Advertisements
Advertisements
प्रश्न
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.
Advertisements
उत्तर
Since, the normal to the plane is equally inclined to the axes
∴ cos α = cos β = cos ϒ
⇒ cos2α + cos2α + cos2α = 1
⇒ 3 cos2α = 1
⇒ cos α = `1/sqrt(3)`
⇒ cos α = cos β = cos ϒ = `1/sqrt(3)`
So, the normal is `vec"N" = 1/sqrt(3)hat"i" + 1/sqrt(3)hat"j" + 1/sqrt(3)hat"k"`
∴ Equation of the plane is `vec"r" . vec"N"` = d
⇒ `vec"r" . vec"N"/|vec"N"|` = d
⇒ `(vec"r"*(1/sqrt(3)hat"i" + 1/sqrt(3)hat"j" + 1/sqrt(3)hat"k"))/1 = 3sqrt(3)`
⇒ `vec"r"*(1/sqrt(3)hat"i" + 1/sqrt(3)hat"j" + 1/sqrt(3)hat"k") = 3sqrt(3)`
⇒ `(xhat"i" + yhat"j" + zhat"k") * 1/sqrt(3) (hat"i" + hat"j" + hat"k") = 3sqrt(3)`
⇒ x + y + z = `3sqrt(3) * sqrt(3)`
⇒ x + y + z = 9
Hence, the required equation of plane is x + y + z = 9.
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
In following cases, determine the direction cosines of the normal to the plane and the distance from the origin.
5y + 8 = 0
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 YZ-plane
Find the coordinates of the point where the line through the points (3, - 4, - 5) and (2, - 3, 1), crosses the plane determined by the points (1, 2, 3), (4, 2,- 3) and (0, 4, 3)
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.
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.
Find a unit normal vector to the plane x + 2y + 3z − 6 = 0.
Find the equation of a plane which is at a distance of \[3\sqrt{3}\] units from the origin and the normal to which is equally inclined to the coordinate axes.
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 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 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 \[\vec{r} \cdot \left( 2 \hat{i} + \hat{j} + 2 \hat{k} \right) = 5 .\]
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 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 ______.
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`
