Question

the point (0,2,3) lies in (a) xy-plane (b) yz plane (c) xz plane (d) non of these? 2).find the distance between the points p(_3,7,2,)andq(2,4,_1)?​

Answer 1

Answer:

The distance between the given points is $\sqrt{43} \mbox{units}$.

Step-by-step explanation:

1) Given the point (0,2,3)

Given the point in cartesian coordinate systems (x, y, z).

The x-coordinate of the point is 0.

As it has only y-and z-coordinates, the points lies in yz plane.

So, the correct answer is option b.

2) Given the points p(-3, 7, 2) and q(2, 4, -1)

The distance between two points $p(x_1, y_1, z_1)$ and $q(x_2, y_2, z_2)$ is given by

$\sqrt{(x_2- x_1)^2+(y_2- y_1)^2+(z_2- z_1)^2 }$

Using the formula for the given points, p(-3, 7, 2) and q(2, 4, -1),

$\sqrt{(2- (-3))^2+(4- 7)^2+(-1- 2)^2 }$

$=\sqrt{(2+3)^2+(4- 7)^2+(-1- 2)^2 }$

$=\sqrt{5^2+(- 3)^2+(-3)^2 }$

$=\sqrt{25+9+9 } =\sqrt{43 }$

Therefore, the distance between the given points is $\sqrt{43} \mbox{units}$.

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