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(D) Let the first term of the $A.P.$ be $A$ and the common difference be $D$. The terms are given by:$a = A + (p - 1)D$$b = A + (q - 1)D$$c = A + (r -

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$\vec x, \vec y$ are orthogonal vectors

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(D) Let the first term of the $A.P.$ be $A$ and the common difference be $D$. The terms are given by:$a = A + (p - 1)D$$b = A + (q - 1)D$$c = A + (r - 1)D$The dot product $\vec x \cdot \vec y$ is given by:$\vec x \cdot \vec y = (q - r)a + (r - p)b + (p - q)c$Substituting the values of $a, b, c$:$\vec x \cdot \vec y = (q - r)(A + (p - 1)D) + (r - p)(A + (q - 1)D) + (p - q)(A + (r - 1)D)$$= A(q - r + r - p + p - q) + D[(q - r)(p - 1) + (r - p)(q - 1) + (p - q)(r - 1)]$$= A(0) + D[qp - q - rp + r + rq - r - pq + p + pr - p - qr + q]$$= 0 + D[0] = 0$Since the dot product is $0$,the vectors $\vec x$ and $\vec y$ are orthogonal.

If $a$,$b$,$c$ are the $p^{th}$,$q^{th}$,$r^{th}$ terms of an $A.P.$ and $\vec x = (q - r)\hat i + (r - p)\hat j + (p - q)\hat k$ and $\vec y = a\hat i + b\hat j + c\hat k$,then:

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