Question -
Answer -
(a) z = 2
Given:
The equation of theplane, z = 2 or 0x + 0y + z = 2 …. (1)
Direction ratio of thenormal (0, 0, 1)
By using the formula,
√[(0)2 +(0)2 + (1)2] = √1
= 1
Now,
Divide both the sidesof equation (1) by 1, we get
0x/(1) + 0y/(1) + z/1= 2
So this is of the formlx + my + nz = d
Where, l, m, n are thedirection cosines and d is the distance
∴ The direction cosinesare 0, 0, 1
Distance (d) from theorigin is 2 units
(b) x + y + z = 1
Given:
The equation of theplane, x + y + z = 1…. (1)
Direction ratio of thenormal (1, 1, 1)
By using the formula,
√[(1)2 +(1)2 + (1)2] = √3
Now,
Divide both the sidesof equation (1) by √3, weget
x/(√3) + y/(√3) + z/(√3) = 1/√3
So this is of the formlx + my + nz = d
Where, l, m, n are thedirection cosines and d is the distance
∴ The direction cosinesare 1/√3, 1/√3, 1/√3
Distance (d) from theorigin is 1/√3 units
(c) 2x + 3y – z = 5
Given:
The equation of theplane, 2x + 3y – z = 5…. (1)
Direction ratio of thenormal (2, 3, -1)
By using the formula,
√[(2)2 +(3)2 + (-1)2] = √14
Now,
Divide both the sidesof equation (1) by √14, weget
2x/(√14) + 3y/(√14) – z/(√14) = 5/√14
So this is of the formlx + my + nz = d
Where, l, m, n are thedirection cosines and d is the distance
∴ The direction cosinesare 2/√14, 3/√14, -1/√14
Distance (d) from theorigin is 5/√14 units
(d) 5y + 8 = 0
Given:
The equation of theplane, 5y + 8 = 0
-5y = 8 or
0x – 5y + 0z = 8…. (1)
Direction ratio of thenormal (0, -5, 0)
By using the formula,
√[(0)2 +(-5)2 + (0)2] = √25
= 5
Now,
Divide both the sidesof equation (1) by 5, we get
0x/(5) – 5y/(5) –0z/(5) = 8/5
So this is of the formlx + my + nz = d
Where, l, m, n are thedirection cosines and d is the distance
∴ The direction cosinesare 0, -1, 0
Distance (d) from theorigin is 8/5 units