In a triangle ABC, right-angled at the vertex | vedclass

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(D) Given position vectors are $\vec{A} = 3\hat{i} + \hat{j} - \hat{k},$ $\vec{B} = -\hat{i} + 3\hat{j} + p\hat{k},$ and $\vec{C} = 5\hat{i} + q\hat{j

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making an acute angle with the positive direction of $x$-axis

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(D) Given position vectors are $\vec{A} = 3\hat{i} + \hat{j} - \hat{k},$ $\vec{B} = -\hat{i} + 3\hat{j} + p\hat{k},$ and $\vec{C} = 5\hat{i} + q\hat{j} - 4\hat{k}.$Since the triangle is right-angled at $A,$ we have $\vec{AB} \perp \vec{AC},$ which implies $\vec{AB} \cdot \vec{AC} = 0.$First,calculate the vectors $\vec{AB}$ and $\vec{AC}$:$\vec{AB} = \vec{B} - \vec{A} = (-1 - 3)\hat{i} + (3 - 1)\hat{j} + (p - (-1))\hat{k} = -4\hat{i} + 2\hat{j} + (p + 1)\hat{k}.$$\vec{AC} = \vec{C} - \vec{A} = (5 - 3)\hat{i} + (q - 1)\hat{j} + (-4 - (-1))\hat{k} = 2\hat{i} + (q - 1)\hat{j} - 3\hat{k}.$Now,compute the dot product:$\vec{AB} \cdot \vec{AC} = (-4)(2) + (2)(q - 1) + (p + 1)(-3) = 0.$$-8 + 2q - 2 - 3p - 3 = 0.$$-3p + 2q - 13 = 0 \Rightarrow 3p - 2q + 13 = 0.$Replacing $(p, q)$ with $(x, y),$ the line is $3x - 2y + 13 = 0.$Rearranging to slope-intercept form: $2y = 3x + 13 \Rightarrow y = \frac{3}{2}x + \frac{13}{2}.$The slope $m = \frac{3}{2}.$ Since $m > 0,$ the line makes an acute angle with the positive direction of the $x$-axis.

In a triangle $ABC,$ right-angled at the vertex $A,$ if the position vectors of $A, B,$ and $C$ are respectively $3\hat{i} + \hat{j} - \hat{k},$ $-\hat{i} + 3\hat{j} + p\hat{k},$ and $5\hat{i} + q\hat{j} - 4\hat{k},$ then the point $(p, q)$ lies on a line

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