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

Exercise 10.2 · Q1: Magnitudes of i^+j^+k^\hat i+\hat j+\hat k⁠, 2i^−7j^−3k^2\hat i-7\hat j-3\hat k and 13(i^+j^−k^)\frac{1}{\sqrt3}(\hat i+\hat j-\hat k)

Exercise 10.2Components of a vector · Section 10.5.1
The question

Compute the magnitude of the following vectors:

a⃗=i^+j^+k;b⃗=2i^−7j^−3k^;c⃗=13i^+13j^−13k^\begin{aligned} &\vec a = \hat i + \hat j + k; \\ &\vec b = 2\hat i - 7\hat j - 3\hat k; \\ &\vec c = \frac{1}{\sqrt3}\hat i + \frac{1}{\sqrt3}\hat j - \frac{1}{\sqrt3}\hat k \end{aligned}
Exam answer

For r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k⁠, ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2}⁠.

Reading the printed kk as k^\hat k⁠, a⃗=i^+j^+k^\vec a = \hat i + \hat j + \hat k⁠.

∣a⃗∣=12+12+12=3\begin{aligned} |\vec a| &= \sqrt{1^2 + 1^2 + 1^2} \\ &= \sqrt3 \end{aligned} ∣b⃗∣=22+(−7)2+(−3)2=4+49+9=62\begin{aligned} |\vec b| &= \sqrt{2^2 + (-7)^2 + (-3)^2} \\ &= \sqrt{4 + 49 + 9} \\ &= \sqrt{62} \end{aligned}

Each component of c⃗\vec c is ±13\pm\frac{1}{\sqrt3}⁠, whose square is 13\frac13⁠.

∣c⃗∣=13+13+13=1=1\begin{aligned} |\vec c| &= \sqrt{\tfrac13 + \tfrac13 + \tfrac13} \\ &= \sqrt1 = 1 \end{aligned}

∴ ∣a⃗∣=3\therefore\ |\vec a| = \sqrt3⁠, ∣b⃗∣=62|\vec b| = \sqrt{62}⁠, ∣c⃗∣=1|\vec c| = 1⁠.

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

The one idea this problem is built on

The magnitude (length) of a vector in component form comes from Pythagoras applied twice: once across the floor, once up the wall. The result is one formula,

∣xi^+yj^+zk^∣=x2+y2+z2.|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}.

Square every component (a negative component gives a positive square), add, and take the square root.

Given, to find, and the plan
Given
a⃗=i^+j^+k^\vec a = \hat i + \hat j + \hat k (the book prints the last term as kk without its hat; it is the unit vector k^\hat k⁠, as in b⃗\vec b and c⃗\vec c⁠), b⃗=2i^−7j^−3k^\vec b = 2\hat i - 7\hat j - 3\hat k and c⃗=13i^+13j^−13k^\vec c = \frac{1}{\sqrt3}\hat i + \frac{1}{\sqrt3}\hat j - \frac{1}{\sqrt3}\hat k⁠.
To find
∣a⃗∣|\vec a|⁠, ∣b⃗∣|\vec b| and ∣c⃗∣|\vec c|⁠.
Plan
(1) Recall why ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2}⁠. (2) Read off the three components of each vector, with their signs. (3) Square, add and take the root.

Basics you need

1Component form

Every vector can be written r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k⁠; the numbers x,y,zx, y, z are its scalar components (Section 10.5.1). A term with no number in front has component 1, and a minus sign belongs to the component: in −3k^-3\hat k the component is −3-3⁠. Lesson: components, every vector is xi + yj + zk

2Pythagoras twice

The length of the floor diagonal of a box with sides x,yx, y is x2+y2\sqrt{x^2 + y^2}⁠; going up by zz gives x2+y2+z2\sqrt{x^2 + y^2 + z^2}⁠. Basics: use Pythagoras twice, floor, then up

3Squares of negatives and of fractions

(−7)2=49(-7)^2 = 49⁠, (−3)2=9(-3)^2 = 9 and (±13)2=13\left(\pm\frac{1}{\sqrt3}\right)^2 = \frac{1}{3}⁠: squaring removes the sign, and (3)2=3(\sqrt3)^2 = 3⁠.

Solution

  1. The formula for the magnitude

    For r⃗=xi^+yj^+zk^\vec r = x\hat i + y\hat j + z\hat k⁠, the length is

    ∣r⃗∣=x2+y2+z2.|\vec r| = \sqrt{x^2 + y^2 + z^2}.
    The point P(x,y,z)P(x, y, z) is reached from O by going xx along the x-axis, yy along the y-axis and zz up. The right triangle on the floor has legs ∣x∣|x| and ∣y∣|y|⁠, so its hypotenuse is x2+y2\sqrt{x^2 + y^2}⁠. That hypotenuse and the vertical side ∣z∣|z| form a second right triangle, whose hypotenuse OP has length (x2+y2)+z2\sqrt{(x^2 + y^2) + z^2}⁠.
    The book derives this in Section 10.5.1 from the right triangles OQP1OQP_1 and OP1POP_1P of Fig 10.14. Basics: use Pythagoras twice

  2. The magnitude of a⃗\vec a

    The components of a⃗=i^+j^+k^\vec a = \hat i + \hat j + \hat k are 1,1,11, 1, 1⁠. So

    ∣a⃗∣=12+12+12=3.|\vec a| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt3.
    Each unit vector appears once, so each component is 1. Geometrically, a⃗\vec a is the long diagonal of a cube of side 1, and 3\sqrt3 is that diagonal's length.
    i^\hat i⁠, j^\hat j⁠, k^\hat k are the unit vectors along the x, y and z axes (Section 10.5.1). Lesson: components

  3. The magnitude of b⃗\vec b

    The components of b⃗=2i^−7j^−3k^\vec b = 2\hat i - 7\hat j - 3\hat k are 2,−7,−32, -7, -3⁠. So

    ∣b⃗∣=22+(−7)2+(−3)2=4+49+9=62.\begin{aligned} |\vec b| &= \sqrt{2^2 + (-7)^2 + (-3)^2} \\ &= \sqrt{4 + 49 + 9} \\ &= \sqrt{62}. \end{aligned}
    7 2 √53 step 1: the floor √(2² + 7²) = √53 √53 3 √62 step 2: then up √(53 + 3²) = √62
    Pythagoras twice for b⃗\vec b⁠, both triangles drawn to the same scale (26 px per unit). The legs are the sizes of the components, ∣2∣=2|2| = 2⁠, ∣−7∣=7|-7| = 7 and ∣−3∣=3|-3| = 3⁠: the floor triangle gives 22+72=53\sqrt{2^2 + 7^2} = \sqrt{53}⁠, and the upright triangle on that diagonal gives 53+32=62\sqrt{53 + 3^2} = \sqrt{62}⁠.
    The minus signs say that b⃗\vec b goes 7 units in the negative y direction and 3 units in the negative z direction. Direction does not change a length, and the squares (−7)2=49(-7)^2 = 49 and (−3)2=9(-3)^2 = 9 are positive, so the signs drop out. 62 has no square factor (62=2×3162 = 2 \times 31⁠), so 62\sqrt{62} is already in simplest form.
    The square of a negative number is positive: (−7)2=(−7)(−7)=49(-7)^2 = (-7)(-7) = 49⁠. Basics: use Pythagoras twice

  4. The magnitude of c⃗\vec c

    The components of c⃗\vec c are 13,13,−13\frac{1}{\sqrt3}, \frac{1}{\sqrt3}, -\frac{1}{\sqrt3}⁠. Each squares to 1(3)2=13\frac{1}{(\sqrt3)^2} = \frac13⁠, so

    ∣c⃗∣=13+13+13=1=1.\begin{aligned} |\vec c| &= \sqrt{\tfrac13 + \tfrac13 + \tfrac13} \\ &= \sqrt1 = 1. \end{aligned}
    The last component carries the minus sign, but its square is still 13\frac13⁠. Three equal squares of 13\frac13 add to 1, so c⃗\vec c has length exactly 1: it is a unit vector. In fact c⃗=13(i^+j^−k^)\vec c = \frac{1}{\sqrt3}(\hat i + \hat j - \hat k)⁠, the vector i^+j^−k^\hat i + \hat j - \hat k (length 3\sqrt3⁠) divided by its own length.
    A vector whose magnitude is 1 is a unit vector; 1∣a⃗∣a⃗\frac{1}{|\vec a|}\vec a is the unit vector in the direction of a⃗\vec a (Sections 10.3 and 10.5). Lesson: components
Result ∣a⃗∣=3,∣b⃗∣=62,∣c⃗∣=1.\begin{aligned} |\vec a| &= \sqrt3, \\ |\vec b| &= \sqrt{62}, \\ |\vec c| &= 1. \end{aligned}

c⃗\vec c is a unit vector. (The book prints a⃗=i^+j^+k\vec a = \hat i + \hat j + k⁠, without the hat on kk⁠; the answer treats it as k^\hat k⁠, as the key's value 3\sqrt3 does.)

(Textbook answer: ∣a⃗∣=3|\vec a| = \sqrt3⁠, ∣b⃗∣=62|\vec b| = \sqrt{62}⁠, ∣c⃗∣=1|\vec c| = 1⁠.)

Check it with numbers

3≈1.7321\sqrt3 \approx 1.7321⁠, and 1.73212≈3.00021.7321^2 \approx 3.0002⁠. 62≈7.8740\sqrt{62} \approx 7.8740⁠, and 7.87402≈62.07.8740^2 \approx 62.0⁠. By the two triangles: 53≈7.2801\sqrt{53} \approx 7.2801⁠, and 7.28012+32≈53.000+9=62.0007.2801^2 + 3^2 \approx 53.000 + 9 = 62.000⁠.

For c⃗\vec c⁠: 13≈0.57735\frac{1}{\sqrt3} \approx 0.57735⁠, and 0.577352≈0.333330.57735^2 \approx 0.33333⁠; three of these add to 0.99999≈10.99999 \approx 1 (the small gap is rounding).

Common mistakes
  1. Squaring the sign wrongly. (−7)2=49(-7)^2 = 49⁠, not −49-49⁠; writing −72-7^2 without brackets invites this slip.
  2. Adding the components instead of their squares. 2−7−3=−82 - 7 - 3 = -8 is not a length; a length is never negative.
  3. Forgetting the square root. 4+49+9=624 + 49 + 9 = 62 is ∣b⃗∣2|\vec b|^2⁠, not ∣b⃗∣|\vec b|⁠.
  4. Getting ∣c⃗∣=13|\vec c| = \frac{1}{\sqrt3} or 3\sqrt3⁠. Square each component first: 13+13+13=1\frac13 + \frac13 + \frac13 = 1⁠.
What to take away
  1. ∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}⁠: Pythagoras twice.
  2. Signs of components do not affect the magnitude.
  3. A vector whose magnitude works out to 1 is a unit vector; 13(i^+j^−k^)\frac{1}{\sqrt3}(\hat i + \hat j - \hat k) is one.

Every fact used, and where it comes from

FactUsed inSource
Component form xi^+yj^+zk^x\hat i + y\hat j + z\hat k⁠; scalar componentsSteps 2–4Section 10.5.1 (Lesson: components)
∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} by Pythagoras twiceSteps 1–4Section 10.5.1 and Fig 10.14 (Basics: Pythagoras twice)
Unit vector: magnitude 1Step 4Section 10.3
1∣a⃗∣a⃗\frac{1}{|\vec a|}\vec a is the unit vector along a⃗\vec aStep 4Section 10.5
(−x)2=x2(-x)^2 = x^2⁠; (3)2=3(\sqrt3)^2 = 3Steps 3–4Laws of exponents and surds
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 →