Advertisements
Advertisements
Question
Find the perpendicular distance of the origin from the plane 6x – 2y + 3z – 7 = 0.
Advertisements
Solution
The equation of the plane is
6x – 2y + 3z – 7 = 0
∴ its vector equation is
`bar"r".(6hat"i" - 2hat"j" + 3hat"k")` = 7 ...(1)
where `bar"r" = xhat"i" + yhat"j" + zhat"k"`
∴ `bar"n" = 6hat"i" - 2hat"j" + 3hat"k"` is normal to the plane.
`|bar"n"| = sqrt(6^2 + (-2)^2 + 3^2)`
= `sqrt(49)`
= 7
Unit vector along `bar"n"` is
`hat"n" = bar"n"/|bar"n"|= (6hat"i" - 2hat"j" + 3hat"k")/(7)`
Dividing bothsides of (1) by 7, we get
`bar"r".((6hat"i" - 2hat"j" + 3hat"k")/7) = (7)/(7)`
∴ `bar"r".hat"n"`= 1
Comparing with normal form of equation of the plane `hat"r".hat"n" = p` it follows that length of perpendicular from origin is 1 unit.
Alternative Method :
The equation of the plane is 6x – 2y + 3z – 7 = 0
i.e. `(6/(6^2 + (-2)^2 + 3))x - (2/(sqrt(6^2) + (-2)^2 + 3^2))y + ((3)/(sqrt(6^2 + (-2)^2 + 3^2)))z = 7/(sqrt(6^2 + (-2)^2 + 3)`
i.e. `(6)/(7)x -(2)/(7)y + (3)/(7)z = (7)/(7)` = 1
This is the normal form of the equation of plane.
∴ perpendicular distance of the origin frm the plane is p = 1 unit.
APPEARS IN
RELATED QUESTIONS
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`.
Choose correct alternatives :
The length of the perpendicular from (1, 6,3) to the line `x/(1) = (y - 1)/(2) =(z - 2)/(3)`
Choose correct alternatives :
The lines `x/(1) = y/(2) = z/(3) and (x - 1)/(-2) = (y - 2)/(-4) = (z - 3)/(6)` are
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 :
Find the perpendicular distance of the origin from the plane 6x + 2y + 3z - 7 = 0
Solve the following :
Find the coordinates of the foot of the perpendicular drawn from the origin to the plane 2x + 3y + 6z = 49.
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.
The equation of X axis is ______
If the planes 2x – my + z = 3 and 4x – y + 2z = 5 are parallel then m = ______
If the foot of the perpendicular drawn from the origin to the plane is (4, −2, -5), then the equation of the plane is ______
Find the direction ratios of the normal to the plane 2x + 3y + z = 7
If the normal to the plane has direction ratios 2, −1, 2 and it’s perpendicular distance from origin is 6, find its equation
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 equation of a plane containing the line of intersection of the planes 2x - y - 4 = 0 and y + 2z - 4 = 0 and passing through the point (1, 1, 0) is ______
Equation of plane parallel to ZX-plane and passing through the point (0, 5, 0) is ______
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 ______
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 = ______.
Let the line `(x - 2)/3 = (y - 1)/(-5) = (z + 2)/2` lie in the plane x + 3y - αz + β = 0. Then, (α, β) equals ______
The d.r.s of normal to the plane through (1, 0, 0), (0, 1, 0) which makes an angle `pi/4` with plane x + y = 3, are ______.
The equation of the plane passing through a point having position vector`-2hat"i" + 7hat"j" + 5hat"k"` and parallel to the vectors `4hat"i" - hat"j" + 3hat"k"` and `hat"i" + hat"j" + hat"k"` 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 ______.
If the foot of the perpendicular drawn from the origin to the plane is (4, –2, 5), then the equation of the plane is ______.
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.
Find the equation of plane which is at a distance of 4 units from the origin and which is normal to the vector `2hati - 2hatj + hatk`.
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 ______.
Find the vector equation of the line passing through the point (–2, 1, 4) and perpendicular to the plane `barr*(4hati - 5hatj + 7hatk)` = 15
Find the equation of the plane which contains the line of intersection of the planes x + 2y + 4z = 4 and 2x – 3y – z = 9 and which is perpendicular to the plane 4x – 3y + 5z = 10.
Find the point of intersection of the line `(x + 1)/2 = (y - 1)/3 = (z - 2)/1` with the plane x + 2y – z = 6.
The perpendicular distance of the plane `bar r. (3 hat i + 4 hat j + 12 hat k) = 78` from the origin is ______.
The Cartesian equation of a plane through A (7, 8, 6) and parallel to the XY plane is
