Advertisements
Advertisements
प्रश्न
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0
उत्तर
5y + 8 = 0
⇒ 0x − 5y + 0z = 8 … (1)
The direction ratios of normal are 0, −5, and 0.
`:. sqrt(0+(-5)^2 + 0)` = 5
Dividing both sides of equation (1) by 5, we obtain
`-y = 8/5`
This equation is of the form lx + my + nz = d, where l, m, n are the direction cosines of normal to the plane and d is the distance of normal from the origin.
Therefore, the direction cosines of the normal to the plane are 0, −1, and 0 and the distance of normal from the origin is `8/5` units.
APPEARS IN
संबंधित प्रश्न
Find the coordinates of the point where the line through (5, 1, 6) and (3, 4, 1) crosses the ZX − plane.
Find the coordinates of the point where the line through (3, −4, −5) and (2, − 3, 1) crosses the plane 2x + y + z = 7).
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)`
Find the equation of the plane passing through the point (2, 3, 1), given that the direction ratios of the normal to the plane are proportional to 5, 3, 2.
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 2x − 3y − 6z = 14 to the normal form and, hence, find the length of the perpendicular from the origin to the plane. Also, find the direction cosines of the normal to the plane.
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 vector equation of the plane which is at a distance of \[\frac{6}{\sqrt{29}}\] from the origin and its normal vector from the origin is \[2 \hat{i} - 3 \hat{j} + 4 \hat{k} .\] Also, find its Cartesian form.
Find the distance of the plane 2x − 3y + 4z − 6 = 0 from the origin.
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 equation of the plane passing through the points (−1, 2, 0), (2, 2, −1) and parallel to the line \[\frac{x - 1}{1} = \frac{2y + 1}{2} = \frac{z + 1}{- 1}\]
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 a vector normal to the plane \[\vec{r} = l \vec{b} + m \vec{c} .\]
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 \[\vec{r} \cdot \left( 2 \hat{i} + 3 \hat{j} + 4 \hat{k} \right) = 4 .\]
Write the vector equation of the line passing through the point (1, −2, −3) and normal to the plane \[\vec{r} \cdot \left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right) = 5 .\]
The equation of the plane \[\vec{r} = \hat{i} - \hat{j} + \lambda\left( \hat{i} + \hat{j} + \hat{k} \right) + \mu\left( \hat{i} - 2 \hat{j} + 3 \hat{k} \right)\] in scalar product form is
Find the image of the point having position vector `hat"i" + 3hat"j" + 4hat"k"` in the plane `hat"r" * (2hat"i" - hat"j" + hat"k") + 3` = 0.
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 plane 2x – 3y + 6z – 11 = 0 makes an angle sin–1(α) with x-axis. The value of α is equal to ______.
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`
