Q5. Find |a| and |b| if (a + b) · (a − b) = 8 and |a| = 8|b| (NCERT Exercise 10.3, Q6, page 362).
(a + b) · (a − b) = |a|2 − |b|2 = 63|b|2 = 8. So |b| = √(8/63) = 2√2/(3√7) and |a| = 16√2/(3√7).
Q6. Find |x| if, for a unit vector a, (x − a) · (x + a) = 12 (NCERT Exercise 10.3, Q9, page 362).
The product is |x|2 − |a|2 = |x|2 − 1 = 12, so |x| = √13.
Q7. Find λ and μ if (2î + 6ĵ + 27k̂) × (î + λĵ + μk̂) = 0 (NCERT Exercise 10.4, Q5, page 368).
A zero cross product means the vectors are parallel, so 1/2 = λ/6 = μ/27. Hence λ = 3 and μ = 27/2.
Q8. Find the area of the parallelogram with adjacent sides î − ĵ + 3k̂ and 2î − 7ĵ + k̂ (NCERT Exercise 10.4, Q10, page 369).
The cross product is 20î + 5ĵ − 5k̂, of magnitude √(400 + 25 + 25) = √450 = 15√2.
Q9. A unit vector a makes angles π/3 with î and π/4 with ĵ, and an acute angle θ with k̂. Find θ and the components of a (NCERT Exercise 10.4, Q3, page 368).
cos2 θ = 1 − 1/4 − 1/2 = 1/4, and θ is acute, so θ = π/3. The components are the direction cosines: a = (1/2)î + (1/√2)ĵ + (1/2)k̂.
Q10. The scalar product of î + ĵ + k̂ with a unit vector along the sum of 2î + 4ĵ − 5k̂ and λî + 2ĵ + 3k̂ is 1. Find λ (NCERT Miscellaneous Exercise on Chapter 10, Q13, page 372).
The sum is (2 + λ)î + 6ĵ − 2k̂. The condition is (2 + λ + 6 − 2) = √((2 + λ)2 + 40), so (λ + 6)2 = (λ + 2)2 + 40, which gives 8λ = 8 and λ = 1. It satisfies the original unsquared condition: λ + 6 = 7 > 0 and the norm is also 7.
Q11. If a, b, c are mutually perpendicular vectors of equal magnitudes, show that a + b + c is equally inclined to a, b and c (NCERT Miscellaneous Exercise on Chapter 10, Q14, page 372, as intended: see the answer).
Assume the common magnitude k > 0; if all are zero, none of the angles is defined. The dot product of any two different ones is 0, so (a + b + c) · a = k2, and the same for b and c. Also |a + b + c| = √3 k. So the cosine of the angle with each of a, b, c is k2/(√3 k × k) = 1/√3: the three angles are equal. NCERT slip (page 372): the question prints “show that the vector c · d = 15 is equally inclined to a, b and c”; the words “c · d = 15” belong to Q12 just above, and the vector meant is a + b + c. In an exam, prove it for a + b + c.
Q12. Beyond the NCERT text (JEE): show that |a × b|2 + (a · b)2 = |a|2|b|2.
If either vector is zero, both sides are zero. Otherwise, the left side is |a|2|b|2(sin2 θ + cos2 θ), which is |a|2|b|2. It gives |a × b| from a · b without finding the angle.