Let PQR$PQR$ be a triangle such that overrightarrowPQ = -2hati - hatj + 2hatk$\overrightarrow{PQ} = -2\hat{i} - \hat{j} + 2\hat{k}$ and overrightarrowPR = ahati + bhatj - 4hatk$\overrightarrow{PR} = a\hat{i} + b\hat{j} - 4\hat{k}$, a, b in mathbbZ$a, b \in \mathbb{Z}$. Let S$S$ be the point on QR$QR$, which is equidistant from the lines PQ$PQ$ and PR$PR$. If |overrightarrowPR| = 9$|\overrightarrow{PR}| = 9$ and overrightarrowPS = hati - 7hatj + 2hatk$\overrightarrow{PS} = \hat{i} - 7\hat{j} + 2\hat{k}$, 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:
overrightarrowPQ = -2hati - hatj + 2hatk implies |overrightarrowPQ| = sqrt4 + 1 + 4 = 3$\overrightarrow{PQ} = -2\hat{i} - \hat{j} + 2\hat{k} \implies |\overrightarrow{PQ}| = \sqrt{4 + 1 + 4} = 3$.
overrightarrowPR = ahati + bhatj - 4hatk$\overrightarrow{PR} = a\hat{i} + b\hat{j} - 4\hat{k}$ and |overrightarrowPR| = 9$|\overrightarrow{PR}| = 9$.
a^2 + b^2 + (-4)^2 = 81 implies a^2 + b^2 = 65 $$ a^2 + b^2 + (-4)^2 = 81 \implies a^2 + b^2 = 65 $$
Since a, b in mathbbZ$a, b \in \mathbb{Z}$, the only integers solving a^2 + b^2 = 65$a^2 + b^2 = 65$ are combinations of pm 7, pm 4$\pm 7, \pm 4$ or pm 8, pm 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 angle QPR$\angle QPR$.
### Step 1: Compute Angle from Vector Dot Products
Using the dot product formula for angle bisector, the angle between overrightarrowPQ$\overrightarrow{PQ}$ and overrightarrowPS$\overrightarrow{PS}$ must equal the angle between overrightarrowPS$\overrightarrow{PS}$ and overrightarrowPR$\overrightarrow{PR}$. Let the half-angle be theta$\theta$.
cos theta = fracoverrightarrowPQ cdot overrightarrowPS|overrightarrowPQ| |overrightarrowPS|$\cos \theta = \frac{\overrightarrow{PQ} \cdot \overrightarrow{PS}}{|\overrightarrow{PQ}| |\overrightarrow{PS}|}$
Given overrightarrowPS = hati - 7hatj + 2hatk implies |overrightarrowPS| = sqrt1 + 49 + 4 = sqrt54 = 3sqrt6$\overrightarrow{PS} = \hat{i} - 7\hat{j} + 2\hat{k} \implies |\overrightarrow{PS}| = \sqrt{1 + 49 + 4} = \sqrt{54} = 3\sqrt{6}$.
cos theta = frac(-2)(1) + (-1)(-7) + (2)(2)3 times 3sqrt6 = frac-2 + 7 + 49sqrt6 = frac99sqrt6 = frac1sqrt6 $$ \cos \theta = \frac{(-2)(1) + (-1)(-7) + (2)(2)}{3 \times 3\sqrt{6}} = \frac{-2 + 7 + 4}{9\sqrt{6}} = \frac{9}{9\sqrt{6}} = \frac{1}{\sqrt{6}} $$
### Step 2: Equating Dot Products
Now use cos theta$\cos \theta$ with vectors overrightarrowPS$\overrightarrow{PS}$ and overrightarrowPR$\overrightarrow{PR}$:
frac1sqrt6 = fracoverrightarrowPS cdot overrightarrowPR|overrightarrowPS| |overrightarrowPR| $$ \frac{1}{\sqrt{6}} = \frac{\overrightarrow{PS} \cdot \overrightarrow{PR}}{|\overrightarrow{PS}| |\overrightarrow{PR}|} $$ frac1sqrt6 = frac(1)(a) + (-7)(b) + (2)(-4)3sqrt6 times 9 $$ \frac{1}{\sqrt{6}} = \frac{(1)(a) + (-7)(b) + (2)(-4)}{3\sqrt{6} \times 9} $$ frac1sqrt6 = fraca - 7b - 827sqrt6 $$ \frac{1}{\sqrt{6}} = \frac{a - 7b - 8}{27\sqrt{6}} $$ 27 = a - 7b - 8 implies a - 7b = 35 $$ 27 = a - 7b - 8 \implies a - 7b = 35 $$
### Step 3: Solve for a and b
We have a system:
1) a^2 + b^2 = 65$a^2 + b^2 = 65$
2) a - 7b = 35 implies a = 35 + 7b$a - 7b = 35 \implies a = 35 + 7b$
If we substitute b = -4$b = -4$:
a = 35 + 7(-4) = 35 - 28 = 7$a = 35 + 7(-4) = 35 - 28 = 7$.
Check squares: 7^2 + (-4)^2 = 49 + 16 = 65$7^2 + (-4)^2 = 49 + 16 = 65$. Matches perfectly.
So a = 7$a = 7$ and b = -4$b = -4$.
Calculate the required value:
3a - 4b = 3(7) - 4(-4) = 21 + 16 = 37 $$ 3a - 4b = 3(7) - 4(-4) = 21 + 16 = 37 $$
### Pattern Recognition
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 = frac2bcb+c costheta$PS = \frac{2bc}{b+c} \cos\theta$ yields costheta > 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)
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Vector Algebra
Class 11 Mathematics: Straight Lines
Q7jee_main_2026_21_jan_morningCollinearity and Cross Product of Vectors
Let vecc$\vec{c}$ and vecd$\vec{d}$ be vectors such that |vecc+vecd|=sqrt29$|\vec{c}+\vec{d}|=\sqrt{29}$ and vecctimes(2hati+3hatj+4hatk)=(2hati+3hatj+4hatk)timesvecd$\vec{c}\times(2\hat{i}+3\hat{j}+4\hat{k})=(2\hat{i}+3\hat{j}+4\hat{k})\times\vec{d}$ . If lambda_1,lambda_2(lambda_1>lambda_2)$\lambda_{1},\lambda_{2}(\lambda_{1}>\lambda_{2})$ are the possible values of (vecc+vecd).(-7hati+2hatj+3hatk)$(\vec{c}+\vec{d}).(-7\hat{i}+2\hat{j}+3\hat{k})$ , then the equation K^2x^2+(K^2-5K+lambda_1)xy+left(3K+fraclambda_22right)y^2-8x+12y+lambda_2=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
vecu times vecv = -(vecv times vecu)$$\vec{u} \times \vec{v} = -(\vec{v} \times \vec{u})$$
General equation of a circle requires coefficient of x^2$x^2$ = coefficient of y^2$y^2$ and coefficient of xy$xy$ = 0$0$.
### Core Logic
Given vecc times (2hati + 3hatj + 4hatk) = (2hati + 3hatj + 4hatk) times vecd$\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = (2\hat{i} + 3\hat{j} + 4\hat{k}) \times \vec{d}$
Rearranging:
vecc times (2hati + 3hatj + 4hatk) + vecd times (2hati + 3hatj + 4hatk) = vec0$$\vec{c} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) + \vec{d} \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = \vec{0}$$(vecc + vecd) times (2hati + 3hatj + 4hatk) = vec0$$(\vec{c} + \vec{d}) \times (2\hat{i} + 3\hat{j} + 4\hat{k}) = \vec{0}$$
This implies vecc + vecd$\vec{c} + \vec{d}$ is parallel to 2hati + 3hatj + 4hatk$2\hat{i} + 3\hat{j} + 4\hat{k}$.
Let vecc + vecd = lambda (2hati + 3hatj + 4hatk)$\vec{c} + \vec{d} = \lambda (2\hat{i} + 3\hat{j} + 4\hat{k})$.
### Step 1: Find the scaling factor
Given |vecc + vecd| = sqrt29$|\vec{c} + \vec{d}| = \sqrt{29}$|lambda| sqrt2^2 + 3^2 + 4^2 = sqrt29$$|\lambda| \sqrt{2^2 + 3^2 + 4^2} = \sqrt{29}$$|lambda| sqrt4 + 9 + 16 = sqrt29$$|\lambda| \sqrt{4 + 9 + 16} = \sqrt{29}$$|lambda| sqrt29 = sqrt29 Rightarrow lambda = pm 1$$|\lambda| \sqrt{29} = \sqrt{29} \Rightarrow \lambda = \pm 1$$
Therefore, vecc + vecd = pm(2hati + 3hatj + 4hatk)$\vec{c} + \vec{d} = \pm(2\hat{i} + 3\hat{j} + 4\hat{k})$.
### Step 2: Evaluate the dot product limits
We need values of (vecc + vecd) cdot (-7hati + 2hatj + 3hatk)$(\vec{c} + \vec{d}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$.
Dot product = lambda(2(-7) + 3(2) + 4(3)) = lambda(-14 + 6 + 12) = 4lambda$= \lambda(2(-7) + 3(2) + 4(3)) = \lambda(-14 + 6 + 12) = 4\lambda$.
Since lambda = pm 1$\lambda = \pm 1$, the possible values are 4$4$ and -4$-4$.
Given lambda_1 > lambda_2$\lambda_1 > \lambda_2$, we have lambda_1 = 4$\lambda_1 = 4$ and lambda_2 = -4$\lambda_2 = -4$.
### Step 3: Apply circle constraints
Substitute lambda_1$\lambda_1$ and lambda_2$\lambda_2$ into the conic equation:
K^2 x^2 + (K^2 - 5K + 4)xy + left(3K - 2right)y^2 - 8x + 12y - 4 = 0$$K^2 x^2 + (K^2 - 5K + 4)xy + \left(3K - 2\right)y^2 - 8x + 12y - 4 = 0$$
For this to represent a circle:
1) Coefficient of xy$xy$ must be 0$0$:
K^2 - 5K + 4 = 0 Rightarrow K = 1, 4$$K^2 - 5K + 4 = 0 \Rightarrow K = 1, 4$$
2) Coefficient of x^2$x^2$ must equal coefficient of y^2$y^2$:
K^2 = 3K - 2 Rightarrow K^2 - 3K + 2 = 0 Rightarrow K = 1, 2$$K^2 = 3K - 2 \Rightarrow K^2 - 3K + 2 = 0 \Rightarrow K = 1, 2$$
The common value satisfying both conditions is K = 1$K = 1$.
### Pattern Recognition
Cross-product equations mapping to X times A = -Y times A$X \times A = -Y \times A$ perfectly factor out to (X+Y) times A = 0$(X+Y) \times A = 0$, guaranteeing (X+Y)$(X+Y)$ is a scalar multiple of A$A$. This immediately unlocks vector magnitudes.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Vector Algebra
Class 11 Maths: Conic Sections
Q15jee_main_2026_21_jan_eveningCross Product
For a triangle ABC, let vecp=overrightarrowBC,vecq=overrightarrowCA$\vec{p}=\overrightarrow{BC},\vec{q}=\overrightarrow{CA}$ and vecr=overrightarrowBA$\vec{r}=\overrightarrow{BA}$. If |vecp|=2sqrt3,|vecq|=2$|\vec{p}|=2\sqrt{3},|\vec{q}|=2$ and costheta=frac1sqrt3$\cos\theta=\frac{1}{\sqrt{3}}$, where theta$\theta$ is the angle between vecp$\vec{p}$ and vecq$\vec{q}$, then left|vecptimes(vecq-3vecr)right|^2+3left|vecrright|^2$\left|\vec{p}\times(\vec{q}-3\vec{r})\right|^{2}+3\left|\vec{r}\right|^{2}$ is equal to:
A.340$340$
B.220$220$
C.410$410$
D.200$200$
Solution
### Related Formula
textCosine Rule in triangle: cos(pi-theta) = frac|vecp|^2+|vecq|^2-|vecr|^22|vecp||vecq|$$\text{Cosine Rule in triangle: } \cos(\pi-\theta) = \frac{|\vec{p}|^2+|\vec{q}|^2-|\vec{r}|^2}{2|\vec{p}||\vec{q}|}$$textCross product identity: vecA times vecA = 0 quad ; quad |vecA times vecB| = |A||B|sintheta$$\text{Cross product identity: } \vec{A} \times \vec{A} = 0 \quad ; \quad |\vec{A} \times \vec{B}| = |A||B|\sin\theta$$
### Core Logic
Vector triangle diagram for Q15 - JEE Main 2026 Evening
From triangle law of addition, overrightarrowBC + overrightarrowCA = overrightarrowBA implies vecp + vecq = vecr$\overrightarrow{BC} + \overrightarrow{CA} = \overrightarrow{BA} \implies \vec{p} + \vec{q} = \vec{r}$.
The angle between vectors vecp$\vec{p}$ and vecq$\vec{q}$ extended head-to-tail is theta$\theta$. The internal angle of the triangle is pi - theta$\pi - \theta$.
### Step 1: Calculate magnitude of r
Using cosine rule for the side |vecr|$|\vec{r}|$:
cos(pi - theta) = frac|vecp|^2 + |vecq|^2 - |vecr|^22|vecp||vecq|$$\cos(\pi - \theta) = \frac{|\vec{p}|^2 + |\vec{q}|^2 - |\vec{r}|^2}{2|\vec{p}||\vec{q}|}$$
Since costheta = 1/sqrt3$\cos\theta = 1/\sqrt{3}$, cos(pi - theta) = -1/sqrt3$\cos(\pi - \theta) = -1/\sqrt{3}$.
-frac1sqrt3 = frac(2sqrt3)^2 + (2)^2 - |vecr|^22(2sqrt3)(2)$$-\frac{1}{\sqrt{3}} = \frac{(2\sqrt{3})^2 + (2)^2 - |\vec{r}|^2}{2(2\sqrt{3})(2)}$$-frac1sqrt3 = frac12 + 4 - |vecr|^28sqrt3$$-\frac{1}{\sqrt{3}} = \frac{12 + 4 - |\vec{r}|^2}{8\sqrt{3}}$$-8 = 16 - |vecr|^2 implies |vecr|^2 = 24$$-8 = 16 - |\vec{r}|^2 \implies |\vec{r}|^2 = 24$$
### Step 2: Simplify the Cross Product term
We need to evaluate |vecp times (vecq - 3vecr)|^2$|\vec{p} \times (\vec{q} - 3\vec{r})|^2$.
Substitute vecr = vecp + vecq$\vec{r} = \vec{p} + \vec{q}$:
= |vecp times (vecq - 3(vecp + vecq))|^2$$= |\vec{p} \times (\vec{q} - 3(\vec{p} + \vec{q}))|^2$$= |vecp times (-3vecp - 2vecq)|^2$$= |\vec{p} \times (-3\vec{p} - 2\vec{q})|^2$$
Since vecp times vecp = 0$\vec{p} \times \vec{p} = 0$:
= |-2 (vecp times vecq)|^2 = 4 |vecp times vecq|^2$$= |-2 (\vec{p} \times \vec{q})|^2 = 4 |\vec{p} \times \vec{q}|^2$$= 4 ( |vecp|^2 |vecq|^2 sin^2theta )$$= 4 ( |\vec{p}|^2 |\vec{q}|^2 \sin^2\theta )$$
### Step 3: Final Calculation
Given costheta = 1/sqrt3 implies cos^2theta = 1/3 implies sin^2theta = 2/3$\cos\theta = 1/\sqrt{3} \implies \cos^2\theta = 1/3 \implies \sin^2\theta = 2/3$.
4 |vecp times vecq|^2 = 4 (12)(4)left(frac23right) = 4(4)(4)(2) = 128$$4 |\vec{p} \times \vec{q}|^2 = 4 (12)(4)\left(\frac{2}{3}\right) = 4(4)(4)(2) = 128$$
The required expression is:
|vecp times (vecq - 3vecr)|^2 + 3|vecr|^2 = 128 + 3(24) = 128 + 72 = 200$$|\vec{p} \times (\vec{q} - 3\vec{r})|^2 + 3|\vec{r}|^2 = 128 + 3(24) = 128 + 72 = 200$$
### Pattern Recognition
Always convert secondary vectors back to base components using the triangle condition vecp + vecq = vecr$\vec{p} + \vec{q} = \vec{r}$. It zeroes out self-cross products instantly.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Vector Algebra
Q1jee_main_2026_22_january_morningProjection Of A Vector On Another Vector
Let overrightarrowAB=2hati+4hatj-5hatk$\overrightarrow{AB}=2\hat{i}+4\hat{j}-5\hat{k}$ and overrightarrowAD=hati+2hatj+lambda hatk, lambdain mathbbR$\overrightarrow{AD}=\hat{i}+2\hat{j}+\lambda \hat{k}, \lambda\in \mathbb{R}$. Let the projection of the vectorvecv=hati+hatj+hatk$\vec{v}=\hat{i}+\hat{j}+\hat{k}$ on the diagonal overrightarrowAC$\overrightarrow{AC}$ of the parallelogram ABCD be of length one unit. If alpha,beta$\alpha,\beta$, where alpha>beta$\alpha>\beta$, be the roots of the equation lambda^2x^2-6lambda x+5=0$\lambda^{2}x^{2}-6\lambda x+5=0$, then 2alpha-beta$2\alpha-\beta$ is equal to
A.1$1$
B.4$4$
C.3$3$
D.6$6$
Solution
### Related Formula
textProjection of vecv text on veca = fracvecv cdot veca|veca|$$\text{Projection of } \vec{v} \text{ on } \vec{a} = \frac{\vec{v} \cdot \vec{a}}{|\vec{a}|}$$overrightarrowAC = overrightarrowAB + overrightarrowAD$$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}$$
### Core Logic
Using the parallelogram law of vector addition, the diagonal overrightarrowAC$\overrightarrow{AC}$ is given by:
overrightarrowAC = overrightarrowAB + overrightarrowAD = (2hati+4hatj-5hatk) + (hati+2hatj+lambda hatk)$$\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD} = (2\hat{i}+4\hat{j}-5\hat{k}) + (\hat{i}+2\hat{j}+\lambda \hat{k})$$overrightarrowAC = 3hati + 6hatj + (lambda - 5)hatk$$\overrightarrow{AC} = 3\hat{i} + 6\hat{j} + (\lambda - 5)\hat{k}$$
### Step 1: Applying the Projection Condition
The projection of vecv = hati+hatj+hatk$\vec{v} = \hat{i}+\hat{j}+\hat{k}$ on overrightarrowAC$\overrightarrow{AC}$ is 1. Therefore:
fracvecv cdot overrightarrowAC|overrightarrowAC| = 1$$\frac{\vec{v} \cdot \overrightarrow{AC}}{|\overrightarrow{AC}|} = 1$$frac3(1) + 6(1) + (lambda - 5)(1)sqrt3^2 + 6^2 + (lambda - 5)^2 = 1$$\frac{3(1) + 6(1) + (\lambda - 5)(1)}{\sqrt{3^2 + 6^2 + (\lambda - 5)^2}} = 1$$3 + 6 + lambda - 5 = sqrt9 + 36 + (lambda - 5)^2$$3 + 6 + \lambda - 5 = \sqrt{9 + 36 + (\lambda - 5)^2}$$lambda + 4 = sqrt45 + (lambda - 5)^2$$\lambda + 4 = \sqrt{45 + (\lambda - 5)^2}$$
Squaring both sides:
(lambda + 4)^2 = 45 + (lambda - 5)^2$$(\lambda + 4)^2 = 45 + (\lambda - 5)^2$$lambda^2 + 8lambda + 16 = lambda^2 - 10lambda + 25 + 45$$\lambda^2 + 8\lambda + 16 = \lambda^2 - 10\lambda + 25 + 45$$18lambda = 54 implies lambda = 3$$18\lambda = 54 \implies \lambda = 3$$
### Step 2: Solving the Quadratic Equation
Substitute lambda = 3$\lambda = 3$ into the given quadratic equation lambda^2x^2 - 6lambda x + 5 = 0$\lambda^2x^2 - 6\lambda x + 5 = 0$:
9x^2 - 18x + 5 = 0$$9x^2 - 18x + 5 = 0$$(3x - 1)(3x - 5) = 0 implies x = frac13, frac53$$(3x - 1)(3x - 5) = 0 \implies x = \frac{1}{3}, \frac{5}{3}$$
Given alpha > beta$\alpha > \beta$, we have alpha = frac53$\alpha = \frac{5}{3}$ and beta = frac13$\beta = \frac{1}{3}$.
Calculate 2alpha - beta$2\alpha - \beta$:
2left(frac53right) - frac13 = frac10 - 13 = 3$$2\left(\frac{5}{3}\right) - \frac{1}{3} = \frac{10 - 1}{3} = 3$$Projection of A Vector diagram for Q1 - JEE Main 2026 Morning
### Pattern Recognition
Projection questions in 3D geometry often couple a simple dot product identity with an unknown parameter. Squaring a linear equation derived from vector magnitudes often cancels the squared terms, making it easily solvable.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Vector Algebra
Class 11 Maths: Quadratic Equations
Q8jee_main_2026_22_january_eveningVector Dot Product and Cross Product
Let veca = 2hati - hatj + hatk$\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$ and vecb = lambdahatj + 2hatk$\vec{b} = \lambda\hat{j} + 2\hat{k}$, lambda in mathbbZ$\lambda \in \mathbb{Z}$ be two vectors. Let vecc = veca times vecb$\vec{c} = \vec{a} \times \vec{b}$ and vecd$\vec{d}$ be a vector of magnitude 2$2$ in yz-plane. If |vecc| = sqrt53$|\vec{c}| = \sqrt{53}$, then the maximum possible value of (vecc cdot vecd)^2$(\vec{c} \cdot \vec{d})^2$ is equal to:
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