Vector Algebra Ex 10.3 Class 12 NCERT

What Ex 10.3 Is Really Testing

Exercise 10.3 is not just about substituting numbers into a formula. It asks whether you understand how the dot product measures the part of one vector acting in the direction of another. If two vectors make an acute angle, their dot product is positive; if the angle is obtuse, it is negative; and if they are perpendicular, it becomes zero. In component form, for a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k, the scalar product is a · b = a1b1 + a2b2 + a3b3. This algebraic form is powerful because it avoids drawing the vectors. Most NCERT Ex 10.3 questions combine this with |a| = √(a · a), cos θ = (a · b)/(|a||b|), and projection of a on b = (a · b)/|b|.

Key Terms and Formulas

Scalar product or dot product

For two non-zero vectors a and b with angle θ between them, a · b = |a||b| cos θ. The answer is a scalar, not a vector.

Component form of dot product

If a = a1 i + a2 j + a3 k and b = b1 i + b2 j + b3 k, then a · b = a1b1 + a2b2 + a3b3.

Perpendicular vectors

Two non-zero vectors are perpendicular if and only if their dot product is 0.

Projection of a on b

The scalar projection of a on b is (a · b)/|b|. It is the signed length of the shadow of a along b.

Magnitude using dot product

For any vector a, a · a = |a|², so |a| = √(a · a).

Step-by-Step Method for Ex 10.3 Problems

1. Write vectors in component form — Express each vector as xi + yj + zk. Check signs carefully, especially when subtracting vectors or using points such as AB = B - A.
2. Choose the correct dot-product formula — Use a · b = a1b1 + a2b2 + a3b3 when components are known. Use a · b = |a||b| cos θ when magnitudes and angle are given.
3. Use the target condition — For angle, use cos θ = (a · b)/(|a||b|). For perpendicularity, put a · b = 0. For projection, divide the dot product by the magnitude of the base vector.
4. Check interpretation and units — A dot product is scalar. Projection may be negative if the vector points opposite to the base direction. An angle should usually be reported between 0° and 180°.

Worked Examples

• Example 1: Find the angle between a = 2i - j + 2k and b = i + 2j - 2k. Step 1: Compute the dot product: a · b = 2(1) + (-1)(2) + 2(-2) = 2 - 2 - 4 = -4. Step 2: Find magnitudes: |a| = √(2² + (-1)² + 2²) = √9 = 3; |b| = √(1² + 2² + (-2)²) = √9 = 3. Step 3: Use cos θ = (a · b)/(|a||b|) = -4/(3 × 3) = -4/9. Final answer: θ = cos⁻¹(-4/9), an obtuse angle.
• Example 2: Find the projection of a = 3i + 4j + k on b = i - 2j + 2k. Step 1: Dot product: a · b = 3(1) + 4(-2) + 1(2) = 3 - 8 + 2 = -3. Step 2: Magnitude of b: |b| = √(1² + (-2)² + 2²) = √9 = 3. Step 3: Projection of a on b = (a · b)/|b| = -3/3 = -1. Final answer: The scalar projection is -1, meaning a has a component of length 1 opposite to the direction of b.
• Example 3: If a = i + 2j + 3k and b = 3i - j + 2k, find λ such that a is perpendicular to b + λa. Step 1: Perpendicular condition: a · (b + λa) = 0. Step 2: Expand using distributive law: a · b + λ(a · a) = 0. Step 3: Calculate a · b = 1(3) + 2(-1) + 3(2) = 3 - 2 + 6 = 7. Step 4: Calculate a · a = 1² + 2² + 3² = 14. Step 5: 7 + 14λ = 0, so λ = -1/2. Final answer: λ = -1/2.

Board Exam Tips and Common Traps

Do not confuse dot product with cross product. In Ex 10.3, the dot product gives a scalar and is linked to cos θ, not sin θ. When finding projection of a on b, divide by |b|, not by |a|. If a · b = 0, conclude perpendicular only when both vectors are non-zero. For angle questions, first compute cos θ carefully; a negative value simply means the angle is obtuse. In CBSE solutions, write the formula before substituting values because it earns method marks even if arithmetic slips later.

Practice Questions with Solutions

• Q: Find the dot product of a = 4i - 2j + k and b = -i + 3j + 5k. A: Step 1: Use a · b = a1b1 + a2b2 + a3b3. Step 2: Substitute values: a · b = 4(-1) + (-2)(3) + 1(5). Step 3: Simplify: -4 - 6 + 5 = -5. Final answer: a · b = -5.
• Q: Find the angle between a = i + j and b = i - j. A: Step 1: Compute dot product: a · b = 1(1) + 1(-1) = 0. Step 2: Since both vectors are non-zero and a · b = 0, the vectors are perpendicular. Step 3: Perpendicular vectors make an angle of 90°. Final answer: The angle is 90°.
• Q: Find the projection of a = 2i + 3j - 6k on b = i - 2j + 2k. A: Step 1: Compute a · b = 2(1) + 3(-2) + (-6)(2) = 2 - 6 - 12 = -16. Step 2: Compute |b| = √(1² + (-2)² + 2²) = √9 = 3. Step 3: Projection of a on b = (a · b)/|b| = -16/3. Final answer: The scalar projection is -16/3.
• Q: For what value of x are the vectors a = xi + 2j + k and b = 3i - j + 4k perpendicular? A: Step 1: Perpendicular vectors satisfy a · b = 0. Step 2: Compute dot product: a · b = x(3) + 2(-1) + 1(4) = 3x - 2 + 4 = 3x + 2. Step 3: Set 3x + 2 = 0, so 3x = -2 and x = -2/3. Final answer: x = -2/3.

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