Advertisements
Advertisements
Question
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.
Advertisements
Solution
Vertices A and C are (3, –1, 2), (−1, 1, 2) respectively.
Coordinates of the midpoint P between A and C `((3 - 1)/2, (-1 + 1)/2, (2 + 2)/2)` = (1, 0, 2)
Let the coordinates of point D be (x, y, z) and the coordinates of point B be (1, 2, −4).
∴ Mid point of DB `(("x" + 1)/2, ("y" + 2)/2, ("z" - 4)/2)`
The diagonals of a parallelogram divide each other into 2 equal parts.
Therefore, `("x" + 1)/2 = 1, ("y" + 2)/2 = 0, ("z" - 4)/2 = 2`
Hence, the coordinates of point D are (1, –2, 8).
APPEARS IN
RELATED QUESTIONS
Coordinate planes divide the space into ______ octants.
Name the octants in which the following points lie:
(–5, –4, 7)
Find the image of:
(–5, 0, 3) in the xz-plane.
Find the distances of the point P(–4, 3, 5) from the coordinate axes.
Determine the points in zx-plane are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).
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:
(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.
Find the ratio in which the sphere x2 + y2 + z2 = 504 divides the line joining the points (12, –4, 8) and (27, –9, 18).
Write the distance of the point P (2, 3,5) from the xy-plane.
Write the coordinates of the foot of the perpendicular from the point (1, 2, 3) on y-axis.
What is the locus of a point for which y = 0, z = 0?
Find the point on x-axis which is equidistant from the points A (3, 2, 2) and B (5, 5, 4).
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
The perpendicular distance of the point P(3, 3,4) from the x-axis is
The length of the perpendicular drawn from the point P(a, b, c) from z-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.
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 ______.
If a line makes angles `pi/2, 3/4 pi` and `pi/4` with x, y, z axis, respectively, then its direction cosines are ______.
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 angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 0
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.
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 straight lines whose direction cosines are given by 2l + 2m – n = 0 and mn + nl + lm = 0 are at right angles.
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 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.
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.
