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
More Vector Algebra Previous-Year Questions — Page 2
Q21jee_main_2026_22_january_eveningAngle Between Vectors
Let a vector a = √(2) i - j + λ k$\vec{a} = \sqrt{2}\hat{i} - \hat{j} + \lambda\hat{k}$, λ > 0$\lambda > 0$, make an obtuse angle with the vector b = -λ² i + 4√(2) j + 4√(2) k$\vec{b} = -\lambda^2\hat{i} + 4\sqrt{2}\hat{j} + 4\sqrt{2}\hat{k}$ and an angle θ$\theta$, (π)/(6) < θ < (π)/(2)$\frac{\pi}{6} < \theta < \frac{\pi}{2}$, with the positive z-axis. If the set of all possible values of λ$\lambda$ is (α, β) - γ$(\alpha, \beta) - \{\gamma\}$, then α + β + γ$\alpha + \beta + \gamma$ is equal to ____.
Numerical Answer.Answer: 5 to 5
Solution
Related Formula
Cos angle with z-axis: θ = a · k| a|$\cos\theta = \frac{\vec{a} \cdot \hat{k}}{|\vec{a}|}$.
Obtuse angle condition: a · b < 0$\vec{a} \cdot \vec{b} < 0$.
Let a = - i + j + 2 k$\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$, b = i - j - 3 k$\vec{b} = \hat{i} - \hat{j} - 3\hat{k}$, c = a × b$\vec{c} = \vec{a} \times \vec{b}$ and d = c × a$\vec{d} = \vec{c} \times \vec{a}$. Then ( a - b) · d$(\vec{a} - \vec{b}) \cdot \vec{d}$ is equal to:
A.4$4$
B.-4$-4$
C.-2$-2$
D.2$2$
Solution
Related Formula
x × ( y × z) = ( x · z) y - ( x · y) z$$\vec{x} \times (\vec{y} \times \vec{z}) = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{x} \cdot \vec{y})\vec{z}$$
Core Logic
Given c = a × b$\vec{c} = \vec{a} \times \vec{b}$ and d = c × a$\vec{d} = \vec{c} \times \vec{a}$. Substitute c$\vec{c}$ into the expression for d$\vec{d}$:
d = ( a × b) × a$$\vec{d} = (\vec{a} \times \vec{b}) \times \vec{a}$$
d = ( a · a) b - ( a · b) a = a² b - ( a · b) a$$\vec{d} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a} = a^2 \vec{b} - (\vec{a} \cdot \vec{b})\vec{a}$$
Step 1: Calculate Magnitudes and Dot Products
For a = - i + j + 2 k$\vec{a} = -\hat{i} + \hat{j} + 2\hat{k}$ and b = i - j - 3 k$\vec{b} = \hat{i} - \hat{j} - 3\hat{k}$:
Instead of computing cross products sequentially (which is tedious and error-prone), immediately expand nested cross products using the standard vector triple product identity BAC-CAB$BAC-CAB$.
Chapter Mix
Class 12 Maths: Vector Algebra
Q3jee_main_2026_23_january_eveningCross Product and Projection
Let a= i-2 j+3 k, b=2 i+ j- k, c=λ i+ j+ k$\vec{a}=\hat{i}-2\hat{j}+3\hat{k}, \vec{b}=2\hat{i}+\hat{j}-\hat{k}, \vec{c}=\lambda\hat{i}+\hat{j}+\hat{k}$ and v= a× b$\vec{v}=\vec{a}\times\vec{b}$. If v· c=11$\vec{v}\cdot\vec{c}=11$ and the length of the projection of b$\vec{b}$ on c$\vec{c}$ is p$p$, then 9p²$9p^{2}$ is equal to:
A.9$9$
B.6$6$
C.4$4$
D.12$12$
Solution
Related Formula
Length of projection of b on c = | b· c|| c|$$\text{Length of projection of } \vec{b} \text{ on } \vec{c} = \frac{|\vec{b}\cdot\vec{c}|}{|\vec{c}|}$$
Core Logic
First, find v = a × b$\vec{v} = \vec{a} \times \vec{b}$.
Standard sequence: compute cross product to find normal vector v$\vec{v}$, take dot product to deduce missing parameter λ$\lambda$, then substitute into scalar projection formula.
Chapter Mix
Class 12 Maths: Vector Algebra
Q10jee_main_2026_23_january_eveningVector Product
Let a, b, c$\vec{a}, \vec{b}, \vec{c}$ be three vectors such that a× b=2( a× c)$\vec{a}\times\vec{b}=2(\vec{a}\times\vec{c})$. If | a|=1, | b|=4, | c|=2$|\vec{a}|=1, |\vec{b}|=4, |\vec{c}|=2$, and the angle between b$\vec{b}$ and c$\vec{c}$ is 60°$60^{\circ}$, then | a· c|$|\vec{a}\cdot\vec{c}|$ is:
A.2$2$
B.4$4$
C.0$0$
D.1$1$
Solution
Related Formula
u × v = 0 u and v are parallel ($ u = λ v $)$$\vec{u} \times \vec{v} = 0 \implies \vec{u} \text{ and } \vec{v} \text{ are parallel ($ \vec{u} = \lambda\vec{v} $)}$$| x + y|² = | x|² + | y|² + 2 x· y$$|\vec{x} + \vec{y}|^2 = |\vec{x}|^2 + |\vec{y}|^2 + 2\vec{x}\cdot\vec{y}$$
Core Logic
a × b - 2( a × c) = 0$$\vec{a} \times \vec{b} - 2(\vec{a} \times \vec{c}) = 0$$a × ( b - 2 c) = 0$$\vec{a} \times (\vec{b} - 2\vec{c}) = 0$$
This implies that ( b - 2 c)$(\vec{b} - 2\vec{c})$ is collinear with a$\vec{a}$.
So, b - 2 c = λ a$\vec{b} - 2\vec{c} = \lambda \vec{a}$ for some scalar λ$\lambda$.
A cross-product equation structured as a×( X)=0$\vec{a}\times(\vec{X})=0$ immediately gives X = λ a$\vec{X} = \lambda\vec{a}$. Expanding the magnitude squared is the standard method to expose the dot products and solve for λ$\lambda$.
Chapter Mix
Class 12 Maths: Vector Algebra
Q5jee_main_2026_24_january_morningCross and Dot Products
Let a = 2 i + j - 2 k$\vec{a} = 2\hat{i} + \hat{j} - 2\hat{k}$, b = i + j$\vec{b} = \hat{i} + \hat{j}$ and c = a × b$\vec{c} = \vec{a} \times \vec{b}$. Let d$\vec{d}$ be a vector such that| d - a| = √(11)$|\vec{d} - \vec{a}| = \sqrt{11}$, | c × d| = 3$|\vec{c} \times \vec{d}| = 3$ and the angle between c$\vec{c}$ and d$\vec{d}$ is (π)/(4)$\frac{\pi}{4}$. Then a · d$\vec{a} \cdot \vec{d}$ is equal to
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.
