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Show that the vectors 2i vector - j vector+ k vector, 3i vector - 4j vector - 4k vector, i vector - 3j vector - 5 k vector, form a right angled triangle.

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Let \(\vec a=2\hat i-\hat j+\hat k\)

\(\vec b=3\hat i-4\hat j-4\hat k\)

\(\vec c=\hat i-3\hat j-5\hat k\) 

\(\vec a.\vec b=(2\hat i-\hat j+\hat k).(3\hat i-4\hat j-4\hat k)\)

 = 6 + 4 - 4

 = 6 \(\neq0\)

\(\vec b.\vec c=(3\hat i-4\hat j-4\hat k).(\hat i-3\hat j-5\hat k)\)

 = 3 + 12 + 20

 = 35 \(\neq\) 0

\(\vec a.\vec c=(2\hat i-\hat j-\hat k).(\hat i-3\hat j-5\hat k)\)

 = 2 + 3  - 5

 = 0

\(\therefore\) Angle between \(\vec a\) & \(\vec c\) is 90°.

\(\therefore\) \(\vec a,\vec b\)\(\vec c\)will form a right triangle.

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