Question ID: #939
Let PQR be a triangle such that $\vec{PQ}=-2\hat{i}-\hat{j}+2\hat{k}$ and $\vec{PR}=a\hat{i}+b\hat{j}-4\hat{k},$ $a, b\in\mathbb{Z}.$ Let S be the point on QR, which is equidistant from the lines PQ and PR. If $|\vec{PR}|=9$ and $\vec{PS}=\hat{i}-7\hat{j}+2\hat{k}$, then the value of $3a-4b$ is
Solution:
Given $|\vec{PR}| = 9$:
$$\sqrt{a^2 + b^2 + (-4)^2} = 9$$
$$a^2 + b^2 + 16 = 81 \Rightarrow a^2 + b^2 = 65 \quad \dots(1)$$
Since S is equidistant from lines PQ and PR, PS must be the angle bisector of $\angle QPR$.
Thus, $\vec{PS}$ makes equal angles with $\vec{PQ}$ and $\vec{PR}$.
$$\cos \theta = \frac{\vec{PQ} \cdot \vec{PS}}{|\vec{PQ}| |\vec{PS}|} = \frac{\vec{PR} \cdot \vec{PS}}{|\vec{PR}| |\vec{PS}|}$$
Calculate magnitudes:
$|\vec{PQ}| = \sqrt{4+1+4} = 3$.
$|\vec{PR}| = 9$.
$|\vec{PS}| = \sqrt{1+49+4} = \sqrt{54} = 3\sqrt{6}$.
Equating cosines:
$$\frac{\vec{PQ} \cdot \vec{PS}}{3 \cdot 3\sqrt{6}} = \frac{\vec{PR} \cdot \vec{PS}}{9 \cdot 3\sqrt{6}}$$
$$3(\vec{PQ} \cdot \vec{PS}) = \vec{PR} \cdot \vec{PS}$$
$$3((-2)(1) + (-1)(-7) + (2)(2)) = a(1) + b(-7) + (-4)(2)$$
$$3(-2 + 7 + 4) = a – 7b – 8$$
$$3(9) = a – 7b – 8$$
$$27 = a – 7b – 8 \Rightarrow a – 7b = 35 \Rightarrow a = 7b + 35$$
Substitute $a$ into (1):
$$(7b+35)^2 + b^2 = 65$$
$$49b^2 + 490b + 1225 + b^2 = 65$$
$$50b^2 + 490b + 1160 = 0$$
$$5b^2 + 49b + 116 = 0$$
$D = 49^2 – 4(5)(116) = 2401 – 2320 = 81$.
$$b = \frac{-49 \pm 9}{10}$$
$b = -4$ or $b = -5.8$. Since $b \in \mathbb{Z}$, $b = -4$.
Then $a = 7(-4) + 35 = -28 + 35 = 7$.
We need $3a – 4b$:
$$3(7) – 4(-4) = 21 + 16 = 37$$
Ans. 37
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