Advertisements
Advertisements
प्रश्न
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
उत्तर
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"`
Direction ratios of the normal to plane (i) and plane (ii) are (1, 2, 0) and (0, 3, – 1)
Since the line is parallel to both the planes.
∴ 1.a + 2.b + 0.c = 0
⇒ a + 2b + 0c = 0
And 0.a + 3.b – 1.c = 0
⇒ 0.a + 3b – c = 0
So `"a"/(-2 - 0) = (-"b")/(-1 - 0) = "c"/(3 - 0) = lambda`
∴ a = – 2λ, b = λ, c = 3λ
So, the equation of line is `(x - 3)/(-2lambda) = y/lambda = (z - 1)/(3lambda)`
Hence, the required equation is `(x - 3)/(-2) = y/1 = (z - 1)/3`
or in vector form is `(x - 3)hat"i" + yhat"j" + (z - 1)hat"k" = lambda(-2hat"i" + hat"j" + 3hat"k")`
APPEARS IN
संबंधित प्रश्न
Find the vector equation of a plane which is at a distance of 7 units from the origin and normal to the vector.`3hati + 5hatj - 6hatk`
Find the equations of the planes that passes through three points.
(1, 1, 0), (1, 2, 1), (−2, 2, −1)
Find the equation of the plane passing through the point (−1, 3, 2) and perpendicular to each of the planes x + 2y + 3z = 5 and 3x + 3y + z = 0.
If the points (1, 1, p) and (−3, 0, 1) be equidistant from the plane `vecr.(3hati + 4hatj - 12hatk)+ 13 = 0`, then find the value of p.
Find the vector equation of the line passing through (1, 2, 3) and parallel to the planes `vecr = (hati - hatj + 2hatk) = 5`and `vecr.(3hati + hatj + hatk) = 6`
Find the vector equation of a plane passing through a point with position vector \[2 \hat{i} - \hat{j} + \hat{k} \] and perpendicular to the vector \[4 \hat{i} + 2 \hat{j} - 3 \hat{k} .\]
Find the equation of the plane that bisects the line segment joining the points (1, 2, 3) and (3, 4, 5) and is at right angle to it.
Show that the normals to the following pairs of planes are perpendicular to each other.
x − y + z − 2 = 0 and 3x + 2y − z + 4 = 0
Show that the normals to the following pairs of planes are perpendicular to each other.
Find the vector equation of a plane which is at a distance of 5 units from the origin and which is normal to the vector \[\hat{i} - \text{2 } \hat{j} - \text{2 } \hat{k} .\]
Determine the value of λ for which the following planes are perpendicular to each other.
Determine the value of λ for which the following planes are perpendicular to each other.
3x − 6y − 2z = 7 and 2x + y − λz = 5
Find the equation of a plane passing through the point (−1, −1, 2) and perpendicular to the planes 3x + 2y − 3z = 1 and 5x − 4y + z = 5.
Find the equation of the plane passing through the points (2, 2, 1) and (9, 3, 6) and perpendicular to the plane 2x + 6y + 6z = 1.
Find the equation of the plane passing through (a, b, c) and parallel to the plane \[\vec{r} \cdot \left( \hat{i} + \hat{j} + \hat{k} \right) = 2 .\]
Find the equation of the plane through (2, 3, −4) and (1, −1, 3) and parallel to x-axis.
Find the vector equation of the line passing through the point (1, −1, 2) and perpendicular to the plane 2x − y + 3z − 5 = 0.
Find the vector equation of the line passing through (1, 2, 3) and perpendicular to the plane \[\vec{r} \cdot \left( \hat{i} + 2 \hat{j} - 5 \hat{k} \right) + 9 = 0 .\]
If the lines \[\frac{x - 1}{- 3} = \frac{y - 2}{- 2k} = \frac{z - 3}{2} \text{ and }\frac{x - 1}{k} = \frac{y - 2}{1} = \frac{z - 3}{5}\] are perpendicular, find the value of k and, hence, find the equation of the plane containing these lines.
Find the coordinates of the foot of the perpendicular and the perpendicular distance of the point P (3, 2, 1) from the plane 2x − y + z + 1 = 0. Also, find the image of the point in the plane.
Find the length and the foot of perpendicular from the point \[\left( 1, \frac{3}{2}, 2 \right)\] to the plane \[2x - 2y + 4z + 5 = 0\] .
Find the equation of the plane that contains the point (1, –1, 2) and is perpendicular to both the planes 2x + 3y – 2z = 5 and x + 2y – 3z = 8. Hence, find the distance of point P (–2, 5, 5) from the plane obtained
Find the position vector of the foot of perpendicular and the perpendicular distance from the point P with position vector \[2 \hat{i} + 3 \hat{j} + 4 \hat{k} \] to the plane \[\vec{r} . \left( 2 \hat{i} + \hat{j} + 3 \hat{k} \right) - 26 = 0\] Also find image of P in the plane.
Find the distance of the point P (–1, –5, –10) from the point of intersection of the line joining the points A (2, –1, 2) and B (5, 3, 4) with the plane x – y + z = 5.
Write the equation of the plane parallel to XOY- plane and passing through the point (2, −3, 5).
Write the equation of the plane parallel to the YOZ- plane and passing through (−4, 1, 0).
Write the general equation of a plane parallel to X-axis.
Write the intercept cut off by the plane 2x + y − z = 5 on x-axis.
Find the length of the perpendicular drawn from the origin to the plane 2x − 3y + 6z + 21 = 0.
Find the vector equation of the plane, passing through the point (a, b, c) and parallel to the plane \[\vec{r} . \left( \hat{i} + \hat{j} + \hat{k} \right) = 2\]
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).
Find the foot of perpendicular from the point (2, 3, –8) to the line `(4 - x)/2 = y/6 = (1 - z)/3`. Also, find the perpendicular distance from the given point to the line.
Find the length and the foot of perpendicular from the point `(1, 3/2, 2)` to the plane 2x – 2y + 4z + 5 = 0.
`vec"AB" = 3hat"i" - hat"j" + hat"k"` and `vec"CD" = -3hat"i" + 2hat"j" + 4hat"k"` are two vectors. The position vectors of the points A and C are `6hat"i" + 7hat"j" + 4hat"k"` and `-9hat"j" + 2hat"k"`, respectively. Find the position vector of a point P on the line AB and a point Q on the line Cd such that `vec"PQ"` is perpendicular to `vec"AB"` and `vec"CD"` both.
If the foot of perpendicular drawn from the origin to a plane is (5, – 3, – 2), then the equation of plane is `vec"r".(5hat"i" - 3hat"j" - 2hat"k")` = 38.
