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Question

Find the distance of the point P (-4, 3, 5) from the coordinate axes.


Solution

Let PQ be the perpendicular to the xy-plane and QA be perpendicular from Q to the y-axis.

PA will be perpendicular to the x-axis. PA will be perpendicular to the x-axis.

Also, QA = |3| and PQ = |5| Now, distance of P from the x-axis :

Now, distance of P from the x-axis :

PB = =BQ2+QP2=32+52=9+25=34

Similarly, From the right-angled : distance of P from the y-axis :

PA=AQ2+QP2=(4)32+(5)2=16+25=41

Similarly, the length of the perpendicular from P to the z-axis = (4)2+(3)2=16+9=25=5


Mathematics
RD Sharma
Standard XI

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