Let PQR be a triangle such that PQ = -2i - j + 2k and PR = ai + bj - 4k, a, b ∈ Z. Let S be the poin

Q: Wait, why does the point S being equidistant from the lines PQ and PR mean it lies on the angle bisector?

That is a fantastic question! Think about the definition of an angle bisector: it is the locus of all points that are equidistant from the two lines forming the angle. Since S is on the segment QR and is equidistant from the lines PQ and PR, it must satisfy this condition. Therefore, the vector \vec{PS} acts as the angle bisector of \angle QPR.

Q: Why can't we just use the vector addition property for the angle bisector?

You absolutely can! The angle bisector vector can be expressed as a linear combination of the unit vectors along PQ and PR. However, in this specific problem, we are given the coordinates of \vec{PS} and need to find the unknown components of \vec{PR}. Using the dot product to equate the cosines is often more direct when you have the full vector \vec{PS} and need to solve for unknown components in another vector.

Q: Why do we reject the value b = -5.8?

The problem explicitly states that a and b must be integers (a, b \in \mathbb{Z}). When we solved the quadratic equation 5b^2 + 49b + 116 = 0, we obtained two roots. Since -5.8 is not an integer, it violates the problem's constraints, leaving b = -4 as the only valid solution.

Question

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 ......... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing at the vertex $P$ of triangle $PQR$. You look out at the lines $PQ$ and $PR$, and you see a point $S$ resting on the segment $QR$.

The problem states that $S$ is equidistant from the lines $PQ$ and $PR$. This is the hallmark of an angle bisector. When we see the phrase "equidistant from two lines," our mathematical intuition should immediately jump to the geometric definition of an angle bisector.

We are given the vectors:
$$\vec{PQ} = -2\hat{i} - \hat{j} + 2\hat{k}$$
$$\vec{PR} = a\hat{i} + b\hat{j} - 4\hat{k}$$

We are also told that the magnitude $|\vec{PR}| = 9$. This gives us our first solid constraint:
$$\sqrt{a^2 + b^2 + (-4)^2} = 9$$

Squaring both sides, we get $a^2 + b^2 + 16 = 81$, which simplifies to:
$$a^2 + b^2 = 65$$

### The Angle Bisector Property

If $\vec{PS}$ bisects the angle $\angle QPR$, then the angle $	heta$ between $\vec{PQ}$ and $\vec{PS}$ must be exactly equal to the angle between $\vec{PR}$ and $\vec{PS}$. We use the dot product formula:
$$\cos 	heta = \frac{\vec{A} \cdot \vec{B}}{|\vec{A}||\vec{B}|}$$

First, we calculate the magnitudes. For $\vec{PQ}$:
$$|\vec{PQ}| = \sqrt{(-2)^2 + (-1)^2 + 2^2} = \sqrt{9} = 3$$

For $\vec{PS} = \hat{i} - 7\hat{j} + 2\hat{k}$:
$$|\vec{PS}| = \sqrt{1^2 + (-7)^2 + 2^2} = \sqrt{54} = 3\sqrt{6}$$

The dot product $\vec{PQ} \cdot \vec{PS}$ is:
$$\vec{PQ} \cdot \vec{PS} = (-2)(1) + (-1)(-7) + (2)(2) = -2 + 7 + 4 = 9$$

Thus, the cosine of the angle is:
$$\cos 	heta = \frac{9}{3 \cdot 3\sqrt{6}} = \frac{1}{\sqrt{6}}$$

### The Algebraic Dance

Now, we apply this same $\cos 	heta$ to the vectors $\vec{PR}$ and $\vec{PS}$:
$$\frac{\vec{PR} \cdot \vec{PS}}{|\vec{PR}||\vec{PS}|} = \frac{1}{\sqrt{6}}$$

Substituting our values, we get:
$$\frac{a(1) + b(-7) + (-4)(2)}{9 \cdot 3\sqrt{6}} = \frac{1}{\sqrt{6}}$$

Simplifying the numerator, we have $a - 7b - 8$. The denominator $27\sqrt{6}$ allows us to cancel the $\sqrt{6}$ from both sides, leaving us with:
$$a - 7b - 8 = 27 \Rightarrow a - 7b = 35$$

### Final Calculation

We now have a system of equations:
1) $a^2 + b^2 = 65$
2) $a = 7b + 35$

Substituting (2) into (1):
$$(7b + 35)^2 + b^2 = 65$$
$$49b^2 + 490b + 1225 + b^2 = 65$$
$$50b^2 + 490b + 1160 = 0$$

Dividing by 10, we obtain the quadratic:
$$5b^2 + 49b + 116 = 0$$

Solving this quadratic, we find $b = -4$ or $b = -5.8$. Since $b$ must be an integer, we choose $b = -4$. Then:
$$a = 7(-4) + 35 = 7$$

Finally, we calculate the requested value:
$$3a - 4b = 3(7) - 4(-4) = 21 + 16 = 37$$

The final answer is **37**.

Enter Numerical Value:

Visualized Solution

Visualizing the Vectors $P Q, P R, P S$

Applying Magnitude Constraint $∣ P R ∣ = 9$

Setting up the Equation

Simplifying to $a^{2} + b^{2} = 65$

The Equidistant Point Property

Equating the Angles $cos θ$

Calculating Magnitudes $∣ P Q ∣$ and $∣ P S ∣$

Finding $cos θ$

Setting up Equation for $P R$

Deriving the Linear Equation

Solving the System of Equations

Expanding the Quadratic

Finding Integer Solutions for $b$

Final Calculation of $3 a - 4 b$

The Sigma Insight: Scalar (Dot) Product

Analyzing the Setup

The Angle Bisector Property

The Algebraic Dance

Final Calculation

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Enter Numerical Value:

Visualized Solution

Visualizing the Vectors .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } PQ​,PR,PS

Applying Magnitude Constraint .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∣PR∣=9