Multiplying a vector by a scalar, components and the section formula

What is a scalar multiple of a vector?

In plain words: multiplying by λ stretches or shrinks a along its own line by the factor |λ|, and turns it round when λ is negative. So 2a is twice as long as a, −2a is as long but reversed, and (−1)a = −a, the additive inverse, with a + (−a) = 0 (page 346, Fig 10.12). Physics uses the same rule when it multiplies a vector by a number. For any scalar k, k 0 = 0 (page 347).

Unit vector along a. For a ≠ 0, dividing by its own length gives a vector of length 1 in the same direction (page 347):

Components of a vector

The unit vectors along the x-, y- and z-axes are written î, ĵ and k̂ (page 347, Fig 10.13). The position vector of P(x, y, z) is then r = xî + yĵ + zk̂ (page 348):

This form of any vector is called its component form.

NCERT Mathematics Part II (Class 12), Section 10.5.1, page 348

Here x, y, z are the scalar components of r, and xî, yĵ, zk̂ its vector components (page 348). Its length, by Pythagoras twice, is √(x² + y² + z²).

The distributive laws k a + m a = (k + m)a, k(m a) = (km)a and k(a + b) = k a + k b hold (page 349). The components a_₁, a_₂, a_₃ are also direction ratios of a, and if l, m, n are its direction cosines, lî + mĵ + nk̂ is the unit vector along it (Remarks (ii)–(iii), page 349).

NCERT Example 5 (page 350): a = î + 2ĵ and b = 2î + ĵ both have magnitude √5, but they are not equal, because their components differ. NCERT’s own words, from Example 4:

Note that two vectors are equal if and only if their corresponding components are equal.

NCERT Mathematics Part II (Class 12), Section 10.5.1, page 349

Collinear vectors

a and b are collinear exactly when b = λa for some scalar λ (with a ≠ 0). In components, this means the components are proportional: b_₁/a_₁ = b_₂/a_₂ = b_₃/a_₃ = λ (Remark (i), page 349). A ratio with 0 below is read as a condition: if a component of a is 0, the matching component of b must be 0 too.

The vector joining two points

For P_₁(x_₁, y_₁, z_₁) and P_₂(x_₂, y_₂, z_₂), the triangle law in OP_₁P_₂ gives P_₁P_₂ = OP_₂ − OP_₁ (page 351, Fig 10.15):

Its magnitude is the distance P_₁P_₂. NCERT Example 10 (page 352): from P(2, 3, 0) to Q(−1, −2, −4), PQ = −3î − 5ĵ − 4k̂.

The section formula

For distinct P and Q and positive m, n, if P and Q have position vectors a and b, and R divides PQ so that PR : RQ = m : n, then (pages 352–353):

In coordinates this is the section formula in three dimensions, which JEE Main 2027 lists in Unit 11 but the Class 11 book does not teach. The internal formula comes from m(b − r) = n(r − a), that is, m RQ = n PR (page 352). NCERT Example 11 (page 353): with OP = 3a − 2b and OQ = a + b, the point dividing PQ in the ratio 2 : 1 is 5a/3 internally and 4b − a externally.

Solved examples

Common mistakes to avoid

Quick revision summary

• λa is collinear with a, |λa| = |λ| |a|, and it is reversed when λ < 0. The unit vector is â = a/|a|.
• In components, r = xî + yĵ + zk̂ with |r| = √(x^² + y^² + z^²); add, subtract and scale component by component.
• Collinear: b = λa, or proportional components. Joining vector: P_₁P_₂ = OP_₂ − OP_₁.
• Section formula, for R with PR : RQ = m : n: (m b + n a)/(m + n) internally, (m b − n a)/(m − n) externally, (a + b)/2 at the midpoint.

Questions students ask

What is the unit vector in the direction of 3î + 4ĵ?

The unit vector in the direction of 3î + 4ĵ is (3/5)î + (4/5)ĵ: its magnitude is √(9 + 16) = 5, and dividing a non-zero vector by its magnitude, â = a/|a|, keeps the direction and makes the length 1. Its components 3/5 and 4/5 are the cosines of the angles it makes with the x- and y-axes, so it makes cos⁻¹ (3/5) ≈ 53.1° with the x-axis. In the same way 4î − 3ĵ, also of magnitude 5, has the unit vector (4/5)î − (3/5)ĵ, which is perpendicular to the first.

How do you prove that three points are collinear using vectors?

To prove with vectors that three distinct points A, B and C are collinear, show that AB and BC are collinear vectors, one a scalar multiple of the other; since they share the point B, all three points lie on one line. For A(−1, 2, 4), B(1, 0, 5) and C(7, −6, 8), AB = 2î − 2ĵ + k̂ and BC = 6î − 6ĵ + 3k̂, which is 3AB, so the points are collinear. Two other tests agree: equality in the triangle inequality, as |AB| = 3 and |BC| = 9 add up to |AC| = 12, and a zero cross product, AB × BC = 0.

What is the external section formula in vectors, and why must m ≠ n?

The point R dividing PQ externally in the ratio m : n of positive numbers, so that PR : QR = m : n with R on the line PQ outside the segment, has position vector (m b − n a)/(m − n), where a and b are the position vectors of P and Q (NCERT page 353). As in the coordinate form, R lies beyond Q when m > n and beyond P when m < n, and m = n gives a zero denominator, as no point outside the segment is equally far from P and Q. For P(1, 0, 2) and Q(3, 4, 1) in the ratio 3 : 1, R has position vector (3b − a)/2, which is 4î + 6ĵ + (1/2)k̂, a point beyond Q.

Practice questions

Try each question first, then open the answer to check your working.

1. Q1. Find the magnitudes of a = î + ĵ + k̂, b = 2î − 7ĵ − 3k̂ and c = (î + ĵ − k̂)/√3 (NCERT Exercise 10.2, Q1, page 354).

   Show answerHide answer

   |a| = √3, |b| = √(4 + 49 + 9) = √62, and |c| = √(1/3 + 1/3 + 1/3) = 1. (NCERT prints a = î + ĵ + k, without the hat; it is k̂.)

2. Q2. For a = 2î − ĵ + 2k̂ and b = −î + ĵ − k̂, find the unit vector along a + b (NCERT Exercise 10.2, Q9, page 354).

   Show answerHide answer

   a + b = î + k̂, of magnitude √2, so the unit vector is (1/√2)î + (1/√2)k̂.

3. Q3. Find the position vector of the midpoint of the vector joining P(2, 3, 4) and Q(4, 1, −2) (NCERT Exercise 10.2, Q16, page 355).

   Show answerHide answer

   (OP + OQ)/2 = (6î + 4ĵ + 2k̂)/2 = 3î + 2ĵ + k̂.

4. Q4. Find x and y if 2î + 3ĵ and xî + yĵ are equal (NCERT Exercise 10.2, Q4, page 354).

   Show answerHide answer

   Equal vectors have equal components: x = 2 and y = 3.