Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Given coordinates are A(2, 3, 4) and B(4, 5, 8)
Now, the coordinates of the mid-point C are `((2 + 4)/2, (3 + 5)/2, (4 + 8)/2)` = (3, 4, 6)
And, the direction ratios of the normal to the plane = direction ratios of AB
= 4 – 2, 5 – 3, 8 – 4
= (2, 2, 4)
Equation of the plane is
a(x – x1) + b(y – y1) + c(z – z1) = 0
2(x – 3) + 2(y – 4) + 4(z – 6) = 0
2x – 6 + 2y – 8 + 4z – 24 = 0
2x + 2y + 4z = 38
x + y + 2z = 19
Thus, the required equation of plane is x + y + 2z = 1
APPEARS IN
संबंधित प्रश्न
Name the octants in which the following points lie:
(4, –3, 5)
Name the octants in which the following points lie:
(–5, –3, –2)
Name the octants in which the following points lie:
(2, –5, –7)
Name the octants in which the following points lie:
(–7, 2 – 5)
Find the image of:
(–5, 4, –3) in the xz-plane.
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.
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.
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).
Verify the following:
(0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles triangle.
Find the locus of the points which are equidistant from the points (1, 2, 3) and (3, 2, –1).
Write the distance of the point P (2, 3,5) from the xy-plane.
Write the length of the perpendicular drawn from the point P(3, 5, 12) on x-axis.
The ratio in which the line joining (2, 4, 5) and (3, 5, –9) is divided by the yz-plane is
XOZ-plane divides the join of (2, 3, 1) and (6, 7, 1) in the ratio
The perpendicular distance of the point P (6, 7, 8) from xy - plane is
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.
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)
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 length and the foot of perpendicular from the point `(1, 3/2, 2)` to the plane 2x – 2y + 4z + 5 = 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.
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.
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 plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(α) with x-axis. The value of α is equal to ______.
The vector equation of the line through the points (3, 4, –7) and (1, –1, 6) is ______.
The cartesian equation of the plane `vecr * (hati + hatj - hatk)` is ______.
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)))`.
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.
