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Chapter 10 · Vector Algebra

Miscellaneous · Q6: Magnitude 5, parallel to the sum of 2i^+3j^−k^2\hat i+3\hat j-\hat k and i^−2j^+k^\hat i-2\hat j+\hat k

Miscellaneous ExerciseResultant, unit vector, scalar multiple · Sections 10.5 and 10.5.1
The question

Find a vector of magnitude 5 units, and parallel to the resultant of the vectors a⃗=2i^+3j^−k^\vec a = 2\hat i + 3\hat j - \hat k and b⃗=i^−2j^+k^\vec b = \hat i - 2\hat j + \hat k⁠.

Exam answer

Let c⃗\vec c be the resultant of a⃗\vec a and b⃗\vec b⁠. Adding corresponding components,

c⃗=a⃗+b⃗=(2+1)i^+(3−2)j^+(−1+1)k^=3i^+j^\begin{aligned} \vec c = \vec a + \vec b &= (2 + 1)\hat i + (3 - 2)\hat j \\ &\quad + (-1 + 1)\hat k \\ &= 3\hat i + \hat j \end{aligned} ∣c⃗∣=32+12=10|\vec c| = \sqrt{3^2 + 1^2} = \sqrt{10}

The unit vector along c⃗\vec c is

c^=c⃗∣c⃗∣=110(3i^+j^)\hat c = \frac{\vec c}{|\vec c|} = \frac{1}{\sqrt{10}}(3\hat i + \hat j)

So the vector of magnitude 5 parallel to c⃗\vec c is

5c^=510(3i^+j^)=102(3i^+j^)\begin{aligned} 5\hat c &= \frac{5}{\sqrt{10}}(3\hat i + \hat j) \\ &= \frac{\sqrt{10}}{2}(3\hat i + \hat j) \end{aligned}

∴\therefore the required vector is 3102i^+102j^\dfrac{3\sqrt{10}}{2}\hat i + \dfrac{\sqrt{10}}{2}\hat j⁠.

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

The one idea this problem is built on

To get a vector of a chosen length in a given direction, first shrink the direction vector to length 1 (its unit vector), then multiply by the length you want: required vector = 5c^5\hat c⁠.

Given, to find, and the plan
Given
a⃗=2i^+3j^−k^\vec a = 2\hat i + 3\hat j - \hat k⁠, b⃗=i^−2j^+k^\vec b = \hat i - 2\hat j + \hat k⁠.
To find
A vector of magnitude 5 parallel to a⃗+b⃗\vec a + \vec b⁠.
Plan
(1) Find the resultant c⃗=a⃗+b⃗\vec c = \vec a + \vec b⁠. (2) Find ∣c⃗∣|\vec c|⁠. (3) Form c^\hat c⁠. (4) Multiply by 5 and simplify. (5) Check the length is 5.

Basics you need

1Resultant

The resultant of two vectors is their sum; in components, add the matching components (Section 10.5.1, (i)). Lesson: adding arrows, head to tail

2Unit vector in a direction

c^=1∣c⃗∣c⃗\hat c = \frac{1}{|\vec c|}\vec c⁠, and ∣λc⃗∣=∣λ∣∣c⃗∣|\lambda\vec c| = |\lambda||\vec c| (Section 10.5). Lesson: components

3Rationalising

510=51010=102\frac{5}{\sqrt{10}} = \frac{5\sqrt{10}}{10} = \frac{\sqrt{10}}{2}⁠: multiply top and bottom by 10\sqrt{10}⁠, and use 10×10=10\sqrt{10} \times \sqrt{10} = 10⁠.

Solution

  1. The resultant

    c⃗=a⃗+b⃗=(2+1)i^+(3−2)j^+(−1+1)k^=3i^+j^+0k^=3i^+j^.\begin{aligned} \vec c &= \vec a + \vec b \\ &= (2 + 1)\hat i + (3 - 2)\hat j + (-1 + 1)\hat k \\ &= 3\hat i + \hat j + 0\hat k = 3\hat i + \hat j. \end{aligned}
    Vectors in component form add component by component. The k^\hat k parts −1-1 and +1+1 cancel, so the resultant lies in the XY-plane even though a⃗\vec a and b⃗\vec b do not.
    Sum by components (Section 10.5.1, (i)). Lesson: components

  2. Its length

    ∣c⃗∣=32+12+02=10.|\vec c| = \sqrt{3^2 + 1^2 + 0^2} = \sqrt{10}.
    The length formula x2+y2+z2\sqrt{x^2 + y^2 + z^2} with x=3x = 3⁠, y=1y = 1⁠, z=0z = 0⁠.
    ∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  3. The unit vector along the resultant

    c^=1∣c⃗∣c⃗=110(3i^+j^).\hat c = \frac{1}{|\vec c|}\vec c = \frac{1}{\sqrt{10}}(3\hat i + \hat j).
    Dividing a vector by its own length keeps its direction and makes its length 1, since ∣110c⃗∣=110×10=1\left|\frac{1}{\sqrt{10}}\vec c\right| = \frac{1}{\sqrt{10}} \times \sqrt{10} = 1⁠.
    Unit vector in the direction of c⃗\vec c (Section 10.5); the same pattern as Example 7 and Example 8. Lesson: Examples 4–10

  4. Stretch it to length 5

    5c^=510(3i^+j^)=102(3i^+j^)=3102i^+102j^.\begin{aligned} 5\hat c &= \frac{5}{\sqrt{10}}(3\hat i + \hat j) = \frac{\sqrt{10}}{2}(3\hat i + \hat j) \\ &= \frac{3\sqrt{10}}{2}\hat i + \frac{\sqrt{10}}{2}\hat j. \end{aligned}

    So the required vector is 3102i^+102j^\frac{3\sqrt{10}}{2}\hat i + \frac{\sqrt{10}}{2}\hat j.

    O a, √14 b, √6 c = a + b, √10 5ĉ, length 5 5ĉ = (√10/2)(3î + ĵ)
    The triangle of a⃗\vec a⁠, b⃗\vec b and c⃗=a⃗+b⃗\vec c = \vec a + \vec b drawn true to scale in its own plane (∣a⃗∣=14|\vec a| = \sqrt{14}⁠, ∣b⃗∣=6|\vec b| = \sqrt6⁠, ∣c⃗∣=10|\vec c| = \sqrt{10}⁠); the vector 5c^5\hat c of length 5 runs along c⃗\vec c⁠.
    Multiplying by the positive number 5 keeps the direction and makes the length 5×1=55 \times 1 = 5⁠. Rationalising: 510=51010=102\frac{5}{\sqrt{10}} = \frac{5\sqrt{10}}{10} = \frac{\sqrt{10}}{2}⁠.
    ∣λc⃗∣=∣λ∣∣c⃗∣|\lambda\vec c| = |\lambda||\vec c|⁠, same direction for λ>0\lambda > 0 (Section 10.5). Exercise 10.2 Q10 does the same with length 8. Lesson: components, lengths and unit vectors

  5. Check, and the other parallel vector

    (3102)2+(102)2=904+104=25,\left(\tfrac{3\sqrt{10}}{2}\right)^2 + \left(\tfrac{\sqrt{10}}{2}\right)^2 = \tfrac{90}{4} + \tfrac{10}{4} = 25,

    so the length is 25=5\sqrt{25} = 5⁠. The opposite vector −5c^=−3102i^−102j^-5\hat c = -\frac{3\sqrt{10}}{2}\hat i - \frac{\sqrt{10}}{2}\hat j also has length 5 and is parallel (collinear) to c⃗\vec c⁠, pointing the other way.

    "Parallel" allows either direction along the line of c⃗\vec c⁠. The question asks for "a vector", so one answer is enough; the natural one, and the one in the answer key, points the same way as the resultant.
    Collinear vectors: b⃗=λa⃗\vec b = \lambda\vec a for some scalar λ\lambda (Section 10.5.1, Remarks (i)). Lesson: components
Result

The required vector is 5c^=3102i^+102j^5\hat c = \frac{3\sqrt{10}}{2}\hat i + \frac{\sqrt{10}}{2}\hat j (and −5c^-5\hat c is the parallel vector of length 5 pointing the other way).

(Textbook answer: 3210 i^+102j^\frac32\sqrt{10}\,\hat i + \frac{\sqrt{10}}{2}\hat j⁠.)

Check it with numbers

3102≈4.743416\frac{3\sqrt{10}}{2} \approx 4.743416 and 102≈1.581139\frac{\sqrt{10}}{2} \approx 1.581139⁠. Their ratio is exactly 3, the same as 3:13 : 1 in c⃗\vec c⁠, so the vector is along c⃗\vec c⁠.

10≈3.162278\sqrt{10} \approx 3.162278⁠, so the answer is about 5÷3.162278≈1.5811395 \div 3.162278 \approx 1.581139 times c⃗\vec c⁠.

Common mistakes
  1. Multiplying c⃗\vec c by 5 directly. 5c⃗=15i^+5j^5\vec c = 15\hat i + 5\hat j has length 5105\sqrt{10}⁠, not 5. First make the unit vector.
  2. Sign slips in the sum. −1+1=0-1 + 1 = 0 for k^\hat k and 3−2=13 - 2 = 1 for j^\hat j⁠.
  3. Dropping the root. ∣c⃗∣=10|\vec c| = \sqrt{10}⁠, not 10.
What to take away
  1. Vector of length LL along c⃗\vec c⁠: Lc^=L∣c⃗∣c⃗L\hat c = \frac{L}{|\vec c|}\vec c⁠.
  2. The resultant of two vectors is their sum.
  3. Check the answer by squaring and adding its components.

Every fact used, and where it comes from

FactUsed inSource
Sum (resultant) by componentsStep 1Section 10.5.1, (i) (Lesson: adding arrows, head to tail)
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Steps 2, 5Section 10.5.1 (Basics: use Pythagoras twice)
c^=1∣c⃗∣c⃗\hat c = \frac{1}{|\vec c|}\vec c⁠; ∣λc⃗∣=∣λ∣∣c⃗∣|\lambda\vec c| = |\lambda||\vec c|Steps 3, 4Section 10.5
Collinear vectors b⃗=λa⃗\vec b = \lambda\vec aStep 5Section 10.5.1, Remarks (i)
The answerResultNCERT answer key, Miscellaneous Exercise on Chapter 10, Q6 (Lesson: put it all together)
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