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

Example 7: Vector of magnitude 7 along i^−2j^\hat i-2\hat j

ExamplesUnit vector, multiplication by a scalar · Sections 10.5 and 10.5.1
The question

Find a vector in the direction of vector a⃗=i^−2j^\vec a = \hat i - 2\hat j that has magnitude 7 units.

Exam answer ∣a⃗∣=12+(−2)2=5|\vec a| = \sqrt{1^2 + (-2)^2} = \sqrt5 a^=1∣a⃗∣a⃗=15(i^−2j^)=15i^−25j^\begin{aligned} \hat a &= \frac{1}{|\vec a|}\vec a = \frac{1}{\sqrt5}(\hat i - 2\hat j) \\ &= \frac{1}{\sqrt5}\hat i - \frac{2}{\sqrt5}\hat j \end{aligned}

Since ∣a^∣=1|\hat a| = 1⁠, the vector 7a^7\hat a has magnitude 7 and the direction of a⃗\vec a⁠.

7a^=7(15i^−25j^)=75i^−145j^\begin{aligned} 7\hat a &= 7\Big(\frac{1}{\sqrt5}\hat i - \frac{2}{\sqrt5}\hat j\Big) \\ &= \frac{7}{\sqrt5}\hat i - \frac{14}{\sqrt5}\hat j \end{aligned}

∴\therefore the required vector is 75i^−145j^\dfrac{7}{\sqrt5}\hat i - \dfrac{14}{\sqrt5}\hat j⁠.

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

The one idea this problem is built on

A vector of any chosen length kk along a⃗\vec a is ka^k\hat a⁠: first shrink (or stretch) a⃗\vec a to length 1 by dividing by ∣a⃗∣|\vec a|⁠, then multiply by kk⁠. The unit vector a^\hat a carries the direction; the number k=7k = 7 supplies the length.

Given, to find, and the plan
Given
a⃗=i^−2j^\vec a = \hat i - 2\hat j⁠; required magnitude 7 units.
To find
A vector of magnitude 7 in the direction of a⃗\vec a⁠.
Plan
(1) Compute ∣a⃗∣|\vec a|⁠. (2) Form a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec a⁠. (3) Multiply by 7. (4) Check the length is 7 and the direction is that of a⃗\vec a⁠.

Basics you need

1Magnitude from components

∣xi^+yj^∣=x2+y2|x\hat i + y\hat j| = \sqrt{x^2 + y^2}⁠; a negative component still gives a positive square: (−2)2=4(-2)^2 = 4 (Section 10.5.1). Basics: use Pythagoras twice

2The unit vector

For a⃗≠0⃗\vec a \ne \vec 0⁠, a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec a has magnitude 1 and the direction of a⃗\vec a (Section 10.5). Lesson: six kinds of vectors worth naming

3Multiplying by a scalar

∣λa⃗∣=∣λ∣ ∣a⃗∣|\lambda\vec a| = |\lambda|\,|\vec a|⁠, and for λ>0\lambda > 0 the direction is unchanged; in components, every component is multiplied by λ\lambda (Sections 10.5 and 10.5.1 (iv)). Lesson: components

Solution

  1. Find the magnitude of a⃗\vec a

    ∣a⃗∣=12+(−2)2=1+4=5.|\vec a| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt5.
    a⃗=i^−2j^\vec a = \hat i - 2\hat j has components 1 and −2-2 (and 0 along k^\hat k⁠). The minus sign tells the direction along the y-axis; it disappears on squaring, as a length must be positive.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  2. Form the unit vector

    a^=1∣a⃗∣a⃗=15(i^−2j^)=15i^−25j^.\begin{aligned} \hat a &= \frac{1}{|\vec a|}\vec a = \frac{1}{\sqrt5}(\hat i - 2\hat j) \\ &= \frac{1}{\sqrt5}\hat i - \frac{2}{\sqrt5}\hat j. \end{aligned}
    Dividing by the positive number 5\sqrt5 keeps the direction of a⃗\vec a and makes the length 15⋅5=1\frac{1}{\sqrt5}\cdot\sqrt5 = 1⁠. Now the direction is stored in a vector of length exactly 1, ready to be scaled to any length.
    a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec a (Section 10.5). Lesson: components

  3. Scale to length 7

    7a^=7(15i^−25j^)=75i^−145j^.\begin{aligned} 7\hat a &= 7\Big(\frac{1}{\sqrt5}\hat i - \frac{2}{\sqrt5}\hat j\Big) \\ &= \frac{7}{\sqrt5}\hat i - \frac{14}{\sqrt5}\hat j. \end{aligned}

    This is the required vector.

    ∣7a^∣=7∣a^∣=7×1=7|7\hat a| = 7|\hat a| = 7 \times 1 = 7⁠, and since 7>07 > 0⁠, 7a^7\hat a has the same direction as a^\hat a⁠, which is the direction of a⃗\vec a⁠. Multiplying by 7 multiplies each component by 7.
    ∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a|⁠; λa⃗=(λa1)i^+(λa2)j^+(λa3)k^\lambda\vec a = (\lambda a_1)\hat i + (\lambda a_2)\hat j + (\lambda a_3)\hat k (Sections 10.5 and 10.5.1 (iv)). Lesson: components

  4. Check the length and the direction

    ∣75i^−145j^∣2=495+1965=2455=49,\begin{aligned} \Big|\frac{7}{\sqrt5}\hat i - \frac{14}{\sqrt5}\hat j\Big|^2 &= \frac{49}{5} + \frac{196}{5} \\ &= \frac{245}{5} = 49, \end{aligned}

    so the length is 49=7\sqrt{49} = 7⁠. And 75i^−145j^=75(i^−2j^)=75 a⃗\frac{7}{\sqrt5}\hat i - \frac{14}{\sqrt5}\hat j = \frac{7}{\sqrt5}(\hat i - 2\hat j) = \frac{7}{\sqrt5}\,\vec a⁠, a positive multiple of a⃗\vec a⁠.

    x y O â, length 1 a = î − 2ĵ, length √5 7â, length 7 (7/√5, −14/√5)
    With equal scales: a^\hat a (length 1), a⃗=i^−2j^\vec a = \hat i - 2\hat j (length 5≈2.24\sqrt5 \approx 2.24⁠) and 7a^7\hat a (length 7, split into 7 unit parts) all lie along one ray from O.
    Both conditions of the question are now confirmed directly on the answer: magnitude 7, and the direction of a⃗\vec a (a positive multiple of it). A negative multiple, such as −7a^-7\hat a⁠, would also have length 7 but would point the opposite way.
    λa⃗\lambda\vec a has the direction of a⃗\vec a when λ>0\lambda > 0 and the opposite direction when λ<0\lambda < 0 (Section 10.5). Lesson: Examples 4–10
Result

The required vector is 7a^=75i^−145j^7\hat a = \frac{7}{\sqrt5}\hat i - \frac{14}{\sqrt5}\hat j (equivalently 755i^−1455j^\frac{7\sqrt5}{5}\hat i - \frac{14\sqrt5}{5}\hat j⁠).

(Textbook answer: 7a^=7(15i^−25j^)=75i^−145j^7\hat a = 7\left(\frac{1}{\sqrt5}\hat i - \frac{2}{\sqrt5}\hat j\right) = \frac{7}{\sqrt5}\hat i - \frac{14}{\sqrt5}\hat j⁠.)

Check it with numbers

75≈3.130495\frac{7}{\sqrt5} \approx 3.130495 and 145≈6.260990\frac{14}{\sqrt5} \approx 6.260990⁠, so the answer is about 3.130495i^−6.260990j^3.130495\hat i - 6.260990\hat j⁠. Its length: 3.1304952+6.2609902≈9.8+39.2=49=7\sqrt{3.130495^2 + 6.260990^2} \approx \sqrt{9.8 + 39.2} = \sqrt{49} = 7⁠.

Direction: the ratio of the components is −6.260990÷3.130495=−2-6.260990 \div 3.130495 = -2⁠, the same as in a⃗=i^−2j^\vec a = \hat i - 2\hat j⁠; both have a positive i^\hat i part, so they point the same way, not opposite ways.

Common mistakes
  1. Writing 7a⃗=7i^−14j^7\vec a = 7\hat i - 14\hat j⁠. That has the right direction but length 75≈15.657\sqrt5 \approx 15.65⁠, not 7. Multiply the unit vector by 7.
  2. Losing the sign. The j^\hat j component stays negative: −145-\frac{14}{\sqrt5}⁠.
  3. ∣a⃗∣=1−4|\vec a| = \sqrt{1 - 4}⁠. Squares are added: (−2)2=+4(-2)^2 = +4⁠.
What to take away
  1. A vector of magnitude kk in the direction of a⃗\vec a is ka^=k∣a⃗∣a⃗k\hat a = \frac{k}{|\vec a|}\vec a⁠.
  2. The unit vector carries the direction; the scalar carries the length.
  3. Check the answer: its length must be kk⁠, and it must be a positive multiple of a⃗\vec a⁠.

Every fact used, and where it comes from

FactUsed inSource
∣xi^+yj^+zk^∣=x2+y2+z2|x\hat i + y\hat j + z\hat k| = \sqrt{x^2 + y^2 + z^2}Steps 1, 4Section 10.5.1 (Basics: use Pythagoras twice)
a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec aStep 2Section 10.5
∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a|⁠; direction kept for λ>0\lambda > 0⁠, reversed for λ<0\lambda < 0Steps 3, 4Section 10.5
λa⃗\lambda\vec a by componentsStep 3Section 10.5.1 (iv) (Lesson: components)
The book's working and answerResultExample 7, Section 10.5.1 (Lesson: Examples 4–10)
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 →