Question

Distance between the points A(2,3,1) and B(1,-2,0) is
1)3units (b) 3root3units (c)root 3units (d)none of these

Answer 1

Answer:

(b) 3√3 units

Step-by-step explanation:

Given that,

There are two points in 3-d plane such that,

• A = (2,3,1)
• B = (1,-2,0)

To find the distance between them.

We know that,

Distance between two points having coordinate (a,b,c) and (d,e,f) is given by,

• $\sqrt{ {(d - a)}^{2} + {(e - b)}^{2} + {(f - c)}^{2} }$

Therefore, we will get,

$= > d = \sqrt{ {(2 - 1)}^{2} + {(3 + 2)}^{2} + {(1 - 0)}^{2} } \\ \\ = > d = \sqrt{ {1}^{2} + {5}^{2} + {1}^{2} } \\ \\ = > d = \sqrt{1 + 25 + 1} \\ \\ = > d = \sqrt{27} \\ \\ = > d = \sqrt{9 \times 3} \\ \\ = > d = \sqrt{9} \times \sqrt{3} \\ \\ = > d = 3 \sqrt{3}$

Hence, the Correct Answer is (b)3√3 units.

Answer 2

Given ,

The two points are A(2,3,1) and B(1,-2,0)

We know that , the distance between two points (x1,y1,z1) and (x2,y2,z2) is given by

$\sf \fbox{\sf D= \sqrt{ {( x_{2} - x_{1} )}^{2} + {(y_{2} - y_{1} )}^{2} + {(z_{2} - z_{1}) }^{2} }\: \: \: }$

Thus ,

$\sf \Rightarrow D = \sqrt{ {(2 - 1)}^{2} + {(3 + 2)}^{2} + {(1 - 0)}^{2} } \\ \\\sf \Rightarrow D = \sqrt{ {(1)}^{2} + {(5)}^{2} + {(1)}^{2} } \\ \\\sf \Rightarrow D = \sqrt{1 + 25 + 1} \\ \\\sf \Rightarrow D= \sqrt{27} \\ \\\sf \Rightarrow D = 3 \sqrt{3} \: \: units</p><p>$

$\therefore \sf \bold{ \underline{The \: distance \: is \: 3 \sqrt{3} \: \: units }}$