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Introduction to Three Dimensional Geometry
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Solution:
Let P(0,7,10),Q(−1,6,6) and R(−4,9,6) be the vertices of a triangle
Here, PQ=1+1+16=32 QR=9+9+0=32 PR=16+4+16=6
Now, PQ2+QR2=(32)2+(32)2=36=(PR)2
Therefore, ΔPQR is a right angled triangle at Q. Also, OQ= OR. Hence, ΔPQR is a right angled isosceles triangle.