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Provide solution for RD Sharma maths class 12 chapter 26 Direction Cosines and Direction Ratios exercise Fill in the blanks question 2

Answers (1)

Final Answer : \sqrt{a^{2}+b^{2}}

 

Hint:

Use Distance Formula
 

Given:

Point P(a,b,c)

To Find: 

Distance of point P from z -axis

Solution:

The distance of the point (a,b,c)  from z-axis will be the perpendicular distance from point (a,b,c) to z-axis whose co-ordinates are (0,0,c)
Distance formula is
\begin{aligned} &d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\\\ &d=\sqrt{(0-a)^{2}+(0-b)^{2}+(c-c)^{2}} \end{aligned}\begin{aligned} &d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\\\ &d=\sqrt{(0-a)^{2}+(0-b)^{2}+(c-c)^{2}} \end{aligned}\begin{aligned} &d=\sqrt{\left(x_{2}-x_{1}\right)^{2}+\left(y_{2}-y_{1}\right)^{2}+\left(z_{2}-z_{1}\right)^{2}} \\ &d=\sqrt{(0-a)^{2}+(0-b)^{2}+(c-c)^{2}} \end{aligned}

where \left(x_{1}, y_{1}, z_{1}\right)=(a, b, c) \text { and }\left(x_{2}, y_{2}, z_{2}\right)=(0,0, c)
d=\sqrt{a^{2}+b^{2}}

Therefore distance of the point (a,b,c)  from z axis is  \sqrt{a^{2}+b^{2}}

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