Relative to a fixed origin 0, the point A has position vector $\hat{i} - 3 \hat{j} + 2 \hat{k}$ and the point B has position vector $- 2 \hat{i} + 2 \hat{j} - \hat{k}$ The points A and B lie on a straight-line 1. The point C has position vector $2 \hat{i} + p \hat{j} - 4 \hat{k}$ with respect to O, where p is a constant. Given that AC is erpendicular to 1. Find the value of p.

Question

Relative to a fixed origin 0, the point A has position vector $\hat{i} - 3 \hat{j} + 2 \hat{k}$ and the point B has position vector $- 2 \hat{i} + 2 \hat{j} - \hat{k}$ The points A and B lie on a straight-line 1. The point C has position vector $2 \hat{i} + p \hat{j} - 4 \hat{k}$  with respect to O, where p is a constant. Given that AC is erpendicular to 1. Find the value of p.

Accepted Answer

ঘ. p =-6

ব্যাখ্যা: A ও B বিন্দুগামী সরলরেখা $l$-এর দিক নির্দেশক ভেক্টর:
$\vec{AB} = \vec{OB} - \vec{OA} = (-2\hat{i} + 2\hat{j} - \hat{k}) - (\hat{i} - 3\hat{j} + 2\hat{k}) = -3\hat{i} + 5\hat{j} - 3\hat{k}$

আবার, $\vec{AC} = \vec{OC} - \vec{OA} = (2\hat{i} + p\hat{j} - 4\hat{k}) - (\hat{i} - 3\hat{j} + 2\hat{k}) = \hat{i} + (p + 3)\hat{j} - 6\hat{k}$

যেহেতু AC রেখাটি $l$ রেখার ওপর লম্ব, তাই এদের ডট গুণন শূন্য হবে:
$\vec{AC} \cdot \vec{AB} = 0$
$\Rightarrow 1(-3) + (p + 3)(5) + (-6)(-3) = 0$
$\Rightarrow -3 + 5p + 15 + 18 = 0$
$\Rightarrow 5p + 30 = 0$
$\Rightarrow p = -6$।

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