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Chapter 10 · Vector Algebra

Exercise 10.3 · Q3: Projection of i^−j^\hat i-\hat j on i^+j^\hat i+\hat j

Exercise 10.3Projection of a vector on a vector · Section 10.6.2
The question

Find the projection of the vector i^−j^\hat i - \hat j on the vector i^+j^\hat i + \hat j⁠.

Exam answer

Let a⃗=i^−j^\vec a = \hat i - \hat j and b⃗=i^+j^\vec b = \hat i + \hat j⁠. The projection of a⃗\vec a on b⃗\vec b is a⃗⋅b⃗∣b⃗∣\frac{\vec a\cdot\vec b}{|\vec b|}⁠.

a⃗⋅b⃗=(1)(1)+(−1)(1)=1−1=0\begin{aligned} \vec a\cdot\vec b &= (1)(1) + (-1)(1) \\ &= 1 - 1 = 0 \end{aligned} ∣b⃗∣=12+12=2|\vec b| = \sqrt{1^2 + 1^2} = \sqrt2 Projection=02=0\text{Projection} = \frac{0}{\sqrt2} = 0

∴\therefore the projection of i^−j^\hat i - \hat j on i^+j^\hat i + \hat j is 0.

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

The one idea this problem is built on

The projection of a⃗\vec a on b⃗\vec b is the signed length of the shadow of a⃗\vec a on the line of b⃗\vec b⁠:

∣a⃗∣cos⁡θ=a⃗⋅b⃗∣b⃗∣.|\vec a|\cos\theta = \frac{\vec a\cdot\vec b}{|\vec b|}.

Here the scalar product is 0: the two vectors are perpendicular, so the shadow of a⃗\vec a on the line of b⃗\vec b shrinks to a single point and the projection is 0.

Given, to find, and the plan
Given
a⃗=i^−j^\vec a = \hat i - \hat j⁠, b⃗=i^+j^\vec b = \hat i + \hat j⁠.
To find
The projection of a⃗\vec a on b⃗\vec b⁠.
Plan
(1) Compute a⃗⋅b⃗\vec a\cdot\vec b⁠. (2) Compute ∣b⃗∣|\vec b|⁠. (3) Divide. (4) Explain the zero: the vectors are perpendicular.

Basics you need

1Projection on a vector

The projection of a⃗\vec a on b⃗\vec b is a⃗⋅b^=a⃗⋅b⃗∣b⃗∣=1∣b⃗∣(a⃗⋅b⃗)\vec a\cdot\hat b = \vec a\cdot\frac{\vec b}{|\vec b|} = \frac{1}{|\vec b|}(\vec a\cdot\vec b) (Section 10.6.2, Observation 2, Fig 10.20). Lesson: the dot product

2Scalar product in components

a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3⁠; a missing component is 0 (Section 10.6.1). Lesson: Examples 13–21

3Perpendicular vectors

For nonzero a⃗\vec a⁠, b⃗\vec b⁠: a⃗⋅b⃗=0  ⟺  a⃗⊥b⃗\vec a\cdot\vec b = 0 \iff \vec a \perp \vec b (Section 10.6.1, Observation 2). Basics: cos θ is a shadow

Solution

  1. The scalar product

    a⃗⋅b⃗=(1)(1)+(−1)(1)+(0)(0)=1−1+0=0.\begin{aligned} \vec a\cdot\vec b &= (1)(1) + (-1)(1) + (0)(0) \\ &= 1 - 1 + 0 = 0. \end{aligned}
    Both vectors have no k^\hat k-part, so their k^\hat k-components are 0 and that product adds nothing. The i^\hat i-components give 1×1=11 \times 1 = 1⁠, and the j^\hat j-components give (−1)×1=−1(-1) \times 1 = -1⁠; they cancel exactly.
    a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3 (Section 10.6.1). Lesson: the dot product

  2. The length of b⃗\vec b

    ∣b⃗∣=12+12=2.|\vec b| = \sqrt{1^2 + 1^2} = \sqrt2.
    We project on b⃗\vec b⁠, so we divide by the length of b⃗\vec b⁠, the vector that fixes the line. It is not zero, so the division is allowed.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  3. Divide

    a⃗⋅b⃗∣b⃗∣=02=0.\frac{\vec a\cdot\vec b}{|\vec b|} = \frac{0}{\sqrt2} = 0.

    So the projection of i^−j^\hat i - \hat j on i^+j^\hat i + \hat j is 0.

    X Y l O a = i − j b = i + j a · b = 1 − 1 = 0 so a and b meet at 90° shadow of a on l: only the point O
    Both vectors lie in the xy-plane, drawn with one scale on both axes. a⃗=i^−j^\vec a = \hat i - \hat j points down and to the right at 45∘45^\circ below OX, and b⃗=i^+j^\vec b = \hat i + \hat j points up and to the right at 45∘45^\circ above OX, so they meet at 90∘90^\circ⁠. Dropping a perpendicular from the tip of a⃗\vec a to the line ll of b⃗\vec b lands at O itself: the shadow has length 0.
    Zero divided by any nonzero number is zero. The length ∣b⃗∣=2|\vec b| = \sqrt2 does not change the answer, but the formula still needs it, and it is not zero.
    Projection of a⃗\vec a on b⃗\vec b (Section 10.6.2, Observation 2). Lesson: Examples 13–21

  4. Why the projection is zero

    Both vectors are nonzero and a⃗⋅b⃗=0\vec a\cdot\vec b = 0⁠, so a⃗⊥b⃗\vec a \perp \vec b⁠: the angle between them is θ=π2\theta = \frac{\pi}{2}⁠. Then the projection is ∣a⃗∣cos⁡π2=2×0=0|\vec a|\cos\frac{\pi}{2} = \sqrt2 \times 0 = 0⁠.

    A projection is the length of a shadow cast at right angles onto a line. A vector that is itself at right angles to the line casts no shadow along it, only a point, which is what the book says for θ=π2\theta = \frac{\pi}{2}⁠: the projection vector is the zero vector. A zero projection does not mean a⃗\vec a is zero; ∣a⃗∣=2|\vec a| = \sqrt2⁠.
    Section 10.6.1, Observation 2; Section 10.6.2, Observation 4. Lesson: the dot product
Result

The projection of i^−j^\hat i - \hat j on i^+j^\hat i + \hat j is 0 (the two vectors are perpendicular).

(Textbook answer: 0.)

Check it with numbers

The unit vector along b⃗\vec b is b^=12(i^+j^)\hat b = \frac{1}{\sqrt2}(\hat i + \hat j) ≈0.707107 i^+0.707107 j^\approx 0.707107\,\hat i + 0.707107\,\hat j⁠, and a⃗⋅b^≈(1)(0.707107)\vec a\cdot\hat b \approx (1)(0.707107) +(−1)(0.707107)=0+ (-1)(0.707107) = 0⁠.

The angle: cos⁡θ=02×2=0\cos\theta = \frac{0}{\sqrt2 \times \sqrt2} = 0⁠, so θ=90∘\theta = 90^\circ⁠. In the xy-plane, a⃗\vec a makes −45∘-45^\circ with OX and b⃗\vec b makes +45∘+45^\circ⁠, and 45∘−(−45∘)=90∘45^\circ - (-45^\circ) = 90^\circ⁠. Both have length 2≈1.414214\sqrt2 \approx 1.414214⁠.

The other way round, the projection of b⃗\vec b on a⃗\vec a is 02=0\frac{0}{\sqrt2} = 0 too.

Common mistakes
  1. Losing the minus sign. Taking a⃗⋅b⃗=1+1=2\vec a\cdot\vec b = 1 + 1 = 2 gives 22=2\frac{2}{\sqrt2} = \sqrt2⁠, which would mean a⃗\vec a lies along b⃗\vec b⁠.
  2. Thinking a zero projection means a zero vector. a⃗\vec a has length 2\sqrt2⁠; it is the shadow that has length 0.
  3. Dividing by ∣a⃗∣|\vec a|⁠. Here it happens to give 0 as well, but the projection on b⃗\vec b divides by ∣b⃗∣|\vec b|⁠.
What to take away
  1. Projection of a⃗\vec a on b⃗\vec b⁠: a⃗⋅b⃗∣b⃗∣\frac{\vec a\cdot\vec b}{|\vec b|}⁠.
  2. The projection is 0 exactly when a⃗⋅b⃗=0\vec a\cdot\vec b = 0⁠, that is, when nonzero vectors are perpendicular.
  3. Missing components count as 0 in the scalar product.

Every fact used, and where it comes from

FactUsed inSource
a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3Step 1Section 10.6.1 (Lesson: the dot product)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Step 2Section 10.5.1
Projection of a⃗\vec a on b⃗\vec b is 1∣b⃗∣(a⃗⋅b⃗)\frac{1}{|\vec b|}(\vec a\cdot\vec b)Step 3Section 10.6.2, Observation 2, Fig 10.20
a⃗⋅b⃗=0  ⟺  a⃗⊥b⃗\vec a\cdot\vec b = 0 \iff \vec a \perp \vec b (nonzero vectors)Step 4Section 10.6.1, Observation 2
θ=π2\theta = \frac{\pi}{2} gives the zero projection vectorStep 4Section 10.6.2, Observation 4 (Basics: cos θ is a shadow)
The answerResultNCERT answer key, Exercise 10.3 Q3 (Lesson: the dot product)
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