Vectors — Class 12 (Advanced)

Introduction — memory-first summary

These advanced vectors notes explain vector algebra and geometry clearly: position vectors, components, scalar (dot) product, vector (cross) product, projections, equations of lines & planes, and 3D geometry applications. Designed for Class 12 boards and competitive exam revision (Police Bharti, Railway, SSC, NEET quick topics).

Why this page?

• Memory-first layout: definition → theorem → proof → diagram → example → mnemonic.
• SVG diagrams for geometric intuition (parallelogram, triangle method, vector projection).
• SEO optimized: unique title & meta, JSON-LD for Course/Breadcrumb/FAQ, H1/H2/H3 structure.

Core definitions & notation

1. Position vector: If P(x,y,z) then vector OP = r = xi + yj + zk.
2. Equal vectors: same magnitude and direction: a = b ⇔ a_x=b_x, a_y=b_y, a_z=b_z.
3. Scalar (dot) product: a⋅b = |a||b|cosθ = a_x b_x + a_y b_y + a_z b_z.
4. Vector (cross) product: a×b is orthogonal to a and b, magnitude |a×b|=|a||b|sinθ, direction by right-hand rule.
5. Projection: Component of a along b: comp_b(a) = (a⋅b)/|b|.

Theorems & proofs (compact)

Theorem — Dot product properties

Commutative: a⋅b=b⋅a. Distributive: a⋅(b+c)=a⋅b + a⋅c. Proof: express vectors in components and use algebra.

Theorem — Cross product properties

Anti-commutative: a×b = - (b×a). Distributive over addition. a×a=0. Proof by determinants of components.

SVG Diagrams (parallelogram, projection, right-hand rule)

Formula summary table

परिचय — सारांश

हे notes Vectors चे संपूर्ण chapter स्पष्ट करतात: dot, cross, projection, रेषा व समतलाची समीकरणे, आणि 3D geometry चे application. परीक्षेसाठी उपयुक्त (SSC, Railway, Police Bharti, NEET).

8 Solved Examples — Step-by-step

Example 1 — Dot product & angle

Problem: Find angle between a = 2i + 3j - k and b = i - j + 2k.

Solution:

1. Compute dot: a⋅b = 2*1 + 3*(-1) + (-1)*2 = 2 -3 -2 = -3.
2. Magnitudes: |a|=√(4+9+1)=√14, |b|=√(1+1+4)=√6.
3. cosθ = (-3)/(√14 √6) = -3/√84 = -3/(2√21). So θ = arccos(that).

Example 2 — Projection

Problem: Project a = 3i + 4j on b = 2i + j.

Solution:

1. a⋅b = 6 + 4 = 10.
2. |b|^2 = 4 +1 =5. So projection vector = (10/5) b = 2 b = 4i +2j.

Example 3 — Cross product & area

Problem: Find area of parallelogram formed by a = i + 2j + k and b = 2i - j + 3k.

Solution:

1. Compute cross via determinant: a×b = |i j k; 1 2 1; 2 -1 3| = i(2*3 -1*(-1)) - j(1*3 -1*2) + k(1*(-1)-2*2) = i(6+1) - j(3-2) + k(-1-4) = 7i -1j -5k.
2. Magnitude = √(49+1+25)=√75 = 5√3. Area = 5√3.

Example 4 — Equation of line through point in direction b

Problem: Find equation of line through (1,2,3) with direction vector b=2i+3j-4k.

Solution: Parametric form: r = (i+2j+3k) + t(2i+3j-4k). Component form: (x-1)/2 = (y-2)/3 = (z-3)/-4.

Example 5 — Distance from point to line

Problem: Distance from point P(3, -1, 2) to line r = (1,0,0) + t(1,1,1).

Solution:

1. Vector from point on line A(1,0,0) to P: AP = (2, -1, 2).
2. Direction b = (1,1,1). Distance = |AP × b| / |b|.
3. AP × b = |i j k; 2 -1 2; 1 1 1| = i((-1)*1 -2*1) - j(2*1 -2*1) + k(2*1 -(-1)*1) = i(-1-2) - j(2-2) + k(2+1) = -3i +0j +3k.
4. |AP × b| = √(9+0+9)=√18=3√2. |b|=√3. Distance = (3√2)/√3 = 3√(2/3) = 3√6/3 = √6 (simplify to √6).

Example 6 — Angle between planes (normals)

Problem: Find angle between planes 2x - y + z = 3 and x + y - 2z = 1.

Solution:

1. Normals n1=(2,-1,1), n2=(1,1,-2). Angle between planes = angle between normals.
2. cosθ = (n1⋅n2)/(|n1||n2|) = (2*1 + -1*1 +1*-2) / (√6 √6) = (2 -1 -2)/(6) = -1/6. So θ = arccos(-1/6).

Example 7 — Closest point on line to given point

Problem: Closest point from P(2,1,0) to line r = (1,0,1) + t(1,2,-1).

Solution:

1. Let A=(1,0,1), b=(1,2,-1). Vector AP = (1,1,-1).
2. Project AP onto b: t* = (AP⋅b)/|b|^2 = (1*1 +1*2 + -1*-1)/(1+4+1) = (1+2+1)/6 =4/6=2/3.
3. Closest point = A + t*b = (1,0,1) + (2/3)(1,2,-1) = (1+2/3, 0+4/3, 1-2/3) = (5/3,4/3,1/3).

Example 8 — Volume of parallelepiped

Problem: Vectors a=(1,2,3), b=(0,1,2), c=(2,0,1). Find volume.

Solution:

1. Volume = |a⋅(b×c)|. Compute b×c = |i j k; 0 1 2; 2 0 1| = i(1*1 -2*0) - j(0*1 -2*2) + k(0*0 -1*2) = i(1) - j( -4 ) + k(-2) = i +4j -2k.
2. a⋅(b×c) = 1*1 + 2*4 + 3*(-2) = 1 +8 -6 =3. Volume = |3| = 3.

Practice Questions

1. Find equation of plane through points A(1,0,0), B(0,1,0), C(0,0,1).
2. Solve for t if |a + t b| = 5 where a=(1,2,1), b=(2,-1,2).
3. Show that vectors (1,2,3), (4,5,6), (7,8,9) are coplanar or not.
4. Find distance between skew lines r = (1,2,3) + s(1,0,1) and r = (0,1,2) + t(0,1,1).
5. Find projection of (3, -2, 1) on (1,1,1).

Answer Key (brief)

1. Plane: x + y + z = 1.
2. Square both sides → quadratic in t, solve for t (two roots possible).
3. Compute triple scalar product; determinant = 0 ⇒ coplanar (these give 0 so yes coplanar).
4. Distance between skew lines: |(d1 × d2)⋅(r2 - r1)| / |d1 × d2| → compute per data.
5. Projection = ((3-2+1)/3) (1,1,1) = (2/3)(1,1,1) → vector (2/3,2/3,2/3).

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