Q5. Find a vector in the direction of 5î − ĵ + 2k̂ which has magnitude 8 units (NCERT Exercise 10.2, Q10, page 354).
The magnitude is √(25 + 1 + 4) = √30, so the vector is (8/√30)(5î − ĵ + 2k̂), that is, (40î − 8ĵ + 16k̂)/√30.
Q6. Find the direction cosines of î + 2ĵ + 3k̂, and show that î + ĵ + k̂ is equally inclined to the axes OX, OY and OZ (NCERT Exercise 10.2, Q12 and Q14, page 354).
√(1 + 4 + 9) = √14, so the direction cosines are 1/√14, 2/√14, 3/√14. For î + ĵ + k̂ all three are 1/√3, so the three angles are equal.
Q7. Find the scalar and vector components of the vector with initial point (2, 1) and terminal point (−5, 7) (NCERT Exercise 10.2, Q5, page 354).
The vector is (−5 − 2)î + (7 − 1)ĵ = −7î + 6ĵ. Scalar components −7 and 6; vector components −7î and 6ĵ.
Q8. Find the value of x for which x(î + ĵ + k̂) is a unit vector (NCERT Miscellaneous Exercise on Chapter 10, Q5, page 372).
|x(î + ĵ + k̂)| = |x|√3 = 1, so x = ±1/√3.
Q9. Write down a unit vector in the XY-plane making an angle of 30° with the positive direction of the x-axis (NCERT Miscellaneous Exercise on Chapter 10, Q1, page 372).
cos 30° î + sin 30° ĵ, that is, (√3/2)î + (1/2)ĵ.
Q10. A girl walks 4 km towards west, then 3 km in a direction 30° east of north, and stops. Find her displacement from her starting point (NCERT Miscellaneous Exercise on Chapter 10, Q3, page 372).
Take î east and ĵ north. The first walk is −4î; the second is 3 sin 30° î + 3 cos 30° ĵ, that is, (3/2)î + (3√3/2)ĵ. The displacement is −(5/2)î + (3√3/2)ĵ, of length √(25/4 + 27/4) = √13 km.
Q11. P and Q have position vectors 2a + b and a − 3b. Find R dividing PQ externally in the ratio 1 : 2, and show that P is the midpoint of RQ (NCERT Miscellaneous Exercise on Chapter 10, Q9, page 372).
With m = 1 and n = 2 in the external formula, the top is (a − 3b) − 2(2a + b), which is −3a − 5b, and the bottom is 1 − 2 = −1. So r = 3a + 5b. The midpoint of RQ is half of (3a + 5b) + (a − 3b), which is 2a + b, the position vector of P.
Q12. Show that the points with position vectors 3î − 4ĵ − 4k̂, 2î − ĵ + k̂ and î − 3ĵ − 5k̂ form a right-angled triangle (NCERT Exercise 10.2, Q17, page 355).
Call them A, B, C. AB = −î + 3ĵ + 5k̂, BC = −î − 2ĵ − 6k̂ and CA = 2î − ĵ + k̂, with squared lengths 35, 41 and 6. Since 35 + 6 = 41, the angle at A is a right angle.