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\(\vec {OA} = \hat i + 2 \hat j\)

\(\vec {OB} = \hat i + \hat j\)

\(\vec {OC} = \hat i - 2 \hat j\)

\(\vec {AB} = \vec {OB} - \vec {OA} = (\hat i + \hat j) - (\hat i + 2\hat j) = -\hat j\)

\(\vec {BC} = \vec{OC} - \vec{OB} = ( \hat i - 2\hat j)-( \hat i + \hat j) = - 3\hat j\)

\(\vec {AB} \times \vec {BC} = -\hat j \times -3\hat j= 3(\hat j \times \hat j) = 3 \times\vec 0 = \vec 0\)

Hence, zero vector is perpendicular to the plane in which \(\vec {AB}\) and \(\vec {BC}\) present

or zero vector is perpendicular to both \(\vec {AB}\) & \(\vec {BC}\) .

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