Let AB = 2î + 4ĵ – 5k̂ and AD = î + 2ĵ + λk̂, λ ∈ ℝ. Let the projection of the vector v = î + ĵ + k̂ on the diagonal AC of the parallelogram ABCD be of length one unit. If α, β, where α > β, be the roots of the equation λ²x² – 6λx + 5 = 0, then 2α – β is equal t

The projection of vector $\vec{a}$ onto vector $\vec{b}$ is the component of $\vec{a}$ in the direction of $\vec{b}$. It represents how much of $\vec{a}$ lies along $\vec{b}$. The scalar projection (signed length) is given by $\text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|}$. The vector projection is $\text{proj}_{\vec{b}} \vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2} \vec{b}$. Key properties include: projection is always in the direction of $\vec{b}$ (or opposite if negative), the magnitude of projection is $|\text{proj}_{\vec{b}} \vec{a}| = \frac{|\vec{a} \cdot \vec{b}|}{|\vec{b}|}$, and if $\vec{a} \perp \vec{b}$, then projection is zero. The dot product $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$ where $\theta$ is the angle between vectors. Therefore, $\text{proj}_{\vec{b}} \vec{a} = |\vec{a}|\cos\theta$. Geometrically, drop a perpendicular from the tip of $\vec{a}$ onto the line containing $\vec{b}$; the signed distance from origin to this point is the scalar projection. Applications include finding work done by a force ($W = \vec{F} \cdot \vec{d}$), decomposing vectors into components parallel and perpendicular to a given direction, and solving optimization problems in physics and engineering. In JEE problems, projection often appears with parallelograms, tetrahedrons, and coordinate geometry, requiring combination with other vector operations like cross product and triple scalar product.

In vector notation, a parallelogram $ABCD$ with vertex $A$ and adjacent sides $\overrightarrow{AB}$ and $\overrightarrow{AD}$ has powerful properties. The diagonal vectors are $\overrightarrow{AC} = \overrightarrow{AB} + \overrightarrow{AD}$ (main diagonal) and $\overrightarrow{BD} = \overrightarrow{AD} – \overrightarrow{AB}$ (other diagonal). This follows from the parallelogram law of vector addition. The diagonals bisect each other: if $O$ is the intersection point, then $\overrightarrow{AO} = \frac{1}{2}\overrightarrow{AC}$ and $\overrightarrow{BO} = \frac{1}{2}\overrightarrow{BD}$. The area of parallelogram is $|\overrightarrow{AB} \times \overrightarrow{AD}|$, which equals the magnitude of the cross product of adjacent sides. For a rhombus (all sides equal), $|\overrightarrow{AB}| = |\overrightarrow{AD}|$. For a rectangle, $\overrightarrow{AB} \perp \overrightarrow{AD}$, meaning $\overrightarrow{AB} \cdot \overrightarrow{AD} = 0$. The diagonals of a rectangle are equal: $|\overrightarrow{AC}| = |\overrightarrow{BD}|$. The midpoint of diagonal $AC$ is at position $\vec{A} + \frac{1}{2}(\overrightarrow{AB} + \overrightarrow{AD})$. The opposite sides are parallel and equal: $\overrightarrow{AB} = \overrightarrow{DC}$ and $\overrightarrow{AD} = \overrightarrow{BC}$. Understanding these vector relationships is crucial for solving geometry problems efficiently in JEE, particularly when combined with dot products, cross products, and projections.

The dot product (scalar product) of two vectors $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$ is defined as $\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$. Geometrically, $\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos\theta$ where $\theta$ is the angle between the vectors. The dot product is commutative: $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$, and distributive: $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$. Important properties: $\vec{a} \cdot \vec{a} = |\vec{a}|^2$ (magnitude squared), $\vec{a} \perp \vec{b}$ if and only if $\vec{a} \cdot \vec{b} = 0$ (perpendicularity test), and the angle between vectors is $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}$. For unit vectors: $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$ and $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$. Applications include finding work ($W = \vec{F} \cdot \vec{s}$), power ($P = \vec{F} \cdot \vec{v}$), determining if vectors are orthogonal, computing projections, and solving geometric problems involving angles and distances. The dot product is scalar-valued, unlike the cross product which is vector-valued. Mastering dot product manipulation is essential for JEE vector problems.

For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, Vieta’s formulas give: sum of roots $\alpha + \beta = -\frac{b}{a}$ and product of roots $\alpha\beta = \frac{c}{a}$. These relationships are derived from the factored form $(x – \alpha)(x – \beta) = x^2 – (\alpha + \beta)x + \alpha\beta$. When asked to find expressions like $2\alpha – \beta$, $\alpha^2 + \beta^2$, or $\frac{1}{\alpha} + \frac{1}{\beta}$, use Vieta’s formulas rather than solving explicitly. Common derived results: $\alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta$, $\alpha^3 + \beta^3 = (\alpha + \beta)^3 – 3\alpha\beta(\alpha + \beta)$, $|\alpha – \beta| = \sqrt{(\alpha + \beta)^2 – 4\alpha\beta}$, and $\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta}$. The discriminant $\Delta = b^2 – 4ac$ determines the nature of roots: $\Delta > 0$ gives two distinct real roots, $\Delta = 0$ gives one repeated real root (equal roots), $\Delta < 0$ gives two complex conjugate roots. For finding individual roots when needed, use the quadratic formula: $\alpha, \beta = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$. In JEE problems, Vieta's formulas save significant time and reduce calculation errors compared to explicit root-finding.

The magnitude (or length or norm) of a vector $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ is given by $|\vec{a}| = \sqrt{a_1^2 + a_2^2 + a_3^2}$. This is derived from the three-dimensional distance formula and represents the Euclidean length of the vector. In two dimensions, $|\vec{a}| = \sqrt{a_1^2 + a_2^2}$. Properties include: $|\vec{a}| \geq 0$ with equality if and only if $\vec{a} = \vec{0}$, $|k\vec{a}| = |k||\vec{a}|$ for scalar $k$, triangle inequality $|\vec{a} + \vec{b}| \leq |\vec{a}| + |\vec{b}|$, and reverse triangle inequality $||\vec{a}| – |\vec{b}|| \leq |\vec{a} – \vec{b}|$. A unit vector in the direction of $\vec{a}$ is $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$, which has magnitude 1. The distance between two points with position vectors $\vec{a}$ and $\vec{b}$ is $|\vec{b} – \vec{a}|$. For finding magnitude squared, use $|\vec{a}|^2 = \vec{a} \cdot \vec{a} = a_1^2 + a_2^2 + a_3^2$, which avoids taking square roots until necessary. In physics, magnitude represents quantities like speed (magnitude of velocity), force magnitude, and displacement. JEE problems often require computing magnitudes of vector sums, differences, cross products, and projections, making this a fundamental operation in vector algebra.

Vector Projection Formula Parallelogram Diagonal Dot Product Vieta’s Formulas Magnitude of Vectors Quadratic Roots