JEE Advanced Vectors & 3D Geometry Cheat Sheet

This comprehensive cheat sheet is designed to be your ultimate companion for mastering Vectors and 3D Geometry for the JEE Advanced examination. These topics are not only frequently tested but also foundational for higher-level physics and engineering concepts. A solid understanding here can significantly boost your overall score.

JEE Advanced Vectors & 3D Geometry: The Ultimate Cheat Sheet

Exam Weightage & Importance

Vectors and 3D Geometry consistently contribute a significant portion to the JEE Advanced Mathematics paper. Typically, you can expect 2-3 questions, sometimes even 4, directly from these topics, accounting for 8-12% of the total mathematics marks. Many physics problems (mechanics, electromagnetism) also rely heavily on vector operations. Mastering this chapter is crucial not just for direct marks but also for developing a strong conceptual base.

These topics demand a blend of spatial visualization, algebraic manipulation, and geometric understanding. While formulas are important, their application in various problem contexts is paramount.

Vectors: The Foundational Concepts

Vectors are quantities possessing both magnitude and direction. They are represented geometrically as directed line segments.

Basic Definitions & Representation

1. Scalar Quantity: A quantity characterized only by magnitude (e.g., mass, distance, speed).
2. Vector Quantity: A quantity characterized by both magnitude and direction (e.g., displacement, velocity, force).
3. Position Vector: The vector $\vec{OA}$ of a point $A$ with respect to the origin $O$. If $A=(x,y,z)$, then $\vec{OA} = x\hat{i} + y\hat{j} + z\hat{k}$.
4. Magnitude of a Vector: If $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$, its magnitude is denoted as $|\vec{a}|$ and calculated as: $$|\vec{a}| = \sqrt{x^2 + y^2 + z^2}$$
5. Direction Cosines: For $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$, the direction cosines are $l = \frac{x}{|\vec{a}|}$, $m = \frac{y}{|\vec{a}|}$, $n = \frac{z}{|\vec{a}|}$. They are the cosines of the angles the vector makes with the positive $x, y, z$ axes, respectively. $$l = \cos\alpha, m = \cos\beta, n = \cos\gamma$$ Key Property: $l^2 + m^2 + n^2 = 1$.
6. Direction Ratios: Any set of numbers proportional to the direction cosines are called direction ratios (e.g., $x,y,z$ for $\vec{a} = x\hat{i} + y\hat{j} + z\hat{k}$). If $a,b,c$ are direction ratios, then $l = \frac{a}{\sqrt{a^2+b^2+c^2}}$, $m = \frac{b}{\sqrt{a^2+b^2+c^2}}$, $n = \frac{c}{\sqrt{a^2+b^2+c^2}}$.
7. Unit Vector: A vector with magnitude 1. The unit vector in the direction of $\vec{a}$ is $\hat{a} = \frac{\vec{a}}{|\vec{a}|}$.
8. Null Vector (Zero Vector): A vector with zero magnitude and arbitrary direction, denoted by $\vec{0}$.
9. Collinear Vectors: Vectors that are parallel to the same line. If $\vec{a}$ and $\vec{b}$ are collinear, then $\vec{a} = \lambda \vec{b}$ for some scalar $\lambda$.
10. Coplanar Vectors: Vectors that lie in the same plane.

Vector Algebra Operations

1. Vector Addition (Triangle Law / Parallelogram Law): If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, then $$\vec{a} + \vec{b} = (a_1+b_1)\hat{i} + (a_2+b_2)\hat{j} + (a_3+b_3)\hat{k}$$
2. Vector Subtraction: $$\vec{a} - \vec{b} = (a_1-b_1)\hat{i} + (a_2-b_2)\hat{j} + (a_3-b_3)\hat{k}$$
3. Scalar Multiplication: $$k\vec{a} = (ka_1)\hat{i} + (ka_2)\hat{j} + (ka_3)\hat{k}$$

Dot Product (Scalar Product)

The dot product of two vectors $\vec{a}$ and $\vec{b}$ is defined as: $$\vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos\theta$$ where $\theta$ is the angle between them ($0 \le \theta \le \pi$).

In component form: If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, then $$\vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3$$

Properties of Dot Product:

• $\vec{a} \cdot \vec{b} = \vec{b} \cdot \vec{a}$ (Commutative)
• $\vec{a} \cdot (\vec{b} + \vec{c}) = \vec{a} \cdot \vec{b} + \vec{a} \cdot \vec{c}$ (Distributive)
• $\vec{a} \cdot \vec{a} = |\vec{a}|^2$
• If $\vec{a} \perp \vec{b}$, then $\vec{a} \cdot \vec{b} = 0$ (and vice-versa for non-zero vectors).
• $\hat{i} \cdot \hat{i} = \hat{j} \cdot \hat{j} = \hat{k} \cdot \hat{k} = 1$
• $\hat{i} \cdot \hat{j} = \hat{j} \cdot \hat{k} = \hat{k} \cdot \hat{i} = 0$
• Angle between two vectors: $\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$.
• Projection of $\vec{a}$ on $\vec{b}$: $\text{proj}_{\vec{b}}\vec{a} = \frac{\vec{a} \cdot \vec{b}}{|\vec{b}|} = \vec{a} \cdot \hat{b}$. This is a scalar value.
• Vector projection of $\vec{a}$ on $\vec{b}$: $(\frac{\vec{a} \cdot \vec{b}}{|\vec{b}|^2})\vec{b}$. This is a vector.

Cross Product (Vector Product)

The cross product of two vectors $\vec{a}$ and $\vec{b}$ is a vector defined as: $$\vec{a} \times \vec{b} = |\vec{a}| |\vec{b}| \sin\theta \hat{n}$$ where $\theta$ is the angle between them ($0 \le \theta \le \pi$) and $\hat{n}$ is a unit vector perpendicular to both $\vec{a}$ and $\vec{b}$ such that $\vec{a}, \vec{b}, \hat{n}$ form a right-handed system.

In component form (using determinant): If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$ and $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, then $$\vec{a} \times \vec{b} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix} = (a_2b_3 - a_3b_2)\hat{i} - (a_1b_3 - a_3b_1)\hat{j} + (a_1b_2 - a_2b_1)\hat{k}$$

Properties of Cross Product:

• $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$ (Anti-commutative)
• $\vec{a} \times (\vec{b} + \vec{c}) = \vec{a} \times \vec{b} + \vec{a} \times \vec{c}$ (Distributive)
• If $\vec{a} || \vec{b}$, then $\vec{a} \times \vec{b} = \vec{0}$ (and vice-versa for non-zero vectors).
• $\vec{a} \times \vec{a} = \vec{0}$
• $\hat{i} \times \hat{i} = \hat{j} \times \hat{j} = \hat{k} \times \hat{k} = \vec{0}$
• $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$ (Cyclic order)
• $\hat{j} \times \hat{i} = -\hat{k}$, $\hat{k} \times \hat{j} = -\hat{i}$, $\hat{i} \times \hat{k} = -\hat{j}$ (Anti-cyclic order)
• $|\vec{a} \times \vec{b}|^2 = |\vec{a}|^2 |\vec{b}|^2 - (\vec{a} \cdot \vec{b})^2$ (Lagrange's Identity)
• Geometric interpretation: $|\vec{a} \times \vec{b}|$ represents the area of the parallelogram formed by adjacent sides $\vec{a}$ and $\vec{b}$.
• Area of a triangle with adjacent sides $\vec{a}$ and $\vec{b}$ is $\frac{1}{2} |\vec{a} \times \vec{b}|$.
• Area of a triangle with vertices $A, B, C$ is $\frac{1}{2} |\vec{AB} \times \vec{AC}|$.

Scalar Triple Product (Box Product)

The scalar triple product of three vectors $\vec{a}, \vec{b}, \vec{c}$ is defined as $\vec{a} \cdot (\vec{b} \times \vec{c})$, denoted as $[\vec{a} \vec{b} \vec{c}]$.

In component form: If $\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}$, $\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}$, $\vec{c} = c_1\hat{i} + c_2\hat{j} + c_3\hat{k}$, then $$[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \ c_1 & c_2 & c_3 \end{vmatrix}$$

Properties of Scalar Triple Product:

• $[\vec{a} \vec{b} \vec{c}] = [\vec{b} \vec{c} \vec{a}] = [\vec{c} \vec{a} \vec{b}]$ (Cyclic permutation of vectors does not change the value).
• $[\vec{a} \vec{b} \vec{c}] = -[\vec{b} \vec{a} \vec{c}]$ (Interchanging any two vectors changes the sign).
• $\vec{a} \cdot (\vec{b} \times \vec{c}) = (\vec{a} \times \vec{b}) \cdot \vec{c}$ (Dot and cross can be interchanged if the cyclic order is maintained).
• If two vectors are identical or collinear, the STP is zero. E.g., $[\vec{a} \vec{a} \vec{b}] = 0$.
• Geometric interpretation: $|[\vec{a} \vec{b} \vec{c}]|$ represents the volume of the parallelepiped formed by adjacent edges $\vec{a}, \vec{b}, \vec{c}$.
• Volume of a tetrahedron with adjacent edges $\vec{a}, \vec{b}, \vec{c}$ is $\frac{1}{6} |[\vec{a} \vec{b} \vec{c}]|$.
• Condition for coplanarity: Three vectors $\vec{a}, \vec{b}, \vec{c}$ are coplanar if and only if $[\vec{a} \vec{b} \vec{c}] = 0$.
• Four points $A, B, C, D$ are coplanar if $\vec{AB}, \vec{AC}, \vec{AD}$ are coplanar, i.e., $[\vec{AB} \vec{AC} \vec{AD}] = 0$.

Vector Triple Product (VTP)

The vector triple product of three vectors $\vec{a}, \vec{b}, \vec{c}$ is $\vec{a} \times (\vec{b} \times \vec{c})$.

Formula: $$\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$$ This is often remembered as "BAC - CAB".

Properties of Vector Triple Product:

• It is a vector coplanar with $\vec{b}$ and $\vec{c}$.
• $\vec{a} \times (\vec{b} \times \vec{c}) \ne (\vec{a} \times \vec{b}) \times \vec{c}$ (Not associative). Note: $(\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}$.

Vector Identities (Advanced)

• $(\vec{a} \times \vec{b}) \cdot (\vec{c} \times \vec{d}) = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{d}) - (\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})$ (Lagrange's identity for four vectors)
• $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) = [\vec{a} \vec{b} \vec{d}]\vec{c} - [\vec{a} \vec{b} \vec{c}]\vec{d} = [\vec{a} \vec{c} \vec{d}]\vec{b} - [\vec{b} \vec{c} \vec{d}]\vec{a}$

3D Geometry: Extending to Space

3D Geometry uses coordinate systems to describe points, lines, and planes in three-dimensional space.

Coordinates and Distances

1. Coordinates of a Point: A point $P$ is represented as $(x, y, z)$.
2. Distance Formula: The distance between two points $P_1(x_1, y_1, z_1)$ and $P_2(x_2, y_2, z_2)$ is $$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}$$
3. Section Formula:
   • Internal division: Point $P(x,y,z)$ dividing $P_1P_2$ in ratio $m:n$ is $$P = \left(\frac{nx_1+mx_2}{m+n}, \frac{ny_1+my_2}{m+n}, \frac{nz_1+mz_2}{m+n}\right)$$
   • External division: Point $P(x,y,z)$ dividing $P_1P_2$ in ratio $m:n$ is $$P = \left(\frac{nx_1-mx_2}{n-m}, \frac{ny_1-my_2}{n-m}, \frac{nz_1-mz_2}{n-m}\right)$$
   • Mid-point: $P = \left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}, \frac{z_1+z_2}{2}\right)$
4. Centroid of a Triangle: For vertices $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$, the centroid is $$G = \left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}, \frac{z_1+z_2+z_3}{3}\right)$$
5. Centroid of a Tetrahedron: For vertices $(x_1, y_1, z_1), \dots, (x_4, y_4, z_4)$, the centroid is $$G = \left(\frac{\sum x_i}{4}, \frac{\sum y_i}{4}, \frac{\sum z_i}{4}\right)$$

Lines in 3D

A line in 3D space can be represented in various forms.

1. Vector Form:

   • Passing through a point $\vec{a}$ and parallel to vector $\vec{b}$: $$\vec{r} = \vec{a} + \lambda\vec{b}$$ where $\lambda$ is a scalar parameter.
   • Passing through two points $\vec{a}$ and $\vec{b}$: $$\vec{r} = \vec{a} + \lambda(\vec{b} - \vec{a})$$ or $\vec{r} = (1-\lambda)\vec{a} + \lambda\vec{b}$.

2. Cartesian Form:

   • Passing through a point $(x_1, y_1, z_1)$ and having direction ratios $(a, b, c)$: $$\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c}$$
   • Passing through two points $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$: $$\frac{x-x_1}{x_2-x_1} = \frac{y-y_1}{y_2-y_1} = \frac{z-z_1}{z_2-z_1}$$

Properties of Lines:

• Angle between two lines: If lines $L_1$ and $L_2$ have direction ratios $(a_1, b_1, c_1)$ and $(a_2, b_2, c_2)$, respectively, or are parallel to vectors $\vec{b_1}$ and $\vec{b_2}$, the angle $\theta$ between them is given by $$\cos\theta = \frac{a_1a_2 + b_1b_2 + c_1c_2}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}} = \frac{|\vec{b_1} \cdot \vec{b_2}|}{|\vec{b_1}| |\vec{b_2}|}$$ (The absolute value gives the acute angle).
• Condition for Perpendicularity: $a_1a_2 + b_1b_2 + c_1c_2 = 0$ (or $\vec{b_1} \cdot \vec{b_2} = 0$).
• Condition for Parallelism: $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ (or $\vec{b_1} = k\vec{b_2}$).
• Shortest Distance between two skew lines: Let the lines be $\vec{r_1} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r_2} = \vec{a_2} + \mu\vec{b_2}$. $$SD = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$$ If $\vec{b_1} \times \vec{b_2} = \vec{0}$, lines are parallel.
• Shortest Distance between two parallel lines: Let the lines be $\vec{r_1} = \vec{a_1} + \lambda\vec{b}$ and $\vec{r_2} = \vec{a_2} + \mu\vec{b}$. $$SD = \left| \frac{(\vec{a_2} - \vec{a_1}) \times \vec{b}}{|\vec{b}|} \right|$$
• Point of Intersection of Lines: Equate components of $\vec{r_1}$ and $\vec{r_2}$ and solve for $\lambda$ and $\mu$. If a consistent solution exists, lines intersect.
• Foot of perpendicular from a point to a line: If a point is $P(x_0, y_0, z_0)$ and line is $\frac{x-x_1}{a} = \frac{y-y_1}{b} = \frac{z-z_1}{c} = k$. Any point on the line is $(x_1+ak, y_1+bk, z_1+ck)$. Let this be $Q$. Vector $\vec{PQ}$ must be perpendicular to the direction vector of the line. So, $\vec{PQ} \cdot (a\hat{i}+b\hat{j}+c\hat{k})=0$. Solve for $k$.
• Image of a point in a line: If $Q$ is the foot of perpendicular from $P$ to the line, and $P'$ is the image, then $Q$ is the midpoint of $PP'$.

Planes in 3D

A plane can be represented in various forms.

1. Vector Form:

   • Normal form: $\vec{r} \cdot \hat{n} = d$, where $\hat{n}$ is the unit normal vector to the plane from origin, and $d$ is the perpendicular distance from the origin to the plane.
   • General form: $\vec{r} \cdot \vec{n} = D$, where $\vec{n}$ is a normal vector to the plane.
   • Passing through a point $\vec{a}$ and normal to vector $\vec{n}$: $(\vec{r} - \vec{a}) \cdot \vec{n} = 0 \implies \vec{r} \cdot \vec{n} = \vec{a} \cdot \vec{n}$.
   • Passing through three non-collinear points $\vec{a}, \vec{b}, \vec{c}$: $(\vec{r} - \vec{a}) \cdot [(\vec{b} - \vec{a}) \times (\vec{c} - \vec{a})] = 0$.

2. Cartesian Form:

   • General equation: $Ax + By + Cz + D = 0$, where $A, B, C$ are direction ratios of the normal to the plane.
   • Normal form: $lx + my + nz = p$, where $l,m,n$ are direction cosines of the normal and $p$ is the perpendicular distance from the origin.
   • Equation of a plane passing through a point $(x_1, y_1, z_1)$ and normal to direction ratios $(A, B, C)$: $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$.
   • Intercept form: $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$, where $a,b,c$ are $x,y,z$ intercepts respectively.
   • Equation of a plane passing through three non-collinear points $(x_1, y_1, z_1), (x_2, y_2, z_2), (x_3, y_3, z_3)$: $$ \begin{vmatrix} x-x_1 & y-y_1 & z-z_1 \ x_2-x_1 & y_2-y_1 & z_2-z_1 \ x_3-x_1 & y_3-y_1 & z_3-z_1 \end{vmatrix} = 0 $$

Properties of Planes:

• Angle between two planes: Let their normals be $\vec{n_1}$ and $\vec{n_2}$. The angle $\theta$ between planes is the angle between their normals. $$\cos\theta = \frac{|\vec{n_1} \cdot \vec{n_2}|}{|\vec{n_1}| |\vec{n_2}|}$$ If $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$, then $\vec{n_1}=(A_1,B_1,C_1)$ and $\vec{n_2}=(A_2,B_2,C_2)$.
• Condition for Perpendicularity: $\vec{n_1} \cdot \vec{n_2} = 0 \implies A_1A_2+B_1B_2+C_1C_2=0$.
• Condition for Parallelism: $\vec{n_1} = k\vec{n_2} \implies \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2}$.
• Distance from a point to a plane: Distance from $P(x_1, y_1, z_1)$ to $Ax+By+Cz+D=0$ is $$d = \frac{|Ax_1+By_1+Cz_1+D|}{\sqrt{A^2+B^2+C^2}}$$
• Distance between two parallel planes: For $Ax+By+Cz+D_1=0$ and $Ax+By+Cz+D_2=0$, $$d = \frac{|D_1-D_2|}{\sqrt{A^2+B^2+C^2}}$$
• Equation of a plane passing through the intersection of two planes: If $P_1: A_1x+B_1y+C_1z+D_1=0$ and $P_2: A_2x+B_2y+C_2z+D_2=0$, then the family of planes passing through their intersection is $P_1 + \lambda P_2 = 0$. Vector form: $\vec{r} \cdot \vec{n_1} - D_1 + \lambda(\vec{r} \cdot \vec{n_2} - D_2) = 0$.

Line and Plane

1. Angle between a line and a plane: If line $\vec{r} = \vec{a} + \lambda\vec{b}$ and plane $\vec{r} \cdot \vec{n} = D$. The angle $\phi$ between the line and the plane is related to the angle $\theta$ between the line's direction vector $\vec{b}$ and the plane's normal vector $\vec{n}$ by $\phi = 90^\circ - \theta$. So, $\sin\phi = \cos\theta = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$. Cartesian: $\sin\phi = \frac{|Aa+Bb+Cc|}{\sqrt{A^2+B^2+C^2}\sqrt{a^2+b^2+c^2}}$.
2. Condition for a line to be parallel to a plane: $\vec{b} \cdot \vec{n} = 0$. (The line's direction vector is perpendicular to the plane's normal).
3. Condition for a line to be perpendicular to a plane: $\vec{b} || \vec{n} \implies \vec{b} = k\vec{n}$.
4. Condition for a line to lie in a plane:
   • The line is parallel to the plane ($\vec{b} \cdot \vec{n} = 0$).
   • A point on the line (e.g., $\vec{a}$) lies in the plane (i.e., $\vec{a} \cdot \vec{n} = D$).
5. Point of Intersection of a line and a plane: Substitute $\vec{r} = \vec{a} + \lambda\vec{b}$ into $\vec{r} \cdot \vec{n} = D$ and solve for $\lambda$. Substitute $\lambda$ back into the line equation to get the point. In Cartesian: Substitute $(x_1+ak, y_1+bk, z_1+ck)$ into $Ax+By+Cz+D=0$.

Sphere

1. Equation of a sphere with center $(h, k, l)$ and radius $R$: $$(x-h)^2 + (y-k)^2 + (z-l)^2 = R^2$$
2. General equation of a sphere: $$x^2 + y^2 + z^2 + 2ux + 2vy + 2wz + d = 0$$ Center is $(-u, -v, -w)$ and radius is $R = \sqrt{u^2+v^2+w^2-d}$. Condition for a real sphere: $u^2+v^2+w^2-d > 0$.
3. Equation of a sphere passing through four points: Substitute points into the general equation and solve the system of equations for $u, v, w, d$.
4. Equation of a sphere with diameter $P_1P_2$: $(x-x_1)(x-x_2) + (y-y_1)(y-y_2) + (z-z_1)(z-z_2) = 0$.
5. Plane section of a sphere: A plane cuts a sphere in a circle (unless it's tangent or too far). Radius of the circle $r = \sqrt{R^2 - D^2}$, where $R$ is sphere radius and $D$ is perpendicular distance from the center of the sphere to the plane. Center of the circle is the foot of the perpendicular from the sphere's center to the plane.

Worked Examples

Example 1: Vector Coplanarity

Problem: Find the value of $\lambda$ for which the vectors $\vec{a} = 2\hat{i} - \hat{j} + \hat{k}$, $\vec{b} = \hat{i} + 2\hat{j} - 3\hat{k}$, and $\vec{c} = 3\hat{i} + \lambda\hat{j} + 5\hat{k}$ are coplanar.

Solution: For three vectors to be coplanar, their scalar triple product must be zero. $$[\vec{a} \vec{b} \vec{c}] = \begin{vmatrix} 2 & -1 & 1 \ 1 & 2 & -3 \ 3 & \lambda & 5 \end{vmatrix} = 0$$ Expand the determinant: $$2((2)(5) - (-3)(\lambda)) - (-1)((1)(5) - (-3)(3)) + 1((1)(\lambda) - (2)(3)) = 0$$ $$2(10 + 3\lambda) + 1(5 + 9) + 1(\lambda - 6) = 0$$ $$20 + 6\lambda + 14 + \lambda - 6 = 0$$ $$7\lambda + 28 = 0$$ $$7\lambda = -28$$ $$\lambda = -4$$ Thus, the vectors are coplanar when $\lambda = -4$.

Example 2: Shortest Distance Between Skew Lines

Problem: Find the shortest distance between the lines: $L_1: \vec{r} = (\hat{i} + 2\hat{j} + \hat{k}) + \lambda(2\hat{i} + \hat{j} + 2\hat{k})$ $L_2: \vec{r} = (2\hat{i} - \hat{j} - \hat{k}) + \mu(\hat{i} - \hat{j} + \hat{k})$

Solution: The general form of skew lines is $\vec{r} = \vec{a_1} + \lambda\vec{b_1}$ and $\vec{r} = \vec{a_2} + \mu\vec{b_2}$. From $L_1$: $\vec{a_1} = \hat{i} + 2\hat{j} + \hat{k}$, $\vec{b_1} = 2\hat{i} + \hat{j} + 2\hat{k}$. From $L_2$: $\vec{a_2} = 2\hat{i} - \hat{j} - \hat{k}$, $\vec{b_2} = \hat{i} - \hat{j} + \hat{k}$.

First, calculate $\vec{a_2} - \vec{a_1}$: $\vec{a_2} - \vec{a_1} = (2\hat{i} - \hat{j} - \hat{k}) - (\hat{i} + 2\hat{j} + \hat{k}) = \hat{i} - 3\hat{j} - 2\hat{k}$.

Next, calculate $\vec{b_1} \times \vec{b_2}$: $$\vec{b_1} \times \vec{b_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 2 & 1 & 2 \ 1 & -1 & 1 \end{vmatrix}$$ $$= \hat{i}(1 \cdot 1 - 2 \cdot (-1)) - \hat{j}(2 \cdot 1 - 2 \cdot 1) + \hat{k}(2 \cdot (-1) - 1 \cdot 1)$$ $$= \hat{i}(1+2) - \hat{j}(2-2) + \hat{k}(-2-1)$$ $$= 3\hat{i} + 0\hat{j} - 3\hat{k} = 3\hat{i} - 3\hat{k}$$

Now, find the magnitude of $\vec{b_1} \times \vec{b_2}$: $$|\vec{b_1} \times \vec{b_2}| = \sqrt{3^2 + (-3)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt{2}.$$

Finally, apply the shortest distance formula: $$SD = \left| \frac{(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2})}{|\vec{b_1} \times \vec{b_2}|} \right|$$ $$(\vec{a_2} - \vec{a_1}) \cdot (\vec{b_1} \times \vec{b_2}) = (\hat{i} - 3\hat{j} - 2\hat{k}) \cdot (3\hat{i} - 3\hat{k})$$ $$= (1)(3) + (-3)(0) + (-2)(-3) = 3 + 0 + 6 = 9$$ $$SD = \left| \frac{9}{3\sqrt{2}} \right| = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$$ The shortest distance is $\frac{3\sqrt{2}}{2}$ units.

Example 3: Finding Image of a Point in a Plane

Problem: Find the image of the point $P(3, 5, 7)$ in the plane $2x + y + z = 16$.

Solution: Let the point be $P(3, 5, 7)$ and the plane be $\Pi: 2x + y + z - 16 = 0$. The normal vector to the plane is $\vec{n} = (2, 1, 1)$. Let $P'(x', y', z')$ be the image of $P$. The line $PP'$ is perpendicular to the plane. The equation of the line $PP'$ passing through $P(3, 5, 7)$ and parallel to $\vec{n}$ is: $$\frac{x-3}{2} = \frac{y-5}{1} = \frac{z-7}{1} = k$$ Any point $Q$ on this line can be written as $(3+2k, 5+k, 7+k)$. This point $Q$ is the foot of the perpendicular from $P$ to the plane. $Q$ lies on the plane, so it must satisfy the plane equation: $$2(3+2k) + (5+k) + (7+k) = 16$$ $$6 + 4k + 5 + k + 7 + k = 16$$ $$18 + 6k = 16$$ $$6k = 16 - 18$$ $$6k = -2$$ $$k = -\frac{1}{3}$$ Now substitute $k$ back into the coordinates of $Q$: $x_Q = 3 + 2(-\frac{1}{3}) = 3 - \frac{2}{3} = \frac{7}{3}$ $y_Q = 5 + (-\frac{1}{3}) = \frac{14}{3}$ $z_Q = 7 + (-\frac{1}{3}) = \frac{20}{3}$ So, the foot of the perpendicular is $Q(\frac{7}{3}, \frac{14}{3}, \frac{20}{3})$.

$Q$ is the midpoint of $PP'$. We have $P(3,5,7)$ and $P'(x',y',z')$. If $Q = (\frac{x+x'}{2}, \frac{y+y'}{2}, \frac{z+z'}{2})$, then: $\frac{3+x'}{2} = \frac{7}{3} \implies 9+3x' = 14 \implies 3x' = 5 \implies x' = \frac{5}{3}$ $\frac{5+y'}{2} = \frac{14}{3} \implies 15+3y' = 28 \implies 3y' = 13 \implies y' = \frac{13}{3}$ $\frac{7+z'}{2} = \frac{20}{3} \implies 21+3z' = 40 \implies 3z' = 19 \implies z' = \frac{19}{3}$ The image of the point $P(3, 5, 7)$ in the plane is $P'(\frac{5}{3}, \frac{13}{3}, \frac{19}{3})$.

Example 4: Area of a Triangle using Vectors

Problem: Find the area of the triangle with vertices $A(1,1,1)$, $B(1,2,3)$, $C(2,3,1)$.

Solution: The area of a triangle with vertices $A, B, C$ is given by $\frac{1}{2} |\vec{AB} \times \vec{AC}|$. First, find the vectors $\vec{AB}$ and $\vec{AC}$: $\vec{AB} = B - A = (1-1)\hat{i} + (2-1)\hat{j} + (3-1)\hat{k} = 0\hat{i} + \hat{j} + 2\hat{k}$ $\vec{AC} = C - A = (2-1)\hat{i} + (3-1)\hat{j} + (1-1)\hat{k} = \hat{i} + 2\hat{j} + 0\hat{k}$

Next, calculate the cross product $\vec{AB} \times \vec{AC}$: $$\vec{AB} \times \vec{AC} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \ 0 & 1 & 2 \ 1 & 2 & 0 \end{vmatrix}$$ $$= \hat{i}(1 \cdot 0 - 2 \cdot 2) - \hat{j}(0 \cdot 0 - 2 \cdot 1) + \hat{k}(0 \cdot 2 - 1 \cdot 1)$$ $$= \hat{i}(-4) - \hat{j}(-2) + \hat{k}(-1)$$ $$= -4\hat{i} + 2\hat{j} - \hat{k}$$

Now, find the magnitude of the cross product: $$|\vec{AB} \times \vec{AC}| = \sqrt{(-4)^2 + 2^2 + (-1)^2} = \sqrt{16 + 4 + 1} = \sqrt{21}$$

Finally, calculate the area of the triangle: Area $= \frac{1}{2} |\vec{AB} \times \vec{AC}| = \frac{\sqrt{21}}{2}$ square units.

Example 5: Finding the Equation of a Plane

Problem: Find the equation of the plane passing through the points $(1, -1, 1)$ and $(2, 1, 3)$ and perpendicular to the plane $x - 2y + 4z = 10$.

Solution: Let the equation of the plane be $A(x-x_1) + B(y-y_1) + C(z-z_1) = 0$. The plane passes through $(1, -1, 1)$, so its equation is $A(x-1) + B(y+1) + C(z-1) = 0$. (Equation 1) Since it also passes through $(2, 1, 3)$, substitute these coordinates into Eq 1: $A(2-1) + B(1+1) + C(3-1) = 0$ $A + 2B + 2C = 0$ (Equation 2)

The plane is perpendicular to $x - 2y + 4z = 10$. The normal vector of this given plane is $\vec{n_1} = (1, -2, 4)$. The normal vector of our required plane is $\vec{n_2} = (A, B, C)$. Since the planes are perpendicular, their normal vectors are perpendicular. So, $\vec{n_1} \cdot \vec{n_2} = 0$: $A(1) + B(-2) + C(4) = 0$ $A - 2B + 4C = 0$ (Equation 3)

We have two equations (2 and 3) for $A, B, C$: $A + 2B + 2C = 0$ $A - 2B + 4C = 0$

Subtract Eq 3 from Eq 2: $(A + 2B + 2C) - (A - 2B + 4C) = 0$ $4B - 2C = 0 \implies 2C = 4B \implies C = 2B$.

Substitute $C = 2B$ into Eq 2: $A + 2B + 2(2B) = 0$ $A + 2B + 4B = 0$ $A + 6B = 0 \implies A = -6B$.

So, we have $A = -6B$, $C = 2B$. Substitute these values into Eq 1 (choosing a non-zero $B$, say $B=1$): If $B=1$, then $A=-6$ and $C=2$. The equation of the plane is: $-6(x-1) + 1(y+1) + 2(z-1) = 0$ $-6x + 6 + y + 1 + 2z - 2 = 0$ $-6x + y + 2z + 5 = 0$ Or, multiplying by -1: $6x - y - 2z - 5 = 0$.

Example 6: Reflection of a Ray of Light

Problem: A ray of light passes through the point $P(1, 2, 3)$ and strikes the plane $x+y+z=12$ at $Q$. The reflected ray passes through $R(3, 2, 1)$. Find the coordinates of $Q$.

Solution: Let the point $P_{reflection}$ be the image of $P(1,2,3)$ in the plane $x+y+z=12$. The normal vector to the plane is $\vec{n} = (1,1,1)$. The line passing through $P(1,2,3)$ and perpendicular to the plane is $\frac{x-1}{1} = \frac{y-2}{1} = \frac{z-3}{1} = k$. Any point on this line is $(1+k, 2+k, 3+k)$. Let this point be $S$, the foot of the perpendicular. $S$ lies on the plane: $(1+k) + (2+k) + (3+k) = 12$ $6 + 3k = 12 \implies 3k = 6 \implies k=2$. So, $S(1+2, 2+2, 3+2) = S(3, 4, 5)$.

If $P'(x', y', z')$ is the image of $P$, then $S$ is the midpoint of $PP'$. $S = (\frac{1+x'}{2}, \frac{2+y'}{2}, \frac{3+z'}{2})$. $\frac{1+x'}{2} = 3 \implies 1+x' = 6 \implies x'=5$. $\frac{2+y'}{2} = 4 \implies 2+y' = 8 \implies y'=6$. $\frac{3+z'}{2} = 5 \implies 3+z' = 10 \implies z'=7$. So the image of $P$ is $P'(5, 6, 7)$.

The reflected ray passes through $R(3, 2, 1)$, and it is the line connecting $P'$ to $R$. The point $Q$ (where the ray strikes the plane) is the intersection of the line $P'R$ and the plane $x+y+z=12$. Equation of line $P'R$ passing through $P'(5,6,7)$ and $R(3,2,1)$: Direction ratios are $(3-5, 2-6, 1-7) = (-2, -4, -6)$, which can be simplified to $(1,2,3)$. So, the equation of the line $P'R$ is: $$\frac{x-5}{1} = \frac{y-6}{2} = \frac{z-7}{3} = m$$ Any point on this line is $(5+m, 6+2m, 7+3m)$. This point is $Q$. Since $Q$ lies on the plane $x+y+z=12$: $(5+m) + (6+2m) + (7+3m) = 12$ $18 + 6m = 12$ $6m = -6 \implies m = -1$. Substitute $m=-1$ into the coordinates of $Q$: $Q = (5+(-1), 6+2(-1), 7+3(-1))$ $Q = (4, 4, 4)$. The coordinates of $Q$ are $(4, 4, 4)$.

Common Mistakes to Avoid

1. Scalar vs. Vector: Confusing dot product results (scalar) with cross product results (vector). Always check the nature of your answer.
2. Order in Cross Product: Remember $\vec{a} \times \vec{b} = -(\vec{b} \times \vec{a})$. Order matters!
3. STP Sign: For coplanarity, $[\vec{a} \vec{b} \vec{c}] = 0$. The sign only matters for orientation or volume of parallelepiped, where you take modulus.
4. Lagrange's Identity for VTP: For $\vec{a} \times (\vec{b} \times \vec{c})$, it's $(\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$. Don't mix up the terms or vectors. "BAC - CAB".
5. Direction Cosines vs. Direction Ratios: Direction cosines $l,m,n$ must satisfy $l^2+m^2+n^2=1$. Direction ratios $a,b,c$ are proportional to $l,m,n$.
6. Angle between Line and Plane: The angle between the line and the plane is $\phi$, while the angle between the line's direction vector and the plane's normal vector is $\theta = 90^\circ - \phi$. So you use $\sin\phi = |\frac{\vec{b} \cdot \vec{n}}{|\vec{b}| |\vec{n}|}|$. Don't use cosine directly.
7. Shortest Distance Formula: Distinguish between skew lines (non-parallel and non-intersecting) and parallel lines. Use the correct formula for each. For parallel lines, the cross product $\vec{b_1} \times \vec{b_2}$ will be zero.
8. Equation of Plane: Carefully use the normal vector and a point on the plane, or three non-collinear points. Ensure you're not using the point as a vector from origin unless explicitly intended.
9. Foot of Perpendicular: When finding the foot of the perpendicular, ensure that the line formed by the point and the foot is perpendicular to the given line/plane. This is a common step for finding images too.
10. Vector Point vs. Position Vector: A point $A=(x,y,z)$ is different from its position vector $\vec{a} = x\hat{i}+y\hat{j}+z\hat{k}$. Operations like $\vec{AB} = B-A$ are effectively $(x_B-x_A)\hat{i} + (y_B-y_A)\hat{j} + (z_B-z_A)\hat{k}$.

Memory Tricks & Mnemonics

• Dot Product: "Dot-Product for Direction-perpendicularity" ($\vec{a} \cdot \vec{b} = 0 \iff \vec{a} \perp \vec{b}$).
• Cross Product: "Cross-Product for Collinearity" ($\vec{a} \times \vec{b} = \vec{0} \iff \vec{a} || \vec{b}$). "Area of Parallelogram = $|\vec{a} \times \vec{b}|$".
• Scalar Triple Product (STP): "Box product for Coplanarity and Volume". $[\vec{a} \vec{b} \vec{c}] = 0 \iff$ a, b, c are coplanar. "$[\vec{a} \vec{b} \vec{c}]$ is Volume".
• Vector Triple Product (VTP): "BAC - CAB". $\vec{a} \times (\vec{b} \times \vec{c}) = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{a} \cdot \vec{b})\vec{c}$. Remember the outer dotted 'a' with the far 'c' first, multiplied by the middle 'b', then subtract the outer 'a' dotted with the middle 'b', multiplied by the far 'c'.
• Direction Cosines: $\cos\alpha$, $\cos\beta$, $\cos\gamma$ are for angles with $x,y,z$ axes respectively.
• Order of basis vectors in cross product: $\hat{i} \times \hat{j} = \hat{k}$, $\hat{j} \times \hat{k} = \hat{i}$, $\hat{k} \times \hat{i} = \hat{j}$. Visualize a cycle $i \rightarrow j \rightarrow k \rightarrow i$. Going with the cycle gives a positive result, against the cycle gives a negative result.
• Line-Plane Angle: "Sine of L-P Angle = Cosine of B-N Angle" (where L=Line, P=Plane, B=line direction Vector, N=Plane Normal Vector). This helps recall $\sin\phi = \frac{|\vec{b} \cdot \vec{n}|}{|\vec{b}| |\vec{n}|}$.

Previous Year Question Patterns (JEE Advanced)

JEE Advanced questions on Vectors and 3D Geometry often combine multiple concepts and require a strong ability to visualize and apply formulas in complex scenarios.

1. Combined Problems: Many questions involve both vectors and 3D geometry concepts. For example, finding the equation of a plane containing a line and a point, or finding the shortest distance between lines that are given in different forms (parametric, symmetric).
2. Geometric Interpretation: Expect questions that heavily rely on the geometric meaning of dot products (perpendicularity, projection), cross products (area, normal vector), and scalar triple products (volume, coplanarity).
   • Areas and Volumes: Calculations of areas of triangles/parallelograms and volumes of tetrahedrons/parallelepipeds.
   • Perpendicularity/Parallelism: Using dot product for perpendicular vectors and cross product for parallel vectors.
3. Line and Plane Interactions:
   • Intersection Points: Finding the point where a line intersects a plane.
   • Image/Foot of Perpendicular: Determining the image of a point in a plane or a line, or the foot of the perpendicular.
   • Angle between Line and Plane / Plane and Plane: Calculating angles using appropriate normal/direction vectors.
   • Equation of Plane: Finding the equation of a plane under various conditions (e.g., passing through a line and a point, passing through two lines, perpendicular to another plane, containing a given line).
4. Shortest Distance: This is a recurring theme, involving both skew and parallel lines. Sometimes, the lines might be presented implicitly, requiring you to extract their vector/Cartesian forms first.
5. Vector Identities: Occasionally, questions test your knowledge of vector triple product or other identities. These are usually in multiple correct options or numerical value questions.
6. Coplanarity: Determining if points or vectors are coplanar using the scalar triple product. This can be combined with unknown constants.
7. Locus Problems: Finding the locus of a point satisfying certain vector/geometric conditions.
8. Properties of Spheres: Though less frequent than lines/planes, questions involving basic sphere properties (equation, center, radius, plane sections) can appear.

Key takeaway for JEE Advanced: Don't just memorize formulas. Understand why they work and how they relate to the underlying geometry. Practice a wide variety of problems, especially those involving multiple steps and different forms of representation (vector vs. Cartesian).

Revision Checklist

Before the exam, ensure you can quickly recall and apply the following:

Vectors:

• Definitions: Scalar, Vector, Position Vector, Magnitude, Unit Vector, Direction Cosines/Ratios.
• Operations: Vector Addition/Subtraction, Scalar Multiplication.
• Dot Product: Definition, component form, properties (commutative, distributive, orthogonality condition), angle, projection.
• Cross Product: Definition, component form (determinant), properties (anti-commutative, distributive, parallelism condition), area applications (triangle, parallelogram).
• Scalar Triple Product (STP): Definition (determinant), properties (cyclic, sign change), geometric interpretation (volume of parallelepiped/tetrahedron), coplanarity condition.
• Vector Triple Product (VTP): BAC-CAB formula, non-associativity, coplanarity with constituent vectors.
• Lagrange's Identity and other identities.

3D Geometry:

• Basic formulas: Distance between points, section formula (internal/external), centroid.
• Lines:
  • Vector and Cartesian equations (point-direction, two-point forms).
  • Angle between two lines (condition for parallel/perpendicular).
  • Shortest distance (skew lines, parallel lines).
  • Foot of perpendicular from a point to a line.
  • Image of a point in a line.
• Planes:
  • Vector and Cartesian equations (normal form, general form, point-normal form, intercept form, three-point form, plane through intersection of two planes).
  • Angle between two planes (condition for parallel/perpendicular).
  • Distance from a point to a plane.
  • Distance between two parallel planes.
• Line and Plane:
  • Angle between a line and a plane (condition for parallel/perpendicular/line lying in plane).
  • Point of intersection.
  • Image of a point in a plane.
• Spheres:
  • Standard and general equations.
  • Center and radius from general equation.
  • Equation from diameter/four points.
  • Plane section of a sphere (radius of circle).

Last Day Strategy

1. Prioritize Concepts: Focus on high-yield and frequently tested areas: Shortest Distance, intersection of lines/planes, image problems, finding equations of planes, and geometric applications of dot/cross products.
2. Formula Revision: Go through this cheat sheet. Ensure you can write down every formula from memory. Pay attention to specific conditions (e.g., non-zero vectors for orthogonality conditions).
3. No New Problems: Avoid trying complex new problems. Stick to revisiting problems you've already solved or quick conceptual checks. This builds confidence and avoids panic.
4. Mental Visualization: For 3D geometry problems, try to mentally visualize the setup. Drawing rough diagrams can be helpful even during the exam to avoid spatial errors.
5. Common Mistake Review: Quickly re-read the "Common Mistakes" section to refresh your memory on pitfalls.
6. Practice Determinants: Many vector and 3D geometry problems involve determinants (cross product, scalar triple product, plane equation). Ensure you can calculate them quickly and accurately.
7. Relax and Trust Your Preparation: A calm mind performs better. Trust the hard work you've put in. Get a good night's sleep.

Good luck with your JEE Advanced journey! Master these concepts, and you'll be well on your way to success.