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Chapter 10 · Vector Algebra

Example 15: a⃗+b⃗⊥a⃗−b⃗\vec a+\vec b\perp\vec a-\vec b for 5i^−j^−3k^5\hat i-\hat j-3\hat k⁠, i^+3j^−5k^\hat i+3\hat j-5\hat k

ExamplesPerpendicular vectors, scalar product · Section 10.6.1
The question

If a⃗=5i^−j^−3k^\vec a = 5\hat i - \hat j - 3\hat k and b⃗=i^+3j^−5k^\vec b = \hat i + 3\hat j - 5\hat k⁠, then show that the vectors a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b are perpendicular.

Exam answer

Two nonzero vectors are perpendicular if their scalar product is zero.

a⃗+b⃗=(5+1)i^+(−1+3)j^+(−3−5)k^=6i^+2j^−8k^\begin{aligned} \vec a + \vec b &= (5 + 1)\hat i + (-1 + 3)\hat j \\ &\quad + (-3 - 5)\hat k \\ &= 6\hat i + 2\hat j - 8\hat k \end{aligned} a⃗−b⃗=(5−1)i^+(−1−3)j^+(−3+5)k^=4i^−4j^+2k^\begin{aligned} \vec a - \vec b &= (5 - 1)\hat i + (-1 - 3)\hat j \\ &\quad + (-3 + 5)\hat k \\ &= 4\hat i - 4\hat j + 2\hat k \end{aligned} (a⃗+b⃗)⋅(a⃗−b⃗)=(6)(4)+(2)(−4)+(−8)(2)=24−8−16=0\begin{aligned} &(\vec a + \vec b)\cdot(\vec a - \vec b) \\ &= (6)(4) + (2)(-4) + (-8)(2) \\ &= 24 - 8 - 16 = 0 \end{aligned}

Both a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b are nonzero, and their scalar product is 0.

∴ a⃗+b⃗\therefore\ \vec a + \vec b and a⃗−b⃗\vec a - \vec b are perpendicular. Hence proved.

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

The one idea this problem is built on

Two nonzero vectors are perpendicular if and only if their scalar product is zero: u⃗⋅v⃗=0  ⟺  u⃗⊥v⃗\vec u\cdot\vec v = 0 \iff \vec u \perp \vec v⁠. So we build a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b in components and show that their scalar product is 0. Behind the numbers: (a⃗+b⃗)⋅(a⃗−b⃗)=∣a⃗∣2−∣b⃗∣2(\vec a + \vec b)\cdot(\vec a - \vec b) = |\vec a|^2 - |\vec b|^2⁠, and here ∣a⃗∣=∣b⃗∣|\vec a| = |\vec b|⁠.

Given, to find, and the plan
Given
a⃗=5i^−j^−3k^\vec a = 5\hat i - \hat j - 3\hat k⁠, b⃗=i^+3j^−5k^\vec b = \hat i + 3\hat j - 5\hat k⁠.
To show
a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b are perpendicular.
Plan
(1) Find a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b⁠. (2) Compute their scalar product. (3) Check both are nonzero and conclude. (4) See why: ∣a⃗∣=∣b⃗∣|\vec a| = |\vec b|⁠.

Basics you need

1Sum and difference in components

a⃗±b⃗=(a1±b1)i^\vec a \pm \vec b = (a_1 \pm b_1)\hat i +(a2±b2)j^+(a3±b3)k^+ (a_2 \pm b_2)\hat j + (a_3 \pm b_3)\hat k (Section 10.5.1). Lesson: components

2Scalar product in components

u⃗⋅v⃗=u1v1+u2v2+u3v3\vec u\cdot\vec v = u_1v_1 + u_2v_2 + u_3v_3 (Section 10.6.1). Lesson: the dot product

3Perpendicular vectors

For nonzero u⃗\vec u⁠, v⃗\vec v⁠: u⃗⋅v⃗=0  ⟺  u⃗⊥v⃗\vec u\cdot\vec v = 0 \iff \vec u \perp \vec v (Section 10.6.1, Observation 2). Lesson: Examples 13–21

Solution

  1. The sum and the difference

    a⃗+b⃗=(5+1)i^+(−1+3)j^+(−3−5)k^=6i^+2j^−8k^,a⃗−b⃗=(5−1)i^+(−1−3)j^+(−3+5)k^=4i^−4j^+2k^.\begin{aligned} \vec a + \vec b &= (5 + 1)\hat i + (-1 + 3)\hat j \\ &\quad + (-3 - 5)\hat k = 6\hat i + 2\hat j - 8\hat k, \\ \vec a - \vec b &= (5 - 1)\hat i + (-1 - 3)\hat j \\ &\quad + (-3 + 5)\hat k = 4\hat i - 4\hat j + 2\hat k. \end{aligned}
    Add or subtract matching components. In the difference every component of b⃗\vec b changes sign: −1−3=−4-1 - 3 = -4 and −3−(−5)=−3+5=2-3 - (-5) = -3 + 5 = 2⁠.
    Sum and difference by components (Section 10.5.1 (i), (ii)). Lesson: components

  2. Their scalar product

    (a⃗+b⃗)⋅(a⃗−b⃗)=(6)(4)+(2)(−4)+(−8)(2)=24−8−16=0.\begin{aligned} &(\vec a + \vec b)\cdot(\vec a - \vec b) \\ &= (6)(4) + (2)(-4) + (-8)(2) \\ &= 24 - 8 - 16 = 0. \end{aligned}
    Multiply matching components and add. The positive product 24 is cancelled exactly by the two negative products −8-8 and −16-16⁠.
    u⃗⋅v⃗=u1v1+u2v2+u3v3\vec u\cdot\vec v = u_1v_1 + u_2v_2 + u_3v_3 (Section 10.6.1). Lesson: the dot product

  3. Conclude

    Both vectors are nonzero (6i^+2j^−8k^6\hat i + 2\hat j - 8\hat k and 4i^−4j^+2k^4\hat i - 4\hat j + 2\hat k have nonzero components), and their scalar product is 0. So a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b are perpendicular.

    The test u⃗⋅v⃗=0  ⟺  u⃗⊥v⃗\vec u\cdot\vec v = 0 \iff \vec u \perp \vec v is stated for nonzero vectors: the scalar product of anything with 0⃗\vec 0 is 0, but the zero vector has no direction, so "perpendicular" would mean nothing. That is why the book begins with "two nonzero vectors".
    Observation 2 (Section 10.6.1). Lesson: Examples 13–21

  4. Why it works: equal lengths

    Expanding with the rules of the scalar product,

    (a⃗+b⃗)⋅(a⃗−b⃗)=a⃗⋅a⃗−a⃗⋅b⃗+b⃗⋅a⃗−b⃗⋅b⃗=∣a⃗∣2−∣b⃗∣2,\begin{aligned} &(\vec a + \vec b)\cdot(\vec a - \vec b) \\ &= \vec a\cdot\vec a - \vec a\cdot\vec b + \vec b\cdot\vec a - \vec b\cdot\vec b \\ &= |\vec a|^2 - |\vec b|^2, \end{aligned}

    and here ∣a⃗∣2=25+1+9=35|\vec a|^2 = 25 + 1 + 9 = 35 and ∣b⃗∣2=1+9+25=35|\vec b|^2 = 1 + 9 + 25 = 35⁠, so the difference is 0.

    O a b a + b a − b 90° |a| = |b| = √35 a rhombus (a + b) · (a − b) = 0
    The parallelogram on a⃗\vec a and b⃗\vec b⁠, drawn in its own plane, true to scale. Its sides are equal (35\sqrt{35} each), so it is a rhombus, and its diagonals a⃗+b⃗\vec a + \vec b (length 104\sqrt{104}⁠) and a⃗−b⃗\vec a - \vec b (length 6) cross at right angles.
    The middle terms cancel because the scalar product is commutative: b⃗⋅a⃗=a⃗⋅b⃗\vec b\cdot\vec a = \vec a\cdot\vec b⁠. Geometrically, a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b are the two diagonals of the parallelogram on a⃗\vec a and b⃗\vec b⁠; with equal sides it is a rhombus, and the diagonals of a rhombus are perpendicular.
    Distributivity and commutativity (Section 10.6.1, Property 1 and Observation 7); a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2 (Observation 3). Basics: dot products obey the same rule
Result

(a⃗+b⃗)⋅(a⃗−b⃗)=0(\vec a + \vec b)\cdot(\vec a - \vec b) = 0 with both vectors nonzero, so a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b are perpendicular.

(Textbook answer: a⃗+b⃗=6i^+2j^−8k^\vec a + \vec b = 6\hat i + 2\hat j - 8\hat k⁠, a⃗−b⃗=4i^−4j^+2k^\vec a - \vec b = 4\hat i - 4\hat j + 2\hat k⁠, and (a⃗+b⃗)⋅(a⃗−b⃗)=24−8−16=0(\vec a + \vec b)\cdot(\vec a - \vec b) = 24 - 8 - 16 = 0⁠; "Hence a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b are perpendicular vectors.")

Check it with numbers

The lengths: ∣a⃗+b⃗∣=36+4+64|\vec a + \vec b| = \sqrt{36 + 4 + 64} =104≈10.198039= \sqrt{104} \approx 10.198039 and ∣a⃗−b⃗∣=16+16+4=6|\vec a - \vec b| = \sqrt{16 + 16 + 4} = 6⁠. For perpendicular diagonals of a rhombus, half of each diagonal and a side form a right triangle: (1042)2+32=26+9=35=∣a⃗∣2\left(\frac{\sqrt{104}}{2}\right)^2 + 3^2 = 26 + 9 = 35 = |\vec a|^2⁠.

The angle between a⃗\vec a and b⃗\vec b themselves is not a right angle: a⃗⋅b⃗=5−3+15=17\vec a\cdot\vec b = 5 - 3 + 15 = 17⁠, so cos⁡θ=1735≈0.485714\cos\theta = \frac{17}{35} \approx 0.485714⁠.

Common mistakes
  1. Sign slips in a⃗−b⃗\vec a - \vec b⁠. −3−(−5)=2-3 - (-5) = 2⁠, not −8-8⁠; bracket the components of b⃗\vec b before subtracting.
  2. Testing a⃗⋅b⃗=0\vec a\cdot\vec b = 0⁠. The question is about a⃗+b⃗\vec a + \vec b and a⃗−b⃗\vec a - \vec b⁠; a⃗⋅b⃗=17\vec a\cdot\vec b = 17 is not zero, and that is fine.
  3. Writing (a⃗+b⃗)⋅(a⃗−b⃗)=a⃗2−b⃗2(\vec a + \vec b)\cdot(\vec a - \vec b) = \vec a^2 - \vec b^2 and stopping. It equals ∣a⃗∣2−∣b⃗∣2|\vec a|^2 - |\vec b|^2⁠, which is zero only because the lengths are equal; that has to be checked.
What to take away
  1. To show two nonzero vectors are perpendicular, show that their scalar product is 0.
  2. (a⃗+b⃗)⋅(a⃗−b⃗)=∣a⃗∣2−∣b⃗∣2(\vec a + \vec b)\cdot(\vec a - \vec b) = |\vec a|^2 - |\vec b|^2⁠.
  3. So a⃗+b⃗⊥a⃗−b⃗\vec a + \vec b \perp \vec a - \vec b exactly when ∣a⃗∣=∣b⃗∣|\vec a| = |\vec b|⁠: the diagonals of a rhombus are perpendicular.

Every fact used, and where it comes from

FactUsed inSource
Sum and difference of vectors by componentsStep 1Section 10.5.1 (Lesson: components)
u⃗⋅v⃗=u1v1+u2v2+u3v3\vec u\cdot\vec v = u_1v_1 + u_2v_2 + u_3v_3Step 2Section 10.6.1 (Lesson: the dot product)
u⃗⋅v⃗=0  ⟺  u⃗⊥v⃗\vec u\cdot\vec v = 0 \iff \vec u \perp \vec v for nonzero vectorsStep 3Section 10.6.1, Observation 2
Distributivity; commutativity; a⃗⋅a⃗=∣a⃗∣2\vec a\cdot\vec a = |\vec a|^2Step 4Section 10.6.1, Property 1, Observations 7 and 3
Parallelogram law; diagonals of a rhombus are perpendicularStep 4Section 10.4; school geometry
The book's working and answerResultExample 15, Section 10.6.2 (Lesson: Examples 13–21)
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