Class 12 Maths, Worked in Full Past papersGet updates
Chapter 10 · Vector Algebra

Example 30: Split β⃗=2i^+j^−3k^\vec\beta=2\hat i+\hat j-3\hat k into parts parallel and perpendicular to α⃗=3i^−j^\vec\alpha=3\hat i-\hat j

ExamplesResolving a vector along and across another · Miscellaneous Examples
The question

If with reference to the right handed system of mutually perpendicular unit vectors i^\hat i⁠, j^\hat j and k^\hat k⁠, α⃗=3i^−j^\vec\alpha = 3\hat i - \hat j⁠, β⃗=2i^+j^−3k^\vec\beta = 2\hat i + \hat j - 3\hat k⁠, then express β⃗\vec\beta in the form β⃗=β⃗1+β⃗2\vec\beta = \vec\beta_1 + \vec\beta_2⁠, where β⃗1\vec\beta_1 is parallel to α⃗\vec\alpha and β⃗2\vec\beta_2 is perpendicular to α⃗\vec\alpha⁠.

Exam answer

Let β⃗1=λα⃗=3λi^−λj^\vec\beta_1 = \lambda\vec\alpha = 3\lambda\hat i - \lambda\hat j⁠, where λ\lambda is a scalar.

β⃗2=β⃗−β⃗1=(2−3λ)i^+(1+λ)j^−3k^\begin{aligned} \vec\beta_2 &= \vec\beta - \vec\beta_1 \\ &= (2 - 3\lambda)\hat i + (1 + \lambda)\hat j - 3\hat k \end{aligned}

Since β⃗2\vec\beta_2 is perpendicular to α⃗\vec\alpha⁠, α⃗⋅β⃗2=0\vec\alpha\cdot\vec\beta_2 = 0⁠:

3(2−3λ)−(1+λ)=0⇒6−9λ−1−λ=0⇒5−10λ=0⇒λ=12\begin{aligned} &3(2 - 3\lambda) - (1 + \lambda) = 0 \\ &\Rightarrow 6 - 9\lambda - 1 - \lambda = 0 \\ &\Rightarrow 5 - 10\lambda = 0 \Rightarrow \lambda = \frac12 \end{aligned} β⃗1=12(3i^−j^)β⃗2=(2−32)i^+(1+12)j^−3k^\begin{aligned} \vec\beta_1 &= \tfrac12(3\hat i - \hat j) \\ \vec\beta_2 &= (2 - \tfrac32)\hat i + (1 + \tfrac12)\hat j - 3\hat k \end{aligned}

∴ β⃗=β⃗1+β⃗2\therefore\ \vec\beta = \vec\beta_1 + \vec\beta_2⁠, where β⃗1=32i^−12j^\vec\beta_1 = \dfrac32\hat i - \dfrac12\hat j and β⃗2=12i^+32j^−3k^\vec\beta_2 = \dfrac12\hat i + \dfrac32\hat j - 3\hat k⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

"Parallel to α⃗\vec\alpha" means "a scalar multiple of α⃗\vec\alpha", so write β⃗1=λα⃗\vec\beta_1 = \lambda\vec\alpha with one unknown λ\lambda⁠. Then β⃗2=β⃗−λα⃗\vec\beta_2 = \vec\beta - \lambda\vec\alpha is forced, and the single condition "β⃗2\vec\beta_2 perpendicular to α⃗\vec\alpha", that is α⃗⋅β⃗2=0\vec\alpha\cdot\vec\beta_2 = 0⁠, fixes λ\lambda⁠.

Given, to find, and the plan
Given
α⃗=3i^−j^\vec\alpha = 3\hat i - \hat j (no k^\hat k part) and β⃗=2i^+j^−3k^\vec\beta = 2\hat i + \hat j - 3\hat k⁠.
To find
β⃗1\vec\beta_1 parallel to α⃗\vec\alpha and β⃗2\vec\beta_2 perpendicular to α⃗\vec\alpha with β⃗1+β⃗2=β⃗\vec\beta_1 + \vec\beta_2 = \vec\beta⁠.
Plan
(1) Put β⃗1=λα⃗\vec\beta_1 = \lambda\vec\alpha⁠. (2) Write β⃗2=β⃗−β⃗1\vec\beta_2 = \vec\beta - \vec\beta_1⁠. (3) Solve α⃗⋅β⃗2=0\vec\alpha\cdot\vec\beta_2 = 0 for λ\lambda⁠. (4) Substitute and check all three conditions. (5) See β⃗1\vec\beta_1 as the projection of β⃗\vec\beta on α⃗\vec\alpha⁠.

Basics you need

1Parallel vectors are scalar multiples

u⃗\vec u is collinear with a nonzero α⃗\vec\alpha exactly when u⃗=λα⃗\vec u = \lambda\vec\alpha for some scalar λ\lambda (Section 10.5.1, Remarks (i)). Lesson: components

2Perpendicular means scalar product zero

α⃗⋅u⃗=0\vec\alpha\cdot\vec u = 0 when u⃗⊥α⃗\vec u \perp \vec\alpha⁠, and α⃗⋅u⃗=α1u1+α2u2+α3u3\vec\alpha\cdot\vec u = \alpha_1u_1 + \alpha_2u_2 + \alpha_3u_3 (Section 10.6.1). Lesson: the dot product

3Projection on a vector

The projection vector of β⃗\vec\beta on α⃗\vec\alpha is α⃗⋅β⃗∣α⃗∣2α⃗\frac{\vec\alpha\cdot\vec\beta}{|\vec\alpha|^2}\vec\alpha⁠, the shadow of β⃗\vec\beta along α⃗\vec\alpha (Section 10.6.2). Basics: cos θ is a shadow

Solution

  1. The part parallel to α⃗\vec\alpha

    β⃗1=λα⃗=λ(3i^−j^)=3λi^−λj^.\vec\beta_1 = \lambda\vec\alpha = \lambda(3\hat i - \hat j) = 3\lambda\hat i - \lambda\hat j.
    Every vector parallel to α⃗\vec\alpha is λα⃗\lambda\vec\alpha for some number λ\lambda (positive for the same direction, negative for the opposite). So the whole problem reduces to finding one number.
    u⃗=λα⃗\vec u = \lambda\vec\alpha means u⃗\vec u is collinear with α⃗\vec\alpha (Section 10.5.1, Remarks (i)). Lesson: components

  2. The rest is the perpendicular part

    β⃗2=β⃗−β⃗1=(2−3λ)i^+(1+λ)j^−3k^.\begin{aligned} \vec\beta_2 &= \vec\beta - \vec\beta_1 \\ &= (2 - 3\lambda)\hat i + (1 + \lambda)\hat j - 3\hat k. \end{aligned}
    Because β⃗1+β⃗2=β⃗\vec\beta_1 + \vec\beta_2 = \vec\beta⁠, once β⃗1\vec\beta_1 is chosen, β⃗2\vec\beta_2 is whatever is left. Subtract component by component: 2−3λ2 - 3\lambda⁠, 1−(−λ)=1+λ1 - (-\lambda) = 1 + \lambda⁠, and −3−0=−3-3 - 0 = -3⁠.
    Difference of vectors by components (Section 10.5.1). Lesson: components

  3. Make it perpendicular to α⃗\vec\alpha

    α⃗⋅β⃗2=3(2−3λ)+(−1)(1+λ)+0(−3)=6−9λ−1−λ=5−10λ.\begin{aligned} \vec\alpha\cdot\vec\beta_2 &= 3(2 - 3\lambda) + (-1)(1 + \lambda) \\ &\quad + 0(-3) \\ &= 6 - 9\lambda - 1 - \lambda \\ &= 5 - 10\lambda. \end{aligned}

    Setting this equal to 0 gives 10λ=510\lambda = 5⁠, so λ=12\lambda = \frac12⁠.

    The k^\hat k part of α⃗\vec\alpha is 0, so the −3k^-3\hat k of β⃗2\vec\beta_2 contributes nothing to the scalar product. The equation is linear in λ\lambda with nonzero coefficient, so there is exactly one λ\lambda⁠: the split is unique.
    u⃗⋅v⃗=0  ⟺  u⃗⊥v⃗\vec u\cdot\vec v = 0 \iff \vec u \perp \vec v for nonzero vectors (Section 10.6.1, Observation 2). Lesson: the dot product

  4. Substitute, and check every condition

    β⃗1=12(3i^−j^)=32i^−12j^,β⃗2=(2−32)i^+(1+12)j^−3k^=12i^+32j^−3k^.\begin{aligned} \vec\beta_1 &= \tfrac12(3\hat i - \hat j) = \tfrac32\hat i - \tfrac12\hat j, \\ \vec\beta_2 &= \left(2 - \tfrac32\right)\hat i + \left(1 + \tfrac12\right)\hat j - 3\hat k \\ &= \tfrac12\hat i + \tfrac32\hat j - 3\hat k. \end{aligned}

    Checks: β⃗1+β⃗2=2i^+j^−3k^=β⃗\vec\beta_1 + \vec\beta_2 = 2\hat i + \hat j - 3\hat k = \vec\beta⁠; β⃗1=12α⃗\vec\beta_1 = \frac12\vec\alpha is parallel to α⃗\vec\alpha⁠; and α⃗⋅β⃗2=32−32+0=0\vec\alpha\cdot\vec\beta_2 = \frac32 - \frac32 + 0 = 0⁠.

    All three requirements of the question are confirmed directly, not just the one used to find λ\lambda⁠. Since λ=12>0\lambda = \frac12 > 0⁠, β⃗1\vec\beta_1 points the same way as α⃗\vec\alpha and is half as long.
    Multiplying a vector by a scalar, by components (Section 10.5.1). Lesson: Examples 26–30

  5. β⃗1\vec\beta_1 is the shadow of β⃗\vec\beta on α⃗\vec\alpha

    α⃗⋅β⃗∣α⃗∣2α⃗=6−1+09+1α⃗=510α⃗=12α⃗=β⃗1.\begin{aligned} \frac{\vec\alpha\cdot\vec\beta}{|\vec\alpha|^2}\vec\alpha &= \frac{6 - 1 + 0}{9 + 1}\vec\alpha \\ &= \frac{5}{10}\vec\alpha = \frac12\vec\alpha = \vec\beta_1. \end{aligned}
    O β₁, length √10/2 α, length √10 β₂, length √46/2 β, length √14 β = β₁ + β₂ β₁ ∥ α, β₂ ⟂ α
    α⃗\vec\alpha and β⃗\vec\beta drawn in their own plane, true to scale: β⃗1\vec\beta_1 lies along α⃗\vec\alpha and is half of it, β⃗2\vec\beta_2 stands at right angles to α⃗\vec\alpha⁠, and together they make β⃗\vec\beta⁠, with ∣β⃗1∣2+∣β⃗2∣2=104+464=14=∣β⃗∣2|\vec\beta_1|^2 + |\vec\beta_2|^2 = \frac{10}{4} + \frac{46}{4} = 14 = |\vec\beta|^2⁠.
    Solving α⃗⋅(β⃗−λα⃗)=0\vec\alpha\cdot(\vec\beta - \lambda\vec\alpha) = 0 in general gives λ=α⃗⋅β⃗∣α⃗∣2\lambda = \frac{\vec\alpha\cdot\vec\beta}{|\vec\alpha|^2}⁠, so β⃗1\vec\beta_1 is always the projection vector of β⃗\vec\beta on α⃗\vec\alpha and β⃗2\vec\beta_2 is what is left over. The words "right handed system" only fix how i^,j^,k^\hat i, \hat j, \hat k are arranged; this split uses scalar products only, so it does not depend on that.
    Projection of a vector on a vector (Section 10.6.2). Basics: cos θ is a shadow
Result

β⃗=β⃗1+β⃗2\vec\beta = \vec\beta_1 + \vec\beta_2 with β⃗1=32i^−12j^\vec\beta_1 = \frac32\hat i - \frac12\hat j (parallel to α⃗\vec\alpha⁠) and β⃗2=12i^+32j^−3k^\vec\beta_2 = \frac12\hat i + \frac32\hat j - 3\hat k (perpendicular to α⃗\vec\alpha⁠).

(Textbook answer: β⃗1=3λi^−λj^\vec\beta_1 = 3\lambda\hat i - \lambda\hat j⁠, β⃗2=(2−3λ)i^+(1+λ)j^−3k^\vec\beta_2 = (2 - 3\lambda)\hat i + (1 + \lambda)\hat j - 3\hat k⁠, 3(2−3λ)−(1+λ)=03(2 - 3\lambda) - (1 + \lambda) = 0⁠, λ=12\lambda = \frac12⁠; therefore β⃗1=32i^−12j^\vec\beta_1 = \frac32\hat i - \frac12\hat j and β⃗2=12i^+32j^−3k^\vec\beta_2 = \frac12\hat i + \frac32\hat j - 3\hat k⁠.)

Check it with numbers

∣β⃗1∣=102≈1.581139|\vec\beta_1| = \frac{\sqrt{10}}{2} \approx 1.581139⁠, ∣β⃗2∣=462≈3.391165|\vec\beta_2| = \frac{\sqrt{46}}{2} \approx 3.391165 and ∣β⃗∣=14≈3.741657|\vec\beta| = \sqrt{14} \approx 3.741657⁠; exactly, 104+464=14\frac{10}{4} + \frac{46}{4} = 14⁠, Pythagoras for the right angle between β⃗1\vec\beta_1 and β⃗2\vec\beta_2⁠.

The projection of β⃗\vec\beta on α⃗\vec\alpha is α⃗⋅β⃗∣α⃗∣=510=102\frac{\vec\alpha\cdot\vec\beta}{|\vec\alpha|} = \frac{5}{\sqrt{10}} = \frac{\sqrt{10}}{2}⁠, the length of β⃗1\vec\beta_1⁠, as it should be.

Common mistakes
  1. Taking β⃗1=α⃗\vec\beta_1 = \vec\alpha⁠. Parallel does not mean equal; β⃗1\vec\beta_1 is λα⃗\lambda\vec\alpha with λ\lambda to be found.
  2. Using the wrong perpendicularity. The condition is α⃗⋅β⃗2=0\vec\alpha\cdot\vec\beta_2 = 0⁠, not β⃗⋅β⃗2=0\vec\beta\cdot\vec\beta_2 = 0 or β⃗1⋅β⃗=0\vec\beta_1\cdot\vec\beta = 0⁠.
  3. Dropping the k^\hat k part. α⃗\vec\alpha has no k^\hat k component, so the whole −3k^-3\hat k of β⃗\vec\beta belongs to β⃗2\vec\beta_2⁠.
What to take away
  1. To split β⃗\vec\beta along and across α⃗\vec\alpha⁠: β⃗1=λα⃗\vec\beta_1 = \lambda\vec\alpha⁠, β⃗2=β⃗−λα⃗\vec\beta_2 = \vec\beta - \lambda\vec\alpha⁠, and α⃗⋅β⃗2=0\vec\alpha\cdot\vec\beta_2 = 0⁠.
  2. The answer is always λ=α⃗⋅β⃗∣α⃗∣2\lambda = \frac{\vec\alpha\cdot\vec\beta}{|\vec\alpha|^2}⁠: β⃗1\vec\beta_1 is the projection vector.
  3. Check by adding the parts and by the scalar product with α⃗\vec\alpha⁠.

Every fact used, and where it comes from

FactUsed inSource
Collinear vectors: u⃗=λα⃗\vec u = \lambda\vec\alphaSteps 1, 4Section 10.5.1, Remarks (i) (Lesson: components)
Sum, difference and scalar multiples by componentsSteps 2, 4Section 10.5.1
u⃗⋅v⃗\vec u\cdot\vec v in components; zero for perpendicular vectorsSteps 3, 4Section 10.6.1 and its Observation 2 (Lesson: the dot product)
Projection of a vector on a vectorStep 5, CheckSection 10.6.2
Right handed system of i^,j^,k^\hat i, \hat j, \hat kStep 5Section 10.6.3, Fig 10.22
The book's working and answerResultExample 30, Miscellaneous Examples (Lesson: Examples 26–30)
Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →