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
Keywords:#equidistant from lines#JEE Main 2026 Morning Q25#Vector Algebra JEE Main 2026#Vector Equations and Triangles JEE Main 2026
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Let a = - i + 2 j + 2 k$\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$ , b = 8 i + 7 j - 3 k$\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$ and c$\vec{c}$ be a vector such that a × c = b$\vec{a} \times \vec{c} = \vec{b}$ . If c. ( i + j + k) = 4$\vec{c}. (\hat{i} + \hat{j} + \hat{k}) = 4$ , then | a + c|²$|\vec{a} + \vec{c}|^2$ is equal to:
Let the unknown vector be c = c₁ i + c₂ j + c₃ k$\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$.
Given a = - i + 2 j + 2 k$\vec{a} = -\hat{i} + 2\hat{j} + 2\hat{k}$ and b = 8 i + 7 j - 3 k$\vec{b} = 8\hat{i} + 7\hat{j} - 3\hat{k}$.
Compute the cross product a × c$\vec{a} \times \vec{c}$:
Whenever a × c = b$\vec{a} \times \vec{c} = \vec{b}$ and a dot product constraint is given, simply expand c$\vec{c}$ algebraically as (c₁, c₂, c₃)$(c_1, c_2, c_3)$. The cross product and dot product create an easily solvable 3x3 linear system.
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Let c$\vec{c}$ and d$\vec{d}$ be vectors such that | c+ d|=√(29)$|\vec{c}+\vec{d}|=\sqrt{29}$ and c×(2 i+3 j+4 k)=(2 i+3 j+4 k)× d$\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$ . If λ₁,λ₂(λ₁>λ₂)$\lambda_{1},\lambda_{2}(\lambda_{1}>\lambda_{2})$ are the possible values of ( c+ d).(-7 i+2 j+3 k)$(\vec{c}+\vec{d}).(-7\hat{i}+2\hat{j}+3\hat{k})$ , then the equation K²x²+(K²-5K+λ₁)xy+(3K+ λ₂2)y²-8x+12y+λ₂=0$K^{2}x^{2}+(K^{2}-5K+\lambda_{1})xy+\left(3K+\frac{\lambda_{2}}{2}\right)y^{2}-8x+12y+\lambda_{2}=0$ represents a circle, for k equal to:
A. 4
B. 1
C. -1
D. 2
Solution
Related Formula
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General equation of a circle requires coefficient of x²$x^2$ = coefficient of y²$y^2$ and coefficient of xy$xy$ = 0$0$.
Core Logic
Given c × (2 i + 3 j + 4 k) = (2 i + 3 j + 4 k) × d$\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \vec{d}$
Rearranging:
This implies c + d$\vec{c} + \vec{d}$ is parallel to 2 i + 3 j + 4 k$2\hat{i} + 3\hat{j} + 4\hat{k}$.
Let c + d = λ (2 i + 3 j + 4 k)$\vec{c} + \vec{d} = \lambda (2\hat{i} + 3\hat{j} + 4\hat{k})$.
Step 1: Find the scaling factor
Given | c + d| = √(29)$|\vec{c} + \vec{d}| = \sqrt{29}$
The common value satisfying both conditions is K = 1$K = 1$.
Pattern Recognition
Cross-product equations mapping to X × A = -Y × A$X \times A = -Y \times A$ perfectly factor out to (X+Y) × A = 0$(X+Y) \times A = 0$, guaranteeing (X+Y)$(X+Y)$ is a scalar multiple of A$A$. This immediately unlocks vector magnitudes.
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C.410$410$
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Solution
Related Formula
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Core Logic
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The angle between vectors p$\vec{p}$ and q$\vec{q}$ extended head-to-tail is θ$\theta$. The internal angle of the triangle is π - θ$\pi - \theta$.
Always convert secondary vectors back to base components using the triangle condition p + q = r$\vec{p} + \vec{q} = \vec{r}$. It zeroes out self-cross products instantly.
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Solution
Related Formula
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Core Logic
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Pattern Recognition
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Practice past-year questions one chapter at a time. Pick an exam → subject → chapter and get every PYQ for that topic — pulled together from all past papers — with the chapter's key formulas alongside.
