Advertisements
Advertisements
प्रश्न
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 ______.
पर्याय
9 sq.units
18 sq.units
27 sq.units
81 sq.units
उत्तर
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 9 sq.units.
Explanation:
We have, `vec(AB) = (2 - 0)hati + (3 - 4)hatj + (-1 -1)hatk = 2hati - hatj - 2hatk`
`vec(BC) = (4 - 2)hati + (5 - 3)hatj + (0 + 1)hatk = 2hati + 2hatj + hatk`
`vec(CD) = (2 - 4)hati + (6 - 5)hatj + (2 - 0)hatk = -2hati + hatj + 2hatk`
`vec(DA) = (0 - 2)hati + (4 - 6)hatj + (1 - 2)hatk = -2hati - 2hatj - hatk`
∴ Area of quadrilateral ABCD = `|vec(AB) xx vec(BC)|`
= `|(hati, hatj, hatk),(2, -1, -2),(2, 2, 1)|`
= `|hati(-1 + 4) - hatj(2 - 4) + hatk(4 + 2)|`
= `|3hati - 6hatj + 6hatk|`
= `sqrt(9 + 36 + 36)`
= 9 sq.units
APPEARS IN
संबंधित प्रश्न
Name the octants in which the following points lie: (5, 2, 3)
Name the octants in which the following points lie:
(–5, –4, 7)
Find the image of:
(–5, 4, –3) in the xz-plane.
A cube of side 5 has one vertex at the point (1, 0, –1), and the three edges from this vertex are, respectively, parallel to the negative x and y axes and positive z-axis. Find the coordinates of the other vertices of the cube.
Determine the points in zx-plane are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).
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.
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 points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
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.
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(3, 4, 5) from z-axis.
Write the length of the perpendicular drawn from the point P(3, 5, 12) on x-axis.
Find the ratio in which the line segment joining the points (2, 4,5) and (3, −5, 4) is divided by the yz-plane.
The ratio in which the line joining (2, 4, 5) and (3, 5, –9) is divided by the yz-plane is
Let (3, 4, –1) and (–1, 2, 3) be the end points of a diameter of a sphere. Then, the radius of the sphere is equal to
XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio
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 `pi/2, 3/4 pi` and `pi/4` with x, y, z axis, respectively, then its direction cosines are ______.
Find the equation of a plane which bisects perpendicularly the line joining the points A(2, 3, 4) and B(4, 5, 8) at right angles.
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.
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.
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 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.
