Let the position vectors of three vertices of a $$$$$$\triangle$$$be$ be $4\vec{p}+\vec{q}-3\vec{r},$, $-5\vec{p}+\vec{q}+2\vec{r}and$ and $2\vec{p}-\vec{q}+2\vec{r}If the position vectors of the orthocenter and the circumcenter of the$ If the position vectors of the orthocenter and the circumcenter of the $\triangleare$ are $\frac{\vec{p}+\vec{q}+\vec{r}}{4}and$ and $\alpha\vec{p}+\beta\vec{q}+\gamma\vec{r}respectively, then$ respectively, then $\alpha+2\beta+5\gamma$ is equal to:

Related Formula

Core Logic

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$\vec{a} \times \vec{c}$:

Step 1: Equate components to find c

Equating this to b$\vec{b}$: 2c₃ - 2c₂ = 8 ⇒ c₃ - c₂ = 4$2c_3 - 2c_2 = 8 \Rightarrow c_3 - c_2 = 4$ c₃ + 2c₁ = 7$c_3 + 2c_1 = 7$ -(c₂ + 2c₁) = -3 ⇒ c₂ + 2c₁ = 3$-(c_2 + 2c_1) = -3 \Rightarrow c_2 + 2c_1 = 3$

We are also given c · ( i + j + k) = 4$\vec{c} \cdot (\hat{i} + \hat{j} + \hat{k}) = 4$:

Step 2: Solve the linear system

From the dot product equation: c₁ = 4 - c₂ - c₃$c_1 = 4 - c_2 - c_3$. Substitute into c₂ + 2c₁ = 3$c_2 + 2c_1 = 3$:

We have the system: c₃ - c₂ = 4 ⇒ c₂ = c₃ - 4$c_3 - c_2 = 4 \Rightarrow c_2 = c_3 - 4$ Substitute into above: (c₃ - 4) + 2c₃ = 5 ⇒ 3c₃ = 9 ⇒ c₃ = 3$(c_3 - 4) + 2c_3 = 5 \Rightarrow 3c_3 = 9 \Rightarrow c_3 = 3$ Then c₂ = 3 - 4 = -1$c_2 = 3 - 4 = -1$. Then c₁ = 4 - (-1) - 3 = 2$c_1 = 4 - (-1) - 3 = 2$. So, c = 2 i - j + 3 k$\vec{c} = 2\hat{i} - \hat{j} + 3\hat{k}$.

Step 3: Evaluate the final expression

Pattern Recognition

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.

Chapter Mix

Class 12 Maths: Vector Algebra

Related Formula

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)$|\vec{c} + \vec{d}| = \sqrt{29}$

Therefore, c + d = ±(2 i + 3 j + 4 k)$\vec{c} + \vec{d} = \pm(2\hat{i} + 3\hat{j} + 4\hat{k})$.

Step 2: Evaluate the dot product limits

We need values of ( c + d) · (-7 i + 2 j + 3 k)$(\vec{c} + \vec{d}) \cdot (-7\hat{i} + 2\hat{j} + 3\hat{k})$. Dot product = λ(2(-7) + 3(2) + 4(3)) = λ(-14 + 6 + 12) = 4λ$= \lambda(2(-7) + 3(2) + 4(3)) = \lambda(-14 + 6 + 12) = 4\lambda$. Since λ = ± 1$\lambda = \pm 1$, the possible values are 4$4$ and -4$-4$. Given λ₁ > λ₂$\lambda_1 > \lambda_2$, we have λ₁ = 4$\lambda_1 = 4$ and λ₂ = -4$\lambda_2 = -4$.

Step 3: Apply circle constraints

Substitute λ₁$\lambda_1$ and λ₂$\lambda_2$ into the conic equation:

For this to represent a circle:

• Coefficient of xy$xy$ must be 0$0$:

• Coefficient of x²$x^2$ must equal coefficient of y²$y^2$:

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.

Chapter Mix

Class 12 Maths: Vector Algebra Class 11 Maths: Conic Sections

Related Formula

Core Logic

Step 1: Calculate magnitude of r

Using cosine rule for the side | r|$|\vec{r}|$:

Since θ = 1/√(3)$\cos\theta = 1/\sqrt{3}$, (π - θ) = -1/√(3)$\cos(\pi - \theta) = -1/\sqrt{3}$.

Step 2: Simplify the Cross Product term

We need to evaluate | p × ( q - 3 r)|²$|\vec{p} \times (\vec{q} - 3\vec{r})|^2$. Substitute r = p + q$\vec{r} = \vec{p} + \vec{q}$:

Since p × p = 0$\vec{p} \times \vec{p} = 0$:

Step 3: Final Calculation

Given θ = 1/√(3) ²θ = 1/3 ²θ = 2/3$\cos\theta = 1/\sqrt{3} \implies \cos^2\theta = 1/3 \implies \sin^2\theta = 2/3$.

The required expression is:

Pattern Recognition

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.

Chapter Mix

Class 12 Maths: Vector Algebra

Related Formula

Core Logic

Using the parallelogram law of vector addition, the diagonal AC$\overrightarrow{AC}$ is given by:

Step 1: Applying the Projection Condition

The projection of v = i+ j+ k$\vec{v} = \hat{i}+\hat{j}+\hat{k}$ on AC$\overrightarrow{AC}$ is 1. Therefore:

Squaring both sides:

Step 2: Solving the Quadratic Equation

Substitute λ = 3$\lambda = 3$ into the given quadratic equation λ²x² - 6λ x + 5 = 0$\lambda^2x^2 - 6\lambda x + 5 = 0$:

Given α > β$\alpha > \beta$, we have α = (5)/(3)$\alpha = \frac{5}{3}$ and β = (1)/(3)$\beta = \frac{1}{3}$.

Calculate 2α - β$2\alpha - \beta$:

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.

Chapter Mix

Class 12 Maths: Vector Algebra Class 11 Maths: Quadratic Equations

Related Formula

Cauchy-Schwarz Inequality for vectors:

Core Logic

Calculate cross product c = a × b$\vec{c} = \vec{a} \times \vec{b}$:

Given | c| = √(53)$|\vec{c}| = \sqrt{53}$:

Since λ in Z$\lambda \in \mathbb{Z}$, we get λ = -3$\lambda = -3$. Thus, c = i - 4 j - 6 k$\vec{c} = \hat{i} - 4\hat{j} - 6\hat{k}$.

Step 1: Maximum Value of Dot Product Squared

Vector d$\vec{d}$ lies in yz-plane: d = y j + z k$\vec{d} = y\hat{j} + z\hat{k}$ with y² + z² = 4$y^2 + z^2 = 4$.

By Cauchy-Schwarz inequality:

Pattern Recognition

Project c$\vec{c}$ onto yz-plane components and apply Cauchy-Schwarz inequality to maximize dot product.

Chapter Mix

Class 12 Maths: Vector Algebra