image of:
(–4, 0, 0) in the xy-plane.
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 an isosceles triangle.
Verify the following:
(5, –1, 1), (7, –4,7), (1, –6,10) and (–1, – 3,4) are the vertices of a rhombus.
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.
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 length of the perpendicular drawn from the point P(3, 5, 12) on x-axis.
The coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz- plane 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)
The coordinates of the foot of the perpendicular drawn from the point (2, 5, 7) on the x-axis are given by ______.
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 ______.
A line makes equal angles with co-ordinate axis. Direction cosines of this line are ______.
If a line makes angles `pi/2, 3/4 pi` and `pi/4` with x, y, z axis, respectively, then its direction cosines are ______.
Prove that the lines x = py + q, z = ry + s and x = p′y + q′, z = r′y + s′ are perpendicular if pp′ + rr′ + 1 = 0.
Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 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.
Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.
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 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 vector equation of the line through the points (3, 4, –7) and (1, –1, 6) is ______.
The intercepts made by the plane 2x – 3y + 5z +4 = 0 on the co-ordinate axis are `-2, 4/3, - 4/5`.
The line `vecr = 2hati - 3hatj - hatk + lambda(hati - hatj + 2hatk)` lies in the plane `vecr.(3hati + hatj - hatk) + 2` = 0.
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.
