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The equation of the plane passing through the point (3, – 3, 1) and perpendicular to the line joining the points (3, 4, – 1) and (2, – 1, 5) is:
The equation of the plane passing through the point (3, – 3, 1) is:
a(x – 3) + b(y + 3) + c(z – 1) = 0 and the direction ratios of the line joining the points
(3, 4, – 1) and (2, – 1, 5) is 2 – 3, – 1 – 4, 5 + 1, i.e., – 1, – 5, 6.
Since the plane is perpendicular to the line whose direction ratios are – 1, – 5, 6, therefore, direction ratios of the normal to the plane is – 1, – 5, 6.
So, required equation of plane is: – 1(x – 3) – 5(y + 3) + 6(z – 1) = 0
i.e., x + 5y – 6z + 18 = 0.
The equation of the plane, which is at a distance of 5 unit from the origin and has as a normal vector, is:
x = 3i  2j  6k
x = ((3)^{2} + (2)^{2} + (6)^{2})
x = (49)^{½}
x = 7
x = x/x
= (3i  2j  6k)/7
The required equation of plane is r.x = d
⇒ r.(3i  2j  6k)/7 = 5
⇒ r.(3i  2j  6k) = 35
Let r = xi + yj + zk
(xi + yj + zk) . (2i + 3j  k) = 10
2x + 3y  k = 10
The length of the perpendicular from the origin to the plane 3x + 2y – 6z = 21 is:
The given point is P(0,0,0) and the given line is 3x+2y6z21=0
Let d be the length of the perpendicular from P(0,0) to the line 3x+2y6z21=0
Then, d= 3*0 + 2*0 + (6)*0 21/[(3)^2 + (2)^{2} + (6)^{2}]^{1/2}
= 21/7
= 3
If l, m, n are the direction cosines of the normal to the plane and p be the perpendicular distance of the plane from the origin, then the equation of the plane is:
Let P(x, y, z) be any point on the plane.
OP = r = xi + yj +zk
Let l,m,n be the direction of cosine of n, then
n  li + mj + nk
As we know that r.n = d
(xi + yj + zk)(li + mj + nk) = d
i.e. lx + my + nz = d
this is the cartesian equation of plane in normal form.
If is the normal from the origin to the plane, and is the unit vector along . P(x, y, z) be any point on the plane and is perpendicular to . Then
NP is perpendicular to ON
NP.ON = NPON cosθ (θ = 90deg)
NPONcos90deg
⇒ 0
The length of the perpendicular from the origin to the plane 2x – 3y + 6z = 21 is:
The angle between two lines whose direction ratios are 1,2,1 and 2,3,4 is:
(1,2,1) and (2, 3, 4)
Cosθ = (a1b1+a2b2+a3b3]/[(a1)^{2}+(a2)^{2} + (a3)^{2})]^{1/2} [(b1)^{2} + (b2)^{2} + (b3)^{2}]^{1/2}
Cosθ = [(1)(2) + (2)(3) + (1)(4)]/ [(1)^2 + (2)^2 + (1)^{2}]^{½} [(2)^{2} + (3)^{2} + (4)^{2}]^{½}
Cosθ = [2  6 + 4]/[1 + 4 + 1]^{½} [4 + 9 + 16]^{½}
Cosθ = 0
θ = 90deg
The equation of the plane passing through the points (2, 1, 0), (3, – 2, – 2) and (3, 1, 7) is:
We know that, the equation of a plane passing through three noncollinear points (x1. y1, z1), (x2 , y2 , z2) and (x3, y3, z3)
{(xx1) (yy1) (zz1)} {(x2x1) (y2y1) (z2z1)} {(x3x2) (y3y2) (z3  z2)}
= {(x2) (y1) (z0)} {(32) (21) (20)} {(32) (11) (70)} = 0
= {(x2) (y1) (z)} {(1) (3) (2)} {(1) (0) (7)}
=> (x2)(21)  (y1)(9) + z(3) = 0
7x + 3y  z = 17
209 videos218 docs139 tests

209 videos218 docs139 tests
