source: ncert
topic: vectors
class_level: class_12
chapter: ch10
difficulty: advanced
Class 12 | Ch10: Vector Algebra VECTOR: quantity with magnitude and direction. Notation: a (bold) or a with arrow. Unit vector: magnitude=1. a_hat=a/|a|. TYPES: Zero vector, Unit vector, Coinitial, Collinear/Parallel, Equal vectors. POSITION VECTOR: of point P(x,y,z): OP=xi+yj+zk. |OP|=sqrt(x^2+y^2+z^2). OPERATIONS: Addition: (a1i+a2j+a3k)+(b1i+b2j+b3k)=(a1+b1)i+(a2+b2)j+(a3+b3)k. Scalar multiplication: k(ai+bj+ck)=kai+kbj+ckk. DOT PRODUCT (Scalar product): a.b=|a||b|cos(theta). a.b=a1b1+a2b2+a3b3. a.b=0 iff perpendicular (or one is zero vector). a.a=|a|^2. i.i=j.j=k.k=1. i.j=j.k=k.i=0. CROSS PRODUCT (Vector product): |a×b|=|a||b|sin(theta). Direction: right-hand rule. a×b=det([[i,j,k],[a1,a2,a3],[b1,b2,b3]]). =(a2b3-a3b2)i-(a1b3-a3b1)j+(a1b2-a2b1)k. a×b=0 iff parallel. a×a=0. i×j=k, j×k=i, k×i=j. j×i=-k. |a×b|=area of parallelogram with sides a and b. (1/2)|a×b|=area of triangle. SCALAR TRIPLE PRODUCT: [a b c]=a.(b×c). Volume of parallelepiped. [a b c]=0 iff coplanar. SOLVED EXAMPLES: Example 1: a=2i+3j-k, b=i-2j+3k. a.b=2(1)+3(-2)+(-1)(3)=2-6-3=-7. |a|=sqrt(4+9+1)=sqrt(14). |b|=sqrt(1+4+9)=sqrt(14). cos(theta)=-7/14=-1/2. theta=120°. Example 2: a×b=det([[i,j,k],[2,3,-1],[1,-2,3]]). =i(9-2)-j(6+1)+k(-4-3)=7i-7j-7k=7(i-j-k). |a×b|=7sqrt(3). Area of parallelogram=7sqrt(3). Example 3: Unit vector along 2i+3j+6k. |v|=sqrt(4+9+36)=sqrt(49)=7. Unit vector=(2i+3j+6k)/7. COMMON MISTAKES: a.b is SCALAR. a×b is VECTOR. i×j=k but j×i=-k (cross product NOT commutative). a×b=0 means PARALLEL (not perpendicular). a.b=0 means PERPENDICULAR.