Let PQR$PQR$ be a triangle such that PQ = -2 i - j + 2 k$\overrightarrow{PQ} = -2\hat{i} - \hat{j} + 2\hat{k}$ and PR = a i + b j - 4 k$\overrightarrow{PR} = a\hat{i} + b\hat{j} - 4\hat{k}$, a, b in Z$a, b \in \mathbb{Z}$. Let S$S$ be the point on QR$QR$, which is equidistant from the lines PQ$PQ$ and PR$PR$. If | PR| = 9$|\overrightarrow{PR}| = 9$ and PS = i - 7 j + 2 k$\overrightarrow{PS} = \hat{i} - 7\hat{j} + 2\hat{k}$, then the value of 3a - 4b$3a - 4b$ is ____.
Numerical Answer Type:
Enter a numerical valueAnswer: 37 to 37+4 marks
Solution & Explanation
Core Logic
Vector Equations and Triangles
Given:
PQ = -2 i - j + 2 k | PQ| = √(4 + 1 + 4) = 3$\overrightarrow{PQ} = -2\hat{i} - \hat{j} + 2\hat{k} \implies |\overrightarrow{PQ}| = \sqrt{4 + 1 + 4} = 3$.
PR = a i + b j - 4 k$\overrightarrow{PR} = a\hat{i} + b\hat{j} - 4\hat{k}$ and | PR| = 9$|\overrightarrow{PR}| = 9$.
Since a, b in Z$a, b \in \mathbb{Z}$, the only integers solving a² + b² = 65$a^2 + b^2 = 65$ are combinations of ± 7, ± 4$\pm 7, \pm 4$ or ± 8, ± 1$\pm 8, \pm 1$.
Since S$S$ is on QR$QR$ and is equidistant from PQ$PQ$ and PR$PR$, PS$PS$ acts as the internal angle bisector of ∠ QPR$\angle QPR$.
Step 1: Compute Angle from Vector Dot Products
Using the dot product formula for angle bisector, the angle between PQ$\overrightarrow{PQ}$ and PS$\overrightarrow{PS}$ must equal the angle between PS$\overrightarrow{PS}$ and PR$\overrightarrow{PR}$. Let the half-angle be θ$\theta$.
θ = PQ · PS| PQ| | PS|$\cos \theta = \frac{\overrightarrow{PQ} \cdot \overrightarrow{PS}}{|\overrightarrow{PQ}| |\overrightarrow{PS}|}$
Given PS = i - 7 j + 2 k | PS| = √(1 + 49 + 4) = √(54) = 3√(6)$\overrightarrow{PS} = \hat{i} - 7\hat{j} + 2\hat{k} \implies |\overrightarrow{PS}| = \sqrt{1 + 49 + 4} = \sqrt{54} = 3\sqrt{6}$.
For points equidistant from two adjacent sides, the line is the internal angle bisector. However, the vector formulation reveals a geometric inconsistency: the actual length of the bisector using formula PS = (2bc)/(b+c) θ$PS = \frac{2bc}{b+c} \cos\theta$ yields θ > 1$\cos\theta > 1$, meaning such a triangle is geometrically impossible in reality. Both NTA and standard logic accept the algebraic projection giving 37.
Our Ans. (Bonus)
NTA Ans. (37)
Chapter Mix
Class 12 Mathematics: Vector Algebra
Class 11 Mathematics: Straight Lines
Let a$\vec{a}$ and b$\vec{b}$ be two vectors such that | a| = 1, | b| = 4$|\vec{a}| = 1, |\vec{b}| = 4$ and a · b = 2$\vec{a} \cdot \vec{b} = 2$. If c = (2 a × b) - 3 b$\vec{c} = (2\vec{a} \times \vec{b}) - 3\vec{b}$ and the angle between b$\vec{b}$ and c$\vec{c}$ is α$\alpha$, then 192 ²α$192\sin^2\alpha$ is equal to
Numerical Answer.Answer: 48 to 48
Solution
Core Logic
b · c = b · ((2 a × b) - 3 b)$$\vec{b} \cdot \vec{c} = \vec{b} \cdot ((2\vec{a} \times \vec{b}) - 3\vec{b})$$|b||c| α = 2( b · ( a × b)) - 3|b|²$$|b||c|\cos\alpha = 2(\vec{b} \cdot (\vec{a} \times \vec{b})) - 3|b|^2$$
Since b · ( a × b) = 0$\vec{b} \cdot (\vec{a} \times \vec{b}) = 0$, we have |b||c| α = -3|b|²$|b||c|\cos\alpha = -3|b|^2$.
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