Keywords:#vector algebra cross product#JEE Main 2026 Morning Q5#Vector Algebra JEE Main 2026#Cross Product and Dot Product Operations JEE Main 2026
More Vector Algebra Previous-Year Questions — Page 5
Q8jee_main_2024_29_january_eveningArea of Quadrilateral and Parallelogram
Let overrightarrowOA = veca, overrightarrowOB = 12veca + 4vecb$\overrightarrow{OA} = \vec{a}, \overrightarrow{OB} = 12\vec{a} + 4\vec{b}$ and overrightarrowOC = vecb$\overrightarrow{OC} = \vec{b}$, where O$O$ is the origin. If S$S$ is the parallelogram with adjacent sides OA$OA$ and OC$OC$, then the ratio of the area of the quadrilateral OABC to the area of S$S$ is equal to
A. 6
B. 10
C. 7
D. 8
Solution
### Related Formula
textArea of Parallelogram S = |veca times vecb|$$\text{Area of Parallelogram } S = |\vec{a} \times \vec{b}|$$textArea of Quadrilateral OABC = textArea(Delta OAB) + textArea(Delta OBC)$$\text{Area of Quadrilateral OABC} = \text{Area}(\Delta OAB) + \text{Area}(\Delta OBC)$$
### Core Logic
Let us compute the component areas using vector cross products:
textArea(Delta OAB) = frac12 |overrightarrowOA times overrightarrowOB| = frac12 |veca times (12veca + 4vecb)| = frac12 |4(veca times vecb)| = 2|veca times vecb|$$\text{Area}(\Delta OAB) = \frac{1}{2} |\overrightarrow{OA} \times \overrightarrow{OB}| = \frac{1}{2} |\vec{a} \times (12\vec{a} + 4\vec{b})| = \frac{1}{2} |4(\vec{a} \times \vec{b})| = 2|\vec{a} \times \vec{b}|$$textArea(Delta OBC) = frac12 |overrightarrowOB times overrightarrowOC| = frac12 |(12veca + 4vecb) times vecb| = frac12 |12(veca times vecb)| = 6|veca times vecb|$$\text{Area}(\Delta OBC) = \frac{1}{2} |\overrightarrow{OB} \times \overrightarrow{OC}| = \frac{1}{2} |(12\vec{a} + 4\vec{b}) \times \vec{b}| = \frac{1}{2} |12(\vec{a} \times \vec{b})| = 6|\vec{a} \times \vec{b}|$$
### Step 1: Finding Total Ratio
Adding both triangles to find the total area of the quadrilateral OABC:
textArea(OABC) = 2|veca times vecb| + 6|veca times vecb| = 8|veca times vecb|$$\text{Area(OABC)} = 2|\vec{a} \times \vec{b}| + 6|\vec{a} \times \vec{b}| = 8|\vec{a} \times \vec{b}|$$Area of Quadrilateral and Parallelogram diagram for Q8 - JEE Main 2024 Evening
Dividing this total by the area of the baseline parallelogram S = |veca times vecb|$S = |\vec{a} \times \vec{b}|$:
textRatio = frac8|veca times vecb||veca times vecb| = 8$$\text{Ratio} = \frac{8|\vec{a} \times \vec{b}|}{|\vec{a} \times \vec{b}|} = 8$$
### Pattern Recognition
Since veca times veca = 0$\vec{a} \times \vec{a} = 0$ and vecb times vecb = 0$\vec{b} \times \vec{b} = 0$, cross product distributions yield terms with purely veca times vecb$\vec{a} \times \vec{b}$, ensuring absolute scalability independent of vectors chosen.
### Evaluation Rubric / Model Answer
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### Chapter Mix
Class 12 Mathematics: Vector Algebra
Q20jee_main_2024_29_january_eveningVector Operations and Angles
Let a unit vector hatmathbfu = xhatmathbfi + yhatmathbfj + zhatmathbfk$\hat{\mathbf{u}} = x\hat{\mathbf{i}} + y\hat{\mathbf{j}} + z\hat{\mathbf{k}}$ make angles fracpi2, fracpi3$\frac{\pi}{2}, \frac{\pi}{3}$ and frac2pi3$\frac{2pi}{3}$ with the vectors frac1sqrt2hatmathbfi + frac1sqrt2hatmathbfk$\frac{1}{\sqrt{2}}\hat{\mathbf{i}} + \frac{1}{\sqrt{2}}\hat{\mathbf{k}}$, frac1sqrt2hatmathrmj + frac1sqrt2hatmathrmk$\frac{1}{\sqrt{2}}\hat{\mathrm{j}} + \frac{1}{\sqrt{2}}\hat{\mathrm{k}}$ and frac1sqrt2hatmathbfi + frac1sqrt2hatmathrmj$\frac{1}{\sqrt{2}}\hat{\mathbf{i}} + \frac{1}{\sqrt{2}}\hat{\mathrm{j}}$ respectively. If \vec{\mathrm{v}} = \frac{1}{\sqrt{2}}\hat{\mathrm{i}} +\frac{1}{\sqrt{2}}\hat{\mathrm{j}} +\frac{1}{\sqrt{2}}\hat{\mathrm{k}}, then |\hat{\mathbf{u}} -\bar{\mathbf{v}} |^2 is equal to
A. \frac{11}{2}
B. \frac{5}{2}
C. 9
D. 7
Solution
### Related Formula
vecA cdot vecB = |vecA| |vecB| cos phi$$\vec{A} \cdot \vec{B} = |\vec{A}| |\vec{B}| \cos \phi$$
### Core Logic
Let the given baseline vectors be vecp_1, vecp_2, vecp_3$\vec{p}_1, \vec{p}_2, \vec{p}_3$. Note that |vecp_1| = |vecp_2| = |vecp_3| = 1$|\vec{p}_1| = |\vec{p}_2| = |\vec{p}_3| = 1$.
1. Angle with vecp_1$\vec{p}_1$ is fracpi2$\frac{\pi}{2}$:
hatu cdot vecp_1 = 0 implies fracxsqrt2 + fraczsqrt2 = 0 implies x + z = 0 quad dots (i)$$\hat{u} \cdot \vec{p}_1 = 0 \implies \frac{x}{\sqrt{2}} + \frac{z}{\sqrt{2}} = 0 \implies x + z = 0 \quad \dots (i)$$
2. Angle with vecp_2$\vec{p}_2$ is fracpi3$\frac{\pi}{3}$:
hatu cdot vecp_2 = cosfracpi3 implies fracysqrt2 + fraczsqrt2 = frac12 implies y + z = frac1sqrt2 quad dots (ii)$$\hat{u} \cdot \vec{p}_2 = \cos\frac{\pi}{3} \implies \frac{y}{\sqrt{2}} + \frac{z}{\sqrt{2}} = \frac{1}{2} \implies y + z = \frac{1}{\sqrt{2}} \quad \dots (ii)$$
3. Angle with vecp_3$\vec{p}_3$ is frac2pi3$\frac{2\pi}{3}$:
hatu cdot vecp_3 = cosfrac2pi3 implies fracxsqrt2 + fracysqrt2 = -frac12 implies x + y = -frac1sqrt2 quad dots (iii)$$\hat{u} \cdot \vec{p}_3 = \cos\frac{2\pi}{3} \implies \frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = -\frac{1}{2} \implies x + y = -\frac{1}{\sqrt{2}} \quad \dots (iii)$$
### Step 1: Finding Vector Coordinates
Subtracting (ii) from (iii):
(x + y) - (y + z) = -frac1sqrt2 - frac1sqrt2 implies x - z = -sqrt2 quad dots (iv)$$(x + y) - (y + z) = -\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} \implies x - z = -\sqrt{2} \quad \dots (iv)$$
Solving (i) and (iv):
2x = -sqrt2 implies x = -frac1sqrt2$$2x = -\sqrt{2} \implies x = -\frac{1}{\sqrt{2}}$$z = frac1sqrt2$$z = \frac{1}{\sqrt{2}}$$
From (ii): y = frac1sqrt2 - frac1sqrt2 = 0$y = \frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} = 0$.
So, hatu = -frac1sqrt2hati + 0hatj + frac1sqrt2hatk$\hat{u} = -\frac{1}{\sqrt{2}}\hat{i} + 0\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$.
### Step 2: Evaluating the Norm Difference
Given vecv = frac1sqrt2hati + frac1sqrt2hatj + frac1sqrt2hatk$\vec{v} = \frac{1}{\sqrt{2}}\hat{i} + \frac{1}{\sqrt{2}}\hat{j} + \frac{1}{\sqrt{2}}\hat{k}$:
hatu - vecv = left(-frac1sqrt2 - frac1sqrt2right)hati + left(0 - frac1sqrt2right)hatj + left(frac1sqrt2 - frac1sqrt2right)hatk$$\hat{u} - \vec{v} = \left(-\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\hat{i} + \left(0 - \frac{1}{\sqrt{2}}\right)\hat{j} + \left(\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}}\right)\hat{k}$$hatu - vecv = -sqrt2hati - frac1sqrt2hatj + 0hatk$$\hat{u} - \vec{v} = -\sqrt{2}\hat{i} - \frac{1}{\sqrt{2}}\hat{j} + 0\hat{k}$$
Evaluating the squared magnitude:
|hatu - vecv|^2 = (-sqrt2)^2 + left(-frac1sqrt2right)^2 = 2 + frac12 = frac52$$|\hat{u} - \vec{v}|^2 = (-\sqrt{2})^2 + \left(-\frac{1}{\sqrt{2}}\right)^2 = 2 + \frac{1}{2} = \frac{5}{2}$$
### Pattern Recognition
Setting up dot products systematically transforms descriptive geometric angles into solvable sets of linear equations.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Vector Algebra
Let veca=hati+2hatj+hatk$\vec{a}=\hat{i}+2\hat{j}+\hat{k}$, vecb=3(hati-hatj+hatk)$\vec{b}=3(\hat{i}-\hat{j}+\hat{k})$. Let vecc$\vec{c}$ be the vector such that vecatimesvecc=vecb$\vec{a}\times\vec{c}=\vec{b}$ and vecacdotvecc=3$\vec{a}\cdot\vec{c}=3$ Then vecacdot((vecctimesvecb)-vecb-vecc)$\vec{a}\cdot((\vec{c}\times\vec{b})-\vec{b}-\vec{c})$ equal to:
A.32$32$
B.24$24$
C.20$20$
D.36$36$
Solution
### Related Formula
vecx cdot (vecy times vecz) = [vecx \ vecy \ vecz] = (vecx times vecy) cdot vecz$$\vec{x} \cdot (\vec{y} \times \vec{z}) = [\vec{x} \ \vec{y} \ \vec{z}] = (\vec{x} \times \vec{y}) \cdot \vec{z}$$
### Core Logic
The expression to evaluate is:
E = veca cdot ((vecc times vecb) - vecb - vecc)$$E = \vec{a} \cdot ((\vec{c} \times \vec{b}) - \vec{b} - \vec{c})$$
Distributing the dot product over the terms gives:
E = veca cdot (vecc times vecb) - veca cdot vecb - veca cdot vecc$$E = \vec{a} \cdot (\vec{c} \times \vec{b}) - \vec{a} \cdot \vec{b} - \vec{a} \cdot \vec{c}$$
### Step 1: Evaluating the Scalar Triple Product
The first term is a scalar triple product:
veca cdot (vecc times vecb) = (veca times vecc) cdot vecb$$\vec{a} \cdot (\vec{c} \times \vec{b}) = (\vec{a} \times \vec{c}) \cdot \vec{b}$$
We are given that veca times vecc = vecb$\vec{a} \times \vec{c} = \vec{b}$. Substitute this in:
(vecb) cdot vecb = |vecb|^2$$(\vec{b}) \cdot \vec{b} = |\vec{b}|^2$$
Given vecb = 3hati - 3hatj + 3hatk$\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$, its magnitude squared is:
|vecb|^2 = 3^2 + (-3)^2 + 3^2 = 9 + 9 + 9 = 27$$|\vec{b}|^2 = 3^2 + (-3)^2 + 3^2 = 9 + 9 + 9 = 27$$
### Step 2: Evaluating the remaining Dot Products
For the second term, calculate veca cdot vecb$\vec{a} \cdot \vec{b}$:
veca = 1hati + 2hatj + 1hatk$$\vec{a} = 1\hat{i} + 2\hat{j} + 1\hat{k}$$vecb = 3hati - 3hatj + 3hatk$$\vec{b} = 3\hat{i} - 3\hat{j} + 3\hat{k}$$veca cdot vecb = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0$$\vec{a} \cdot \vec{b} = (1)(3) + (2)(-3) + (1)(3) = 3 - 6 + 3 = 0$$
For the third term, we are explicitly given:
veca cdot vecc = 3$$\vec{a} \cdot \vec{c} = 3$$
### Step 3: Final Output Calculation
Substitute all individual values back into the expanded expression:
E = 27 - 0 - 3 = 24$$E = 27 - 0 - 3 = 24$$
### Pattern Recognition
When asked to evaluate complex vector expressions containing veca cdot (dots times dots)$\vec{a} \cdot (\dots \times \dots)$, immediately distribute and convert them into Scalar Triple Products [veca \ vecb \ vecc]$[\vec{a} \ \vec{b} \ \vec{c}]$. Cyclic permutations and given cross-product relationships will rapidly collapse the expression into simple magnitudes.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Vector Algebra
Q21jee_main_2024_27_jan_morningDot Product
The least positive integral value of alpha$\alpha$, for which the angle between the vectorsalphahati-2hatj+2hatk$\alpha\hat{i}-2\hat{j}+2\hat{k}$ and alphahati+2alphahatj-2hatk$\alpha\hat{i}+2\alpha\hat{j}-2\hat{k}$ is acute, is:
Numerical Answer.Answer: 5 to 5
Solution
### Related Formula
cos theta = fracveca cdot vecb|veca||vecb|$$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$$
For an acute angle, cos theta > 0$\cos \theta > 0$, which strictly means veca cdot vecb > 0$\vec{a} \cdot \vec{b} > 0$.
### Core Logic
Given vectors vecA = alphahati - 2hatj + 2hatk$\vec{A} = \alpha\hat{i} - 2\hat{j} + 2\hat{k}$ and vecB = alphahati + 2alphahatj - 2hatk$\vec{B} = \alpha\hat{i} + 2\alpha\hat{j} - 2\hat{k}$.
For the angle to be acute, their dot product must be strictly positive:
vecA cdot vecB > 0$$\vec{A} \cdot \vec{B} > 0$$(alpha)(alpha) + (-2)(2alpha) + (2)(-2) > 0$$(\alpha)(\alpha) + (-2)(2\alpha) + (2)(-2) > 0$$alpha^2 - 4alpha - 4 > 0$$\alpha^2 - 4\alpha - 4 > 0$$
### Step 1: Solving the Inequality
Complete the square to find critical points:
alpha^2 - 4alpha + 4 > 8$$\alpha^2 - 4\alpha + 4 > 8$$(alpha - 2)^2 > 8$$(\alpha - 2)^2 > 8$$
Extracting roots gives:
alpha - 2 > 2sqrt2 quad textor quad alpha - 2 < -2sqrt2$$\alpha - 2 > 2\sqrt{2} \quad \text{or} \quad \alpha - 2 < -2\sqrt{2}$$alpha > 2 + 2sqrt2 quad textor quad alpha < 2 - 2sqrt2$$\alpha > 2 + 2\sqrt{2} \quad \text{or} \quad \alpha < 2 - 2\sqrt{2}$$
### Step 2: Finding Least Positive Integer
Approximate the boundary values. Since sqrt2 approx 1.414$\sqrt{2} \approx 1.414$:
2 + 2(1.414) = 4.828$2 + 2(1.414) = 4.828$
The ranges are alpha in (-infty, -0.828) cup (4.828, infty)$\alpha \in (-\infty, -0.828) \cup (4.828, \infty)$.
We need the *least positive integral value* of alpha$\alpha$.
Looking at the interval (4.828, infty)$(4.828, \infty)$, the smallest integer present is 5$5$.
### Pattern Recognition
Acute angle translates directly to a positive dot product. Set up the quadratic inequality, compute numerical bounds of irrational roots, and select the immediate next integer.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Maths: Vector Algebra
Class 11 Maths: Linear Inequalities
Q8jee_main_2024_29_jan_morningCollinear Vectors
Let veca$\vec{a}$, vecb$\vec{b}$ and vecc$\vec{c}$ be three non-zero vectors such that vecb$\vec{b}$ and vecc$\vec{c}$ are non-collinear. If veca+5vecb$\vec{a}+5\vec{b}$ is collinear with vecc$\vec{c}$, vecb+6vecc$\vec{b}+6\vec{c}$ is collinear with veca$\vec{a}$ and veca+alphavecb+betavecc=0$\vec{a}+\alpha\vec{b}+\beta\vec{c}=0$, then alpha+beta$\alpha+\beta$ is equal to
A.35$35$
B.30$30$
C.-30$-30$
D.-25$-25$
Solution
### Related Formula
textIf vecX text and vecY text are collinear, then vecX = lambda vecY text for some scalar lambda.$$\text{If } \vec{X} \text{ and } \vec{Y} \text{ are collinear, then } \vec{X} = \lambda \vec{Y} \text{ for some scalar } \lambda.$$
Linear Independence: If vecu$\vec{u}$ and vecv$\vec{v}$ are non-collinear, then xvecu + yvecv = 0 implies x = 0 text and y = 0$x\vec{u} + y\vec{v} = 0 \implies x = 0 \text{ and } y = 0$.
### Core Logic
Based on the problem statement:
1) veca + 5vecb$\vec{a} + 5\vec{b}$ is collinear with vecc$\vec{c}$:
* quad veca + 5vecb = lambdavecc quad dots(1)$* \quad \vec{a} + 5\vec{b} = \lambda\vec{c} \quad \dots(1)$
2) vecb + 6vecc$\vec{b} + 6\vec{c}$ is collinear with veca$\vec{a}$:
* quad vecb + 6vecc = muveca quad dots(2)$* \quad \vec{b} + 6\vec{c} = \mu\vec{a} \quad \dots(2)$
### Step 1: Eliminate Vector a
From equation (1), isolate veca$\vec{a}$:
veca = lambdavecc - 5vecb$$\vec{a} = \lambda\vec{c} - 5\vec{b}$$
Substitute this into equation (2):
vecb + 6vecc = mu(lambdavecc - 5vecb)$$\vec{b} + 6\vec{c} = \mu(\lambda\vec{c} - 5\vec{b})$$vecb + 6vecc = mulambdavecc - 5muvecb$$\vec{b} + 6\vec{c} = \mu\lambda\vec{c} - 5\mu\vec{b}$$
Rearrange to group coefficients for vecb$\vec{b}$ and vecc$\vec{c}$:
(1 + 5mu)vecb + (6 - mulambda)vecc = 0$$(1 + 5\mu)\vec{b} + (6 - \mu\lambda)\vec{c} = 0$$
### Step 2: Determine Scalars
Since vectors vecb$\vec{b}$ and vecc$\vec{c}$ are given as non-collinear, their linear combination resolving to zero dictates that their scalar coefficients must both individually be zero.
1 + 5mu = 0 Rightarrow mu = -frac15$$1 + 5\mu = 0 \Rightarrow \mu = -\frac{1}{5}$$6 - mulambda = 0 Rightarrow 6 - left(-frac15right)lambda = 0 Rightarrow 6 + fraclambda5 = 0$$6 - \mu\lambda = 0 \Rightarrow 6 - \left(-\frac{1}{5}\right)\lambda = 0 \Rightarrow 6 + \frac{\lambda}{5} = 0$$fraclambda5 = -6 Rightarrow lambda = -30$$\frac{\lambda}{5} = -6 \Rightarrow \lambda = -30$$
### Step 3: Match the Final Equation
Substitute lambda = -30$\lambda = -30$ back into equation (1):
veca + 5vecb = -30vecc$$\vec{a} + 5\vec{b} = -30\vec{c}$$veca + 5vecb + 30vecc = 0$$\vec{a} + 5\vec{b} + 30\vec{c} = 0$$
The problem provides the structure veca + alphavecb + betavecc = 0$\vec{a} + \alpha\vec{b} + \beta\vec{c} = 0$. Comparing the two yields:
alpha = 5, quad beta = 30$$\alpha = 5, \quad \beta = 30$$
Thus:
alpha + beta = 5 + 30 = 35$$\alpha + \beta = 5 + 30 = 35$$
### Pattern Recognition
Double-collinearity equations should always be mapped out by eliminating the "third" vector (veca$\vec{a}$ in this case) and funneling everything into the two known non-collinear vectors. The resulting zero-equation perfectly exposes the scalar unknowns.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Vector Algebra
More Vector Algebra Questions — jee_main_2026_21_jan_morning
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