Advertisements
Advertisements
Question
Find the vector equation of a plane at a distance 6 units from the origin and to which vector `2hat"i" - hat"j" + 2hat"k"` is normal
Solution
Let `bar"n" = 2hat"i" - hat"j" + 2hat"k"`
∴ `hat"n"` is the unit vector along normal
∴ `hat"n" = (bar"n")/|bar"n"|`
= `(2hat"i" - hat"j" + 2hat"k")/sqrt(2^2 + (-1)^2 2 ^2)`
= `(2hat"i" - hat"j" + 2hat"k")/sqrt(4 + 1 + 4)`
= `(2hat"i" - hat"j" + 2hat"k")/3`
and p = 6
Vector equation of plane is `bar"r"*hat"n"` = p
∴ `bar"r"*((2hat"i" - hat"j" + 2hat"k"))/3` = 6
∴ `bar"r"*(2hat"i" - hat"j" + 2hat"k")` = 18
APPEARS IN
RELATED QUESTIONS
Find the perpendicular distance of the point (1, 0, 0) from the line `(x - 1)/(2) = (y + 1)/(-3) = (z + 10)/(8)` Also find the co-ordinates of the foot of the perpendicular.
Find the vector equation of a plane which is at 42 unit distance from the origin and which is normal to the vector `2hati + hatj - 2hatk`.
Find the perpendicular distance of the origin from the plane 6x – 2y + 3z – 7 = 0.
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 6y – 3z = 63.
Reduce the equation `bar"r".(3hat"i" + 4hat"j" + 12hat"k")` to normal form and hence find
(i) the length of the perpendicular from the origin to the plane
(ii) direction cosines of the normal.
Find the vector equation of the plane passing through the point having position vector `hati + hatj + hatk` and perpendicular to the vector `4hati + 5hatj + 6hatk`.
Choose correct alternatives :
The length of the perpendicular from (1, 6,3) to the line `x/(1) = (y - 1)/(2) =(z - 2)/(3)`
The perpendicular distance of the plane 2x + 3y – z = k from the origin is `sqrt(14)` units, the value of k is ______.
Choose correct alternatives :
The equation of the plane passing through (2, -1, 3) and making equal intercepts on the coordinate axes is
Choose correct alternatives :
The direction cosines of the normal to the plane 2x – y + 2z = 3 are ______
Choose correct alternatives :
The equation of the plane passing through the points (1, −1, 1), (3, 2, 4) and parallel to the Y-axis is ______
Solve the following :
Reduce the equation `bar"r".(6hat"i" + 8hat"j" + 24hat"k")` = 13 normal form and hence find
(i) the length of the perpendicular from the origin to the plane.
(ii) direction cosines of the normal.
If the normal to the plane has direction ratios 2, −1, 2 and it’s perpendicular distance from origin is 6, find its equation
Find the perpendicular distance of origin from the plane 6x − 2y + 3z - 7 = 0
Find the equation of the plane passing through the point (7, 8, 6) and parallel to the plane `bar"r"*(6hat"i" + 8hat"j" + 7hat"k")` = 0
If the line `(x - 3)/2 = (y + 2)/-1 = (z + 4)/3` lies in the plane lx + my - z = 9, then l2 + m2 is equal to ______
The equation of the plane passing through the point (– 1, 2, 1) and perpendicular to the line joining the points (– 3, 1, 2) and (2, 3, 4) is ______.
The intercepts of the plane 3x - 4y + 6z = 48 on the co-ordinate axes are ______
Equations of planes parallel to the plane x - 2y + 2z + 4 = 0 which are at a distance of one unit from the point (1, 2, 3) are _______.
Equation of plane parallel to ZX-plane and passing through the point (0, 5, 0) is ______
XY-plane divides the line joining the points A(2, 3, -5) and B(1, -2, -3) in the ratio ______
The equation of the plane through (1, 2, -3) and (2, -2, 1) and parallel to the X-axis is ______
Equation of the plane perpendicular to the line `x/1 = y/2 = z/3` and passing through the point (2, 3, 4) is ______
The equation of the plane through the point (2, -1, -3) and parallel to the lines `(x - 1)/3 = (y + 2)/2 = z/(-4)` and `x/2 = (y - 1)/(-3) = (z - 2)/2` is ______
A plane which passes through the point (3, 2, 0) and the line `(x - 3)/1 = (y - 6)/5, (z - 4)/4` is ______
If the plane passing through the points (1, 2, 3), (2, 3, 1) and (3, 1, 2) is ax + by + cz = d then a + 2b + 3c = ______.
The equation of the plane passing through the intersection of the planes x + 2y + 3z + 4 = 0 and 4x + 3y + 2z + 1 = 0 and the origin is ______.
If the line `(x + 1)/2 = (y - 5)/3 = (z - "p")/6` lies in the plane 3x – 14y + 6z + 49 = 0, then the value of p is ______.
If the mirror image of the point (2, 4, 7) in the plane 3x – y + 4z = 2 is (a, b, c), then 2a + b + 2c is equal to ______.
Let P be a plane Ix + my + nz = 0 containing the line, `(1 - x)/1 = ("y" + 4)/2 = ("z" + 2)/3`. If plane P divides the line segment AB joining points A(–3, –6, 1) and B(2, 4, –3) in ratio k:1 then the value of k is equal to ______.
What will be the equation of plane passing through a point (1, 4, – 2) and parallel to the given plane – 2x + y – 3z = 9?
Reduce the equation `barr*(3hati - 4hatj + 12hatk)` = 3 to the normal form and hence find the length of perpendicular from the origin to the plane.
The coordinates of the foot of the perpendicular from the point P(1, 0, 0) in the line `(x - 1)/2 = (y + 1)/-3 = (z + 10)/8` are ______.
The perpendicular distance of the plane `bar r. (3 hat i + 4 hat j + 12 hat k) = 78` from the origin is ______.
A mobile tower is situated at the top of a hill. Consider the surface on which the tower stands as a plane having points A(1, 0, 2), B(3, –1, 1) and C(1, 2, 1) on it. The mobile tower is tied with three cables from the points A, B and C such that it stands vertically on the ground. The top of the tower is at point P(2, 3, 1) as shown in the figure below. The foot of the perpendicular from the point P on the plane is at the point `Q(43/29, 77/29, 9/29)`.
Answer the following questions.
- Find the equation of the plane containing the points A, B and C.
- Find the equation of the line PQ.
- Calculate the height of the tower.