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 3
Q75jee_main_2025_04_april_eveningProperties of Vectors in Triangles
Let the three sides of a triangle ABC be given by the vectors 2hatmathbfi - hatmathbfj + hatmathbfk$2\hat{\mathbf{i}} - \hat{\mathbf{j}} + \hat{\mathbf{k}}$ , hatmathbfi - 3hatmathbfj - 5hatmathbfk$\hat{\mathbf{i}} - 3\hat{\mathbf{j}} - 5\hat{\mathbf{k}}$ and 3hatmathbfi - mathbf4hatmathbfj - mathbf4hatmathbfk$3\hat{\mathbf{i}} - \mathbf{4}\hat{\mathbf{j}} - \mathbf{4}\hat{\mathbf{k}}$ . Let G be the centroid of the triangle ABC. Then 6leftleft|overlineAGright|^2 + left|overlineBGright|^2 + left|overlineCGright|^2right)$6\left\left|\overline{AG}\right|^2 + \left|\overline{BG}\right|^2 + \left|\overline{CG}\right|^2\right)$ is equal to
Q64jee_main_2025_04_april_morningComponents of Vectors
Consider two vectors vecu = 3hati - hatj$\vec{u} = 3\hat{i} - \hat{j}$ and vecv = 2hati + hatj - lambda hatk$\vec{v} = 2\hat{i} + \hat{j} - \lambda \hat{k}$, where lambda > 0$\lambda > 0$. The angle between them is given by cos^-1left(fracsqrt52sqrt7right)$\cos^{-1}\left(\frac{\sqrt{5}}{2sqrt{7}}\right)$. Let vecv = vecv_1 + vecv_2$\vec{v} = \vec{v}_1 + \vec{v}_2$, where vecv_1$\vec{v}_1$ is parallel to vecu$\vec{u}$ and vecv_2$\vec{v}_2$ is perpendicular to vecu$\vec{u}$. Then the value |vecv_1|^2 + |vecv_2|^2$|\vec{v}_1|^2 + |\vec{v}_2|^2$ is equal to
A.frac232$\frac{23}{2}$
B. 14
C.frac252$\frac{25}{2}$
D. 10
Solution
### Related Formula
By orthogonal vector decomposition (Pythagorean property):
|vecv|^2 = |vecv_1|^2 + |vecv_2|^2 quad textwhen vecv_1 cdot vecv_2 = 0$$|\vec{v}|^2 = |\vec{v}_1|^2 + |\vec{v}_2|^2 \quad \text{when } \vec{v}_1 \cdot \vec{v}_2 = 0$$
### Core Logic
Compute lambda$\lambda$ using dot product formula:
costheta = fracvecu cdot vecv|vecu||vecv| implies fracsqrt52sqrt7 = frac3(2) + (-1)(1)sqrt3^2 + (-1)^2 sqrt2^2 + 1^2 + (-lambda)^2$$\cos\theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}||\vec{v}|} \implies \frac{\sqrt{5}}{2\sqrt{7}} = \frac{3(2) + (-1)(1)}{\sqrt{3^2 + (-1)^2} \sqrt{2^2 + 1^2 + (-\lambda)^2}}$$fracsqrt52sqrt7 = frac5sqrt10sqrt5 + lambda^2 implies frac12sqrt7 = fracsqrt5sqrt10sqrt5 + lambda^2 = frac1sqrt2sqrt5 + lambda^2$$\frac{\sqrt{5}}{2\sqrt{7}} = \frac{5}{\sqrt{10}\sqrt{5 + \lambda^2}} \implies \frac{1}{2\sqrt{7}} = \frac{\sqrt{5}}{\sqrt{10}\sqrt{5 + \lambda^2}} = \frac{1}{\sqrt{2}\sqrt{5 + \lambda^2}}$$
### Step 1: Solve for lambda
Square both sides of equation:
frac128 = frac12(5 + lambda^2) implies 2(5 + lambda^2) = 28 implies 5 + lambda^2 = 14 implies lambda^2 = 9 implies lambda = 3$$\frac{1}{28} = \frac{1}{2(5 + \lambda^2)} \implies 2(5 + \lambda^2) = 28 \implies 5 + \lambda^2 = 14 \implies \lambda^2 = 9 \implies \lambda = 3$$
Since vecv = 2hati + hatj - 3hatk$\vec{v} = 2\hat{i} + \hat{j} - 3\hat{k}$.
### Step 2: Apply Identity
Since components are orthogonal, direct magnitude squared holds:
|vecv_1|^2 + |vecv_2|^2 = |vecv|^2 = 2^2 + 1^2 + (-3)^2 = 4 + 1 + 9 = 14$$|\vec{v}_1|^2 + |\vec{v}_2|^2 = |\vec{v}|^2 = 2^2 + 1^2 + (-3)^2 = 4 + 1 + 9 = 14$$
### Pattern Recognition
Do not waste time explicitly projecting components vecv_1$\vec{v}_1$ and vecv_2$\vec{v}_2$ if only the sum of their squared magnitudes is requested. The scalar length matches the total vector length invariant under any orthogonal basis change.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Vector Algebra
Q52jee_main_2025_07_april_eveningVector Magnitude and Operations
Let veca$\vec{a}$ and vecb$\vec{b}$ be the vectors of the same magnitude such that frac|veca + vecb| + |veca - vecb||veca + vecb| - |veca - vecb| = sqrt2 + 1$\frac{|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|}{|\vec{a} + \vec{b}| - |\vec{a} - \vec{b}|} = \sqrt{2} + 1$. Then frac|veca + vecb|^2|veca|^2$\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2}$ is:
Q56jee_main_2025_24_jan_eveningCentroid, Orthocenter, and Circumcenter
Let the position vectors of three vertices of a \triangle be 4vecp+vecq-3vecr$4\vec{p}+\vec{q}-3\vec{r}$, -5vecp+vecq+2vecr$-5\vec{p}+\vec{q}+2\vec{r}$ and 2vecp-vecq+2vecr$2\vec{p}-\vec{q}+2\vec{r}$ If the position vectors of the orthocenter and the circumcenter of the \triangle are fracvecp+vecq+vecr4$\frac{\vec{p}+\vec{q}+\vec{r}}{4}$ and alphavecp+betavecq+gammavecr$\alpha\vec{p}+\beta\vec{q}+\gamma\vec{r}$ respectively, then alpha+2beta+5gamma$\alpha+2\beta+5\gamma$ is equal to: [cite: 3266, 3267, 3268, 3269, 3270, 3271, 3272]
A.3$3$
B.1$1$
C.6$6$
D.4$4$
Solution
### Related Formula
1. Centroid (G$G$) of a \triangle with vertices A, B, C$A, B, C$ is given by:
vecG = fracvecA + vecB + vecC3$$\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}$$
2. Euler\'s line property: The orthocenter (O$O$), centroid (G$G$), and circumcenter (C$C$) are collinear, and G$G$ divides the segment OC$OC$ internally in the ratio 2:1$2:1$.
### Step 1: Compute the Centroid Vector
Sum the vectors of the three given vertices [cite: 3266, 3268]:
vecA = 4vecp+vecq-3vecr$$\vec{A} = 4\vec{p}+\vec{q}-3\vec{r}$$vecB = -5vecp+vecq+2vecr$$\vec{B} = -5\vec{p}+\vec{q}+2\vec{r}$$vecC = 2vecp-vecq+2vecr$$\vec{C} = 2\vec{p}-\vec{q}+2\vec{r}$$vecG = frac(4 - 5 + 2)vecp + (1 + 1 - 1)vecq + (-3 + 2 + 2)vecr3 = fracvecp + vecq + vecr3$$\vec{G} = \frac{(4 - 5 + 2)\vec{p} + (1 + 1 - 1)\vec{q} + (-3 + 2 + 2)\vec{r}}{3} = \frac{\vec{p} + \vec{q} + \vec{r}}{3}$$
### Step 2: Apply Euler Line Section Ratio
Using the section formula ratio O-G-C$O-G-C$ as 2:1$2:1$ [cite: 3931, 3932]:
Euler Line section diagram for Q56 - JEE Main 2025 EveningvecG = frac2vecC + vecO3 Rightarrow 3vecG = 2vecC + vecO$$\vec{G} = \frac{2\vec{C} + \vec{O}}{3} \Rightarrow 3\vec{G} = 2\vec{C} + \vec{O}$$2vecC = 3vecG - vecO = 3left(fracvecp + vecq + vecr3right) - fracvecp + vecq + vecr4$$2\vec{C} = 3\vec{G} - \vec{O} = 3\left(\frac{\vec{p} + \vec{q} + \vec{r}}{3}\right) - \frac{\vec{p} + \vec{q} + \vec{r}}{4}$$2vecC = (vecp + vecq + vecr) - frac14(vecp + vecq + vecr) = frac34(vecp + vecq + vecr)$$2\vec{C} = (\vec{p} + \vec{q} + \vec{r}) - \frac{1}{4}(\vec{p} + \vec{q} + \vec{r}) = \frac{3}{4}(\vec{p} + \vec{q} + \vec{r})$$vecC = frac38vecp + frac38vecq + frac38vecr$$\vec{C} = \frac{3}{8}\vec{p} + \frac{3}{8}\vec{q} + \frac{3}{8}\vec{r}$$
### Step 3: Coefficient Matching
Compare with the given circumcenter format alphavecp + betavecq + gammavecr$\alpha\vec{p} + \beta\vec{q} + \gamma\vec{r}$ [cite: 3270, 3939]:
alpha = frac38, quad beta = frac38, quad gamma = frac38$$\alpha = \frac{3}{8}, \quad \beta = \frac{3}{8}, \quad \gamma = \frac{3}{8}$$
Calculate alpha + 2beta + 5gamma$\alpha + 2\beta + 5\gamma$ [cite: 3272, 3949]:
frac38 + 2left(frac38right) + 5left(frac38right) = frac3 + 6 + 158 = frac248 = 3$$\frac{3}{8} + 2\left(\frac{3}{8}\right) + 5\left(\frac{3}{8}\right) = \frac{3 + 6 + 15}{8} = \frac{24}{8} = 3$$
### Pattern Recognition
Euler line configuration is universally O-G-C$O-G-C$ in 2:1$2:1$. Remember the mnemonic 'Oil-Gas-Company' or simply 3G = 2C + O$3G = 2C + O$ to prevent swapping structural coefficients under exam stress.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Mathematics: Vector Algebra
Class 11 Mathematics: Properties of Triangles
Q64jee_main_2025_24_jan_eveningVector Triple Product and Projection
Let veca=3hati-hatj+2hatk,$\vec{a}=3\hat{i}-\hat{j}+2\hat{k},$vecb=vecatimes(hati-2hatk)$\vec{b}=\vec{a}\times(\hat{i}-2\hat{k})$ and vecc=vecbtimeshatk$\vec{c}=\vec{b}\times\hat{k}$. Then the projection of vecc-2hatj$\vec{c}-2\hat{j}$ on veca$\vec{a}$ is: [cite: 3358, 3359, 3364]
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