Advertisements
Advertisements
प्रश्न
The equation of the plane containing the two lines
विकल्प
8x + y − 5z − 7 = 0
8x + y + 5z − 7 = 0
8x − y − 5z − 7 = 0
None of these
Advertisements
उत्तर
None of these
\[\frac{x - 1}{2} = \frac{y + 1}{-1} = \frac{z- 0}{3}\] and
\[\frac{x}{- 2} = \frac{y - 2}{- 3} = \frac{z + 1}{- 1}\]
Now, if these two lines lie on a plane, so the direction ratio of lines will be perpendicular to the plane's normal vector.
APPEARS IN
संबंधित प्रश्न
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
z = 2
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
2x + 3y – z = 5
If the coordinates of the points A, B, C, D be (1, 2, 3), (4, 5, 7), (−4, 3, −6) and (2, 9, 2) respectively, then find the angle between the lines AB and CD.
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX − plane.
The planes: 2x − y + 4z = 5 and 5x − 2.5y + 10z = 6 are
(A) Perpendicular
(B) Parallel
(C) intersect y-axis
(C) passes through `(0,0,5/4)`
If the axes are rectangular and P is the point (2, 3, −1), find the equation of the plane through P at right angles to OP.
Reduce the equation \[\vec{r} \cdot \left( \hat{i} - 2 \hat{j} + 2 \hat{k} \right) + 6 = 0\] to normal form and, hence, find the length of the perpendicular from the origin to the plane.
Write the normal form of the equation of the plane 2x − 3y + 6z + 14 = 0.
The direction ratios of the perpendicular from the origin to a plane are 12, −3, 4 and the length of the perpendicular is 5. Find the equation of the plane.
Find a unit normal vector to the plane x + 2y + 3z − 6 = 0.
Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.
Find the equation of the plane which contains the line of intersection of the planes \[x + 2y + 3z - 4 = 0 \text { and } 2x + y - z + 5 = 0\] and whose x-intercept is twice its z-intercept.
Prove that the line of section of the planes 5x + 2y − 4z + 2 = 0 and 2x + 8y + 2z − 1 = 0 is parallel to the plane 4x − 2y − 5z − 2 = 0.
Find the value of λ such that the line \[\frac{x - 2}{6} = \frac{y - 1}{\lambda} = \frac{z + 5}{- 4}\] is perpendicular to the plane 3x − y − 2z = 7.
Find the vector equation of the plane passing through the points (3, 4, 2) and (7, 0, 6) and perpendicular to the plane 2x − 5y − 15 = 0. Also, show that the plane thus obtained contains the line \[\vec{r} = \hat{i} + 3 \hat{j} - 2 \hat{k} + \lambda\left( \hat{i} - \hat{j} + \hat{k} \right) .\]
Write the value of k for which the line \[\frac{x - 1}{2} = \frac{y - 1}{3} = \frac{z - 1}{k}\] is perpendicular to the normal to the plane \[\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right) = 4 .\]
Find the vector equation of a plane which is at a distance of 5 units from the origin and its normal vector is \[2 \hat{i} - 3 \hat{j} + 6 \hat{k} \] .
The equations of x-axis in space are ______.
Find the equation of a plane which is at a distance `3sqrt(3)` units from origin and the normal to which is equally inclined to coordinate axis.
If the line drawn from the point (–2, – 1, – 3) meets a plane at right angle at the point (1, – 3, 3), find the equation of the plane.
The unit vector normal to the plane x + 2y +3z – 6 = 0 is `1/sqrt(14)hat"i" + 2/sqrt(14)hat"j" + 3/sqrt(14)hat"k"`.
Find the vector equation of a plane which is at a distance of 7 units from the origin and which is normal to the vector `3hati + 5hatj - 6hatk`
What will be the cartesian equation of the following plane. `vecr * (hati + hatj - hatk)` = 2
In the following cases find the c9ordinates of foot of perpendicular from the origin `2x + 3y + 4z - 12` = 0
Find the vector and cartesian equations of the planes that passes through (1, 0, – 2) and the normal to the plane is `hati + hatj - hatk`
