Advertisements
Advertisements
Question
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.
Advertisements
Solution
The given planes are
P1: 5x + 3y + 6z + 8 = 0
P2: x + 2y + 3z – 4 = 0
P3: 2x + y – z + 5 = 0
Now, the equation of the plane passing through the line of intersection of P1 and P3 is
(x + 2y + 3z – 4) + λ(2x + y – z + 5) = 0
(1 + 2λ)x + (2 + λ)y + (3 – λ)z – 4 + 5λ = 0 ......(i)
From the question its understood that plane (i) is perpendicular to P1, then
5(1 + 2λ) + 3(2 + λ) + 6(3 – λ) = 0
5 + 10λ + 6 + 3λ + 18 – 6λ = 0
7λ + 29 = 0
λ = `(-29)/7`
Putting the value of the equation (i), we get
`[1 + 2((-29)/7)]x + [2 - 29/7]y + [3 + 29/7]z - 4 + 5((-29)/7)` = 0
⇒ `(-15)/7x - 15/7y + 50/7z - 4 - 145/7` = 0
–15x – 15y + 50z – 28 – 145 = 0
–15x – 15y + 50z – 173 = 0
⇒ 51x + 15y – 50z + 173 = 0
Thus, the required equation of plane is 51x + 15y – 50z + 173 = 0.
APPEARS IN
RELATED QUESTIONS
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.
If the origin is the centroid of the triangle PQR with vertices P (2a, 2, 6), Q (–4, 3b, –10) and R (8, 14, 2c), then find the values of a, b and c.
Name the octants in which the following points lie:
(7, 4, –3)
Find the image of:
(5, 2, –7) in the xy-plane.
Find the image of:
(–4, 0, 0) in the xy-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.
The coordinates of a point are (3, –2, 5). Write down the coordinates of seven points such that the absolute values of their coordinates are the same as those of the coordinates of the given point.
Determine the points in zx-plane are equidistant from the points A(1, –1, 0), B(2, 1, 2) and C(3, 2, –1).
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).
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.
Verify the following:
(0, 7, –10), (1, 6, –6) and (4, 9, –6) are vertices of an isosceles triangle.
Verify the following:
(–1, 2, 1), (1, –2, 5), (4, –7, 8) and (2, –3, 4) are vertices of a parallelogram.
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 plane ax + by + cz + d = 0 divides the line joining the points (x1, y1, z1) and (x2, y2, z2) in the ratio \[- \frac{a x_1 + b y_1 + c z_1 + d}{a x_2 + b y_2 + c z_2 + d}\]
What is the locus of a point for which y = 0, z = 0?
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 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.
Find the direction cosines of the line passing through the points P(2, 3, 5) and Q(–1, 2, 4).
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 ______.
If a line makes an angle of `pi/4` with each of y and z axis, then the angle which it makes with x-axis is ______.
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.
Find the angle between the lines whose direction cosines are given by the equations l + m + n = 0, l2 + m2 – n2 = 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.
If the directions cosines of a line are k, k, k, then ______.
The direction cosines of the vector `(2hati + 2hatj - hatk)` are ______.
The vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` 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`.
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.
