Question

Write the distance of the point P (3, 4, 5) from z-axis______

Answer 1

Hint: We know that distance between points $$P(x_1,y_1,z_1)$$ and $$Q(x_2,y_2,z_2)$$ is given by $$PQ=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2+{(z_1-z_2)}^2}$$ . We also know that any point on the Z-axis has a coordinate $$(0,0,z)$$ . That is X and Y coordinates are zero. We find the value of $$z$$ and then we can find the distance between P and point on a Z-axis.

Complete step-by-step answer:
Let Q be any point on the Z-axis then its coordinates are $$Q(0,0,z)$$ and $$P(3,4,5)$$ .
Look at the below diagram, so that we can understand better.

We need to find the distance between PQ, before that we need the value of the coordinate of Q. That is $$z$$ .
Now the direction ratio of PQ is proportional to $$\Rightarrow (3-0,\text{ 4 - 0, 5 - z) = (3,4,5 - z)}$$ .
We can see from the figure that PQ is perpendicular to the Z-axis then,
  (We know that, if two lines are perpendicular then their product of sum of direction ratios are equal to zero.)
 $$\Rightarrow (3\times 0)+(4\times 0)+(1\times \text{5 - z})=0$$
 $$\Rightarrow 0+0+5-z=0$$
 $$\Rightarrow 5-z=0$$
 $$\Rightarrow z=5$$
Hence the coordinates of the point $$Q(0,0,5)$$ .
Now to find the distance between PQ, $$P(3,4,5)$$ and $$Q(0,0,5)$$
Using the distance formula, that is $$PQ=\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2+{(z_1-z_2)}^2}$$
 $$\Rightarrow PQ=\sqrt{{(3-0)}^2+{(4-0)}^2+{(5-5)}^2}$$
 $$\Rightarrow PQ=\sqrt{9+16+0}$$
 $$\Rightarrow PQ=\sqrt{25}$$
 $$\Rightarrow PQ=5$$
Hence, the distance of the point P (3, 4, 5) from the z-axis is $5$ units.
So, the correct answer is “5 units”.

Note: If the problem is based on the Conic section or three dimensional geometry always express the given problem into the diagram so that we will have a better understanding of the problem. In the above problem we notice that the line PQ is perpendicular to the Z-axis by looking into the diagram. After that all we did is a simple calculation.