If in a right-angled triangle ABC, hypotenuse AC=p, then what is `vec(AB).vec(AC)+vec(BC).vec(BA)+vec(CA).vec(CB)` equal to ?

Options

A. P, Q, R are the vertices of an equilateral triangle
B. `p^(2)`
C. `2p^(2)`
D. `(p^(2))/(2)`

Answer

B

Explanation

`vec(AB).vec(AC)+vec(BC).vec(BA)+vec(CA).vec(CA)`
`=(AB.AC.costheta)+(BC.BA.cos(90-theta))+(CA.CB.cos 90)`
`=(p.x.cos theta)+(y.p.sin theta)+0`
`=p[x cos theta + y sin theta]`
By projection formula:
`p=x cos theta + y cos (90-theta)`
`= x cos theta + y sin theta`
`:. p[x cos theta + y sin theta]=pxxp=p^(2)`.