Let AB=2 i+4 j-5 k$\overrightarrow{AB}=2\hat{i}+4\hat{j}-5\hat{k}$ and AD= i+2 j+λ k, λin R$\overrightarrow{AD}=\hat{i}+2\hat{j}+\lambda \hat{k}, \lambda\in \mathbb{R}$. Let the projection of the vector v= i+ j+ k$\vec{v}=\hat{i}+\hat{j}+\hat{k}$ on the diagonal AC$\overrightarrow{AC}$ of the parallelogram ABCD be of length one unit. If α,β$\alpha,\beta$, where α>β$\alpha>\beta$, be the roots of the equation λ²x²-6λ x+5=0$\lambda^{2}x^{2}-6\lambda x+5=0$, then 2α-β$2\alpha-\beta$ is equal to

Related Formula

Componendo and Dividendo rule states that if (x)/(y) = (p)/(q)$\frac{x}{y} = \frac{p}{q}$, then:

(x+y)/(x-y) = (p+q)/(p-q)$$\frac{x+y}{x-y} = \frac{p+q}{p-q}$$

Core Logic

Given expression:

| a + b| + | a - b|| a + b| - | a - b| = √(2) + 11$$\frac{|\vec{a} + \vec{b}| + |\vec{a} - \vec{b}|}{|\vec{a} + \vec{b}| - |\vec{a} - \vec{b}|} = \frac{\sqrt{2} + 1}{1}$$

Applying Componendo and Dividendo:

2| a + b|2| a - b| = (√(2) + 1) + 1(√(2) + 1) - 1 = √(2) + 2√(2) = 1 + √(2)$$\frac{2|\vec{a} + \vec{b}|}{2|\vec{a} - \vec{b}|} = \frac{(\sqrt{2} + 1) + 1}{(\sqrt{2} + 1) - 1} = \frac{\sqrt{2} + 2}{\sqrt{2}} = 1 + \sqrt{2}$$

Squaring both sides:

| a + b|² = (1 + √(2))² | a - b|²$$|\vec{a} + \vec{b}|^2 = (1 + \sqrt{2})^2 |\vec{a} - \vec{b}|^2$$ | a + b|² = (3 + 2√(2)) | a - b|²$$|\vec{a} + \vec{b}|^2 = (3 + 2\sqrt{2}) |\vec{a} - \vec{b}|^2$$

Step 1: Vector Expansion

Expanding using dot products, keeping in mind that | a| = | b|$|\vec{a}| = |\vec{b}|$:

| a|² + | b|² + 2 a· b = (3 + 2√(2))(| a|² + | b|² - 2 a· b)$$|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b} = (3 + 2\sqrt{2})(|\vec{a}|^2 + |\vec{b}|^2 - 2\vec{a}\cdot\vec{b})$$ 2| a|² + 2 a· b = (3 + 2√(2))(2| a|² - 2 a· b)$$2|\vec{a}|^2 + 2\vec{a}\cdot\vec{b} = (3 + 2\sqrt{2})(2|\vec{a}|^2 - 2\vec{a}\cdot\vec{b})$$ 2| a|² (1 - (3 + 2√(2))) = -2 a· b (1 + 3 + 2√(2))$$2|\vec{a}|^2 (1 - (3 + 2\sqrt{2})) = -2\vec{a}\cdot\vec{b} (1 + 3 + 2\sqrt{2})$$

Simplifying directly leads to:

a· b| a|² = 2 + 2√(2)4 + 2√(2) = 1√(2)$$\frac{\vec{a}\cdot\vec{b}}{|\vec{a}|^2} = \frac{2 + 2\sqrt{2}}{4 + 2\sqrt{2}} = \frac{1}{\sqrt{2}}$$

Step 2: Final Calculation

We need to find | a + b|²| a|²$\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2}$:

| a + b|²| a|² = | a|² + | b|² + 2 a· b| a|² = 1 + 1 + 2 a· b| a|²$$\frac{|\vec{a} + \vec{b}|^2}{|\vec{a}|^2} = \frac{|\vec{a}|^2 + |\vec{b}|^2 + 2\vec{a}\cdot\vec{b}}{|\vec{a}|^2} = 1 + 1 + \frac{2\vec{a}\cdot\vec{b}}{|\vec{a}|^2}$$ = 2 + 2( 1√(2)) = 2 + √(2)$$= 2 + 2\left(\frac{1}{\sqrt{2}}\right) = 2 + \sqrt{2}$$

Pattern Recognition

Whenever symmetric sums and differences like | x|+| y|$|\vec{x}|+|\vec{y}|$ and | x|-| y|$|\vec{x}|-|\vec{y}|$ occur in ratios, Componendo-Dividendo should be applied immediately to isolate the ratio of the individual magnitudes.

Chapter Mix

Class 12 Physics: Vector Algebra Class 12 Mathematics: Vector Algebra

Related Formula

• Centroid (G$G$) of a $$$$$$\triangle$$with vertices $A, B, C$ is given by:

$$ with vertices $A, B, C$ is given by: $$\vec{G} = \frac{\vec{A} + \vec{B} + \vec{C}}{3}

• Euler's line property: The orthocenter ($O$), centroid ($G$), and circumcenter ($C$) are collinear, and $G$ divides the segment $OC$ internally in the ratio $2:1$.

Step 1: Compute the Centroid Vector

Sum the vectors of the three given vertices:

$$

• Euler's line property: The orthocenter ($O$), centroid ($G$), and circumcenter ($C$) are collinear, and $G$ divides the segment $OC$ internally in the ratio $2:1$.

Step 1: Compute the Centroid Vector

Sum the vectors of the three given vertices:

$$\vec{A} = 4\vec{p}+\vec{q}-3\vec{r}$$ $$\vec{B} = -5\vec{p}+\vec{q}+2\vec{r}$$ $$\vec{C} = 2\vec{p}-\vec{q}+2\vec{r}$$ $$\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 $O-G-C$ as $2:1$:

$$

Step 2: Apply Euler Line Section Ratio

Using the section formula ratio $O-G-C$ as $2:1$:

$$\vec{G} = \frac{2\vec{C} + \vec{O}}{3} \Rightarrow 3\vec{G} = 2\vec{C} + \vec{O}$$ $$2\vec{C} = 3\vec{G} - \vec{O} = 3\left(\frac{\vec{p} + \vec{q} + \vec{r}}{3}\right) - \frac{\vec{p} + \vec{q} + \vec{r}}{4}$$ $$2\vec{C} = (\vec{p} + \vec{q} + \vec{r}) - \frac{1}{4}(\vec{p} + \vec{q} + \vec{r}) = \frac{3}{4}(\vec{p} + \vec{q} + \vec{r})$$ $$\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 $α p + β q + γ r$:

$$

Step 3: Coefficient Matching

Compare with the given circumcenter format $\alpha\vec{p} + \beta\vec{q} + \gamma\vec{r}$:

$$\alpha = \frac{3}{8}, \quad \beta = \frac{3}{8}, \quad \gamma = \frac{3}{8}

Calculate $α + 2β + 5γ$:

$$

Calculate $\alpha + 2\beta + 5\gamma$:

$$\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$

Pattern Recognition

Euler line configuration is universally $O-G-Cin$ in $2:1. Remember the mnemonic 'Oil-Gas-Company' or simply$. Remember the mnemonic 'Oil-Gas-Company' or simply $3G = 2C + O$ to prevent swapping structural coefficients under exam stress.

Chapter Mix

Class 12 Mathematics: Vector Algebra Class 11 Mathematics: Properties of Triangles

Related Formula

The scalar projection of vector v$\vec{v}$ onto vector w$\vec{w}$ is calculated as:

Projection = v · w| w|$$\text{Projection} = \frac{\vec{v} \cdot \vec{w}}{|\vec{w}|}$$

Step 1: Calculate b$\vec{b}$

Compute the cross product using standard matrix expansion:

b = a × ( i - 2 k) = vmatrix i & j & k 3 & -1 & 2 1 & 0 & -2 vmatrix$$\vec{b} = \vec{a} \times (\hat{i} - 2\hat{k}) = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ 3 & -1 & 2 \\ 1 & 0 & -2 \end{vmatrix}$$ b = i(2 - 0) - j(-6 - 2) + k(0 - (-1)) = 2 i + 8 j + k$$\vec{b} = \hat{i}(2 - 0) - \hat{j}(-6 - 2) + \hat{k}(0 - (-1)) = 2\hat{i} + 8\hat{j} + \hat{k}$$

Step 2: Calculate c$\vec{c}$ and c - 2 j$\vec{c} - 2\hat{j}$

Perform the second cross product with unit vector k$\hat{k}$:

c = b × k = (2 i + 8 j + k) × k = 2( i × k) + 8( j × k) + 0$$\vec{c} = \vec{b} \times \hat{k} = (2\hat{i} + 8\hat{j} + \hat{k}) \times \hat{k} = 2(\hat{i} \times \hat{k}) + 8(\hat{j} \times \hat{k}) + \vec{0}$$ c = 2(- j) + 8( i) = 8 i - 2 j$$\vec{c} = 2(-\hat{j}) + 8(\hat{i}) = 8\hat{i} - 2\hat{j}$$

Subtract 2 j$2\hat{j}$:

c - 2 j = (8 i - 2 j) - 2 j = 8 i - 4 j$$\vec{c} - 2\hat{j} = (8\hat{i} - 2\hat{j}) - 2\hat{j} = 8\hat{i} - 4\hat{j}$$

Step 3: Compute the projection onto a$\vec{a}$

Using the dot product formula :

Projection = ( c - 2 j) · a| a| = 8, -4, 0 · 3, -1, 2 √(3² + (-1)² + 2²)$$\text{Projection} = \frac{(\vec{c} - 2\hat{j}) \cdot \vec{a}}{|\vec{a}|} = \frac{\langle 8, -4, 0 \rangle \cdot \langle 3, -1, 2 \rangle}{\sqrt{3^2 + (-1)^2 + 2^2}}$$ Projection = 24 + 4 + 0√(9 + 1 + 4) = 28√(14) = 2√(14)$$\text{Projection} = \frac{24 + 4 + 0}{\sqrt{9 + 1 + 4}} = \frac{28}{\sqrt{14}} = 2\sqrt{14}$$

Pattern Recognition

Keep cyclic unit cross products clear: i × k = - j$\hat{i} \times \hat{k} = -\hat{j}$ and j × k = i$\hat{j} \times \hat{k} = \hat{i}$. Missing a negative sign during basic cross multiplications ruins multi-step vector projections easily.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Related Formula

A vector c$\vec{\mathbf{c}}$ coplanar with a$\vec{\mathbf{a}}$ and b$\vec{\mathbf{b}}$ and perpendicular to b$\vec{\mathbf{b}}$ can be expressed using the vector triple product command template:

c = λ ( b × ( a × b)) = λ [( b · b) a - ( a · b) b]$$\vec{\mathbf{c}} = \lambda (\vec{\mathbf{b}} \times (\vec{\mathbf{a}} \times \vec{\mathbf{b}})) = \lambda [(\vec{\mathbf{b}} \cdot \vec{\mathbf{b}})\vec{\mathbf{a}} - (\vec{\mathbf{a}} \cdot \vec{\mathbf{b}})\vec{\mathbf{b}}]$$

Core Logic

First, find the dot products of the given vectors:

b · b = 3² + 1² + (-1)² = 9 + 1 + 1 = 11$$\vec{\mathbf{b}} \cdot \vec{\mathbf{b}} = 3^2 + 1^2 + (-1)^2 = 9 + 1 + 1 = 11$$ a · b = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 - 3 = 2$$\vec{\mathbf{a}} \cdot \vec{\mathbf{b}} = (1)(3) + (2)(1) + (3)(-1) = 3 + 2 - 3 = 2$$

Substituting these values into the expression for c$\vec{\mathbf{c}}$:

c = λ [11 a - 2 b]$$\vec{\mathbf{c}} = \lambda [11\vec{\mathbf{a}} - 2\vec{\mathbf{b}}]$$ c = λ [11( i + 2 j + 3 k) - 2(3 i + j - k)]$$\vec{\mathbf{c}} = \lambda [11(\hat{\mathbf{i}} + 2\hat{\mathbf{j}} + 3\hat{\mathbf{k}}) - 2(3\hat{\mathbf{i}} + \hat{\mathbf{j}} - \hat{\mathbf{k}})]$$ c = λ (5 i + 20 j + 35 k) = 5λ ( i + 4 j + 7 k)$$\vec{\mathbf{c}} = \lambda (5\hat{\mathbf{i}} + 20\hat{\mathbf{j}} + 35\hat{\mathbf{k}}) = 5\lambda (\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}})$$

Step 1: Determine Lambda

Using the given condition a · c = 5$\vec{\mathbf{a}} \cdot \vec{\mathbf{c}} = 5$:

a · [5λ ( i + 4 j + 7 k)] = 5$$\vec{\mathbf{a}} \cdot [5\lambda (\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}})] = 5$$ 5λ (1 · 1 + 2 · 4 + 3 · 7) = 5$$5\lambda (1 \cdot 1 + 2 \cdot 4 + 3 \cdot 7) = 5$$ 5λ (1 + 8 + 21) = 5 30λ = 1 λ = (1)/(30)$$5\lambda (1 + 8 + 21) = 5 \implies 30\lambda = 1 \implies \lambda = \frac{1}{30}$$

Thus, the vector c$\vec{\mathbf{c}}$ is:

c = (5)/(30)( i + 4 j + 7 k) = (1)/(6)( i + 4 j + 7 k)$$\vec{\mathbf{c}} = \frac{5}{30}(\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}}) = \frac{1}{6}(\hat{\mathbf{i}} + 4\hat{\mathbf{j}} + 7\hat{\mathbf{k}})$$

Step 2: Calculate Magnitude

The magnitude of c$\vec{\mathbf{c}}$ is evaluated as:

| c| = √(1² + 4² + 7²)6 = √(1 + 16 + 49)6 = √(66)6 = √((66)/(36)) = √((11)/(6))$$\|\vec{\mathbf{c}}\| = \frac{\sqrt{1^2 + 4^2 + 7^2}}{6} = \frac{\sqrt{1 + 16 + 49}}{6} = \frac{\sqrt{66}}{6} = \sqrt{\frac{66}{36}} = \sqrt{\frac{11}{6}}$$

Pattern Recognition

Whenever a vector is specified to be coplanar with a, b$\vec{\mathbf{a}}, \vec{\mathbf{b}}$ and perpendicular to b$\vec{\mathbf{b}}$, direct setup with the cross-product template b × ( a × b)$\vec{\mathbf{b}} \times (\vec{\mathbf{a}} \times \vec{\mathbf{b}})$ circumvents solving cumbersome linear systems of scalar variables.

Chapter Mix

Class 12 Mathematics: Vector Algebra

Related Formula

The line bisecting the angle between two symmetric vectors passing through the origin in the first quadrant is given by y = x$y = x$ or x - y = 0$x - y = 0$.

Distance from point (x₁, y₁)$(x_1, y_1)$ to line Ax + By + C = 0$Ax + By + C = 0$ is:

d = |Ax₁ + By₁ + C|√(A² + B²)$$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$

Core Logic

Vectors OA = √(3) i+ j$\overline{OA} = \sqrt{3}\hat{i}+\hat{j}$ and OB = i+√(3) j$\overline{OB} = \hat{i}+\sqrt{3}\hat{j}$ are symmetric about the line y = x$y = x$. Therefore, the angle bisector of OA$\overline{OA}$ and OB$\overline{OB}$ is the line x - y = 0$x - y = 0$.

Point C has coordinates (a, 1 - a)$(a, 1 - a)$.

Step 1: Calculate Distance and Solve for a

The perpendicular distance from C(a, 1 - a)$C(a, 1 - a)$ to x - y = 0$x - y = 0$ is:

d = |a - (1 - a)|√(1² + (-1)²) = |2a - 1|√(2)$$d = \frac{|a - (1 - a)|}{\sqrt{1^2 + (-1)^2}} = \frac{|2a - 1|}{\sqrt{2}}$$

Given that this distance is 9√(2)$\frac{9}{\sqrt{2}}$:

|2a - 1|√(2) = 9√(2) |2a - 1| = 9$$\frac{|2a - 1|}{\sqrt{2}} = \frac{9}{\sqrt{2}} \implies |2a - 1| = 9$$

This gives two solutions:

• 2a - 1 = 9 2a = 10 a = 5$2a - 1 = 9 \implies 2a = 10 \implies a = 5$
• 2a - 1 = -9 2a = -8 a = -4$2a - 1 = -9 \implies 2a = -8 \implies a = -4$

Step 2: Find the Sum of Values

Sum of all possible values of a$a$:

Sum = 5 + (-4) = 1$$\text{Sum} = 5 + (-4) = 1$$

Pattern Recognition

Notice that OA$\overline{OA}$ and OB$\overline{OB}$ have swapped coordinates, meaning they are symmetric with respect to y=x$y=x$. Thus, the angle bisector equation is immediate (x-y=0$x-y=0$), simplifying the problem to a standard point-to-line distance calculation.

Chapter Mix

Class 11 Mathematics: Straight Lines Class 12 Mathematics: Vector Algebra