Advertisements
Advertisements
Question
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.
Advertisements
Solution
Let P\[\equiv\](3, 0, −1), Q\[\equiv\](−2, 5, 4)
PE = Distance between the parallel planes ABCP and FQDE
=\[\left| 4 + 1 \right| = 5\]
(These planes are perpendicular to the z-axis)
PA = Distance between the parallel planes ABQF and PCDE
= \[\left| - 2 - 3 \right| = 5\]
(These planes are perpendicular to the x-axis)
Similarly, PC =\[\left| 5 - 0 \right| = 5\]
Thus, the length of the edges of the parallelepiped are 5, 5 and 5
APPEARS IN
RELATED QUESTIONS
Name the octants in which the following points lie:
(4, –3, 5)
Find the image of:
(–5, 4, –3) in the xz-plane.
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.
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.
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).
Are the points A(3, 6, 9), B(10, 20, 30) and C(25, –41, 5), the vertices of a right-angled triangle?
Verify the following:
(0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles 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.
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 distance of the point P (2, 3,5) from the xy-plane.
Write the distance of the point P(3, 4, 5) from z-axis.
The ratio in which the line joining (2, 4, 5) and (3, 5, –9) is divided by the yz-plane is
The ratio in which the line joining the points (a, b, c) and (–a, –c, –b) is divided by the xy-plane is
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 coordinates of the foot of the perpendicular drawn from the point P(3, 4, 5) on the yz- plane are
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
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 image of the point having position vector `hati + 3hatj + 4hatk` in the plane `hatr * (2hati - hatj + hatk)` + 3 = 0.
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 ______.
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 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 straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.
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 cartesian equation of the plane `vecr * (hati + hatj - hatk)` 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 intercepts made by the plane 2x – 3y + 5z +4 = 0 on the co-ordinate axis are `-2, 4/3, - 4/5`.
The angle between the planes `vecr.(2hati - 3hatj + hatk)` = 1 and `vecr.(hati - hatj)` = 4 is `cos^-1((-5)/sqrt(58))`.
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)`.
