Question:
1. Find image of (–2, 3, 5) in YZ plane.
2. Name the octant in which (–5, 4, –3) lies.
3. Find the distance of the point P(4, –3, 5) from XY plane.
4. Find the distance of point P(3, –2,1) from z–axis.
5. Write coordinates of foot of perpendicular from (3, 7, 9) on x axis.
6. Find the distance between points (2, 3, 4) and (–1, 3, –2).
Answer:
1. If a point has coordinate (x, y, z), then the image of this point in the yz-plane has coordinate is (-x, y, -z)
Hense, the image of (–2, 3, 5) in YZ plane is (2, 3, 5)
2. The x-coordinate, y-coordinate, and z-coordinate of point (–5, 4, –3) are negative, positive, and negative respectively.
Therefore, this point lies in octant VI
3. If a point has coordinate (x, y, z), then the image of this point in the xy-plane has coordinate is (x, y, -z)
Hense, the image of point (4, -3, 5) in the xy-plane is (4, -3, -5)
Now, the distance between the points is = √{(4 - 4)2 + ( -3 + 3)2 + (-5 - 5)2 }
= √(-10)2
= √(10)2
= 10
4. Drop perpendicular to the z-axis, it intersects z-axis at the point (0,0,1).
The vector from the point (0,0,1) to the point (3,-2, 1) is perpendicular to the x-axis and its length gives you the distance from the point (3, -2, 1) to the x-axis.
The coordinates of a vector are (3,-2,1).
Now, length is √{32 + (-2)2 + 0} =√(9 + 4) = √13
5. On x-axis, y = 0 and z = 0
So, foot of perpendicular is (3, 0, 0)
6. The distance between points (2, 3, 4) and (–1, 3, –2) = √{(-1 - 2)2 + (3 - 3)2 + (-2 - 4)2 }
= √{(-3)2 + 0 + (-6)2 }
= √{9 + 36 }
= √45
= √(9*5)
= 3√5
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