Relative to a fixed origin 0, the point A has position vector hati-3hatj + 2hatk and the point B has position vector -2hati +2hatj- hatk The points A and B lie on a straight-line 1. The point C has position vector 2hati+ phatj-4hatk with respect to O, where p is a constant. Given that AC is erpendicular to 1. Find the value of p.

Another Explanation (5): ```html

সমাধান:

দেওয়া আছে, \( \vec{OA} = \hat{i} - 3\hat{j} + 2\hat{k} \) এবং \( \vec{OB} = -2\hat{i} + 2\hat{j} - \hat{k} \)। সুতরাং, ভেক্টর \( \vec{AB} = \vec{OB} - \vec{OA} = (-2-1)\hat{i} + (2-(-3))\hat{j} + (-1-2)\hat{k} = -3\hat{i} + 5\hat{j} - 3\hat{k} \)। আরও দেওয়া আছে, \( \vec{OC} = 2\hat{i} + p\hat{j} - 4\hat{k} \)। সুতরাং, ভেক্টর \( \vec{AC} = \vec{OC} - \vec{OA} = (2-1)\hat{i} + (p-(-3))\hat{j} + (-4-2)\hat{k} = \hat{i} + (p+3)\hat{j} - 6\hat{k} \)। যেহেতু \( \vec{AC} \) রেখা \( l \) এর উপর লম্ব, তাই \( \vec{AC} \) ভেক্টর \( \vec{AB} \) ভেক্টরের উপর লম্ব হবে। সুতরাং, \( \vec{AC} \cdot \vec{AB} = 0 \) হবে। \( \vec{AC} \cdot \vec{AB} = (1)(-3) + (p+3)(5) + (-6)(-3) = 0 \) \( -3 + 5p + 15 + 18 = 0 \) \( 5p + 30 = 0 \) \( 5p = -30 \) \( p = -6 \) অতএব, \( p \) এর মান \( -6 \)। 🎉 ```