Advertisements
Advertisements
Question
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.
Advertisements
Solution
Given point is (3, 0, 1) and the equation of planes are
x + 2y = 0 …...(i)
And 3y – z = 0 .....(ii)
Equation of any line l passing through (3, 0, 1) is l: `(x – 3)/a = (y – 0)/b = (z – 1)/c`
Now, the direction ratios of the normal to plane (i) and plane (ii) are (1, 2, 0) and (0, 3, 1).
As the line is parallel to both the planes, we have
1.a + 2.b + 0.c = 0
⇒ a + 2b + 0c = 0
And 0.a + 3.b – 1.c = 0
⇒ 0a + 3b – c = 0
So, `a/(-2 - 0) = (-b)/(-1 - 0) = c/(3 - 0) = lambda`
∴ `a = -2lambda, b = lambda, c = 3lambda`
So, the equation of line is `(x - 3)/(-2lambda) = y/lambda = (z - 1)/(3lambda)`
Thus, the required equation is `(x - 3)/(-2) = y/1 = (z - 1)/3`
or In vetor form is `(x - 3)hati + yhatj + (z - 1)hatk = lambda(-2hati + hatj + 3hatk)`
APPEARS IN
RELATED QUESTIONS
The x-axis and y-axis taken together determine a plane known as_______.
Name the octants in which the following points lie:
(–5, –3, –2)
Name the octants in which the following points lie:
(–7, 2 – 5)
Find the image of:
(5, 2, –7) in the xy-plane.
Find the image of:
(–5, 0, 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.
Determine the point on z-axis which is equidistant from the points (1, 5, 7) and (5, 1, –4).
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.
Show that the points (a, b, c), (b, c, a) and (c, a, b) are the vertices of an equilateral triangle.
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.
The coordinates of the mid-points of sides AB, BC and CA of △ABC are D(1, 2, −3), E(3, 0,1) and F(−1, 1, −4) respectively. Write the coordinates of its centroid.
Write the coordinates of the foot of the perpendicular from the point (1, 2, 3) on y-axis.
Write the length of the perpendicular drawn from the point P(3, 5, 12) on x-axis.
What is the locus of a point for which y = 0, z = 0?
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 the points (a, b, c) and (–a, –c, –b) is divided by the xy-plane is
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
Find the direction cosines of the line passing through the points P(2, 3, 5) and Q(–1, 2, 4).
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 a line makes angles α, β, γ with the positive directions of the coordinate axes, then the value of sin2α + sin2β + sin2γ 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
Find the equation of the plane through the points (2, 1, –1) and (–1, 3, 4), and perpendicular to the plane x – 2y + 4z = 10.
If the directions cosines of a line are k, k, k, then ______.
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 vector equation of the line `(x - 5)/3 = (y + 4)/7 = (z - 6)/2` 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 angle between the planes `vecr.(2hati - 3hatj + hatk)` = 1 and `vecr.(hati - hatj)` = 4 is `cos^-1((-5)/sqrt(58))`.
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.
