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

Miscellaneous · Q5: xx making x(i^+j^+k^)x(\hat i+\hat j+\hat k) a unit vector

Miscellaneous ExerciseUnit vectors, magnitude of a scalar multiple · Sections 10.3 and 10.5
The question

Find the value of xx for which x(i^+j^+k^)x(\hat i + \hat j + \hat k) is a unit vector.

Exam answer

Let r⃗=x(i^+j^+k^)=xi^+xj^+xk^\vec r = x(\hat i + \hat j + \hat k) = x\hat i + x\hat j + x\hat k⁠.

r⃗\vec r is a unit vector, so ∣r⃗∣=1|\vec r| = 1⁠:

x2+x2+x2=13 ∣x∣=1∣x∣=13\begin{aligned} \sqrt{x^2 + x^2 + x^2} &= 1 \\ \sqrt3\,|x| &= 1 \\ |x| &= \frac{1}{\sqrt3} \end{aligned}

∴ x=±13\therefore\ x = \pm\dfrac{1}{\sqrt3}⁠.

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

The one idea this problem is built on

Multiplying a vector by a number xx multiplies its length by ∣x∣|x|⁠. The vector i^+j^+k^\hat i + \hat j + \hat k has length 3\sqrt3⁠, so x(i^+j^+k^)x(\hat i + \hat j + \hat k) has length 3 ∣x∣\sqrt3\,|x|⁠. Setting this equal to 1 gives two values of xx⁠, one for each direction along the line.

Given, to find, and the plan
Given
The vector x(i^+j^+k^)x(\hat i + \hat j + \hat k)⁠, with xx a real number.
To find
Every xx for which it is a unit vector (length 1).
Plan
(1) Write the vector in components. (2) Find its length in terms of xx⁠. (3) Set the length equal to 1 and solve, keeping both signs. (4) Check both values.

Basics you need

1Unit vector

A vector of magnitude 1 (Section 10.3). Lesson: six kinds of vectors

2Length from components

∣ai^+bj^+ck^∣=a2+b2+c2|a\hat i + b\hat j + c\hat k| = \sqrt{a^2 + b^2 + c^2} (Section 10.5.1). Basics: use Pythagoras twice

3Square root of a square

x2=∣x∣\sqrt{x^2} = |x|⁠, not xx⁠: for example (−2)2=2\sqrt{(-2)^2} = 2⁠. So ∣x∣=c|x| = c (with c>0c > 0⁠) means x=cx = c or x=−cx = -c⁠.

Solution

  1. Write the vector in components

    r⃗=x(i^+j^+k^)=xi^+xj^+xk^.\vec r = x(\hat i + \hat j + \hat k) = x\hat i + x\hat j + x\hat k.
    Multiplying a vector by a scalar multiplies each component by that scalar, so all three components are xx⁠.
    Scalar multiple by components (Section 10.5.1, (iv)). Lesson: components

  2. Its length

    ∣r⃗∣=x2+x2+x2=3x2=3 ∣x∣.\begin{aligned} |\vec r| &= \sqrt{x^2 + x^2 + x^2} \\ &= \sqrt{3x^2} = \sqrt3\,|x|. \end{aligned}
    3x2=3 x2\sqrt{3x^2} = \sqrt3\,\sqrt{x^2}⁠, and x2=∣x∣\sqrt{x^2} = |x| because a square root is never negative. This matches the rule ∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a| with λ=x\lambda = x and ∣i^+j^+k^∣=3|\hat i + \hat j + \hat k| = \sqrt3⁠.
    ∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a| (Section 10.5). Basics: use Pythagoras twice

  3. Set the length equal to 1

    3 ∣x∣=1  ⟺  ∣x∣=13  ⟺  x=±13.\begin{aligned} \sqrt3\,|x| = 1 &\iff |x| = \frac{1}{\sqrt3} \\ &\iff x = \pm\frac{1}{\sqrt3}. \end{aligned}

    So x=13x = \frac{1}{\sqrt3} or x=−13x = -\frac{1}{\sqrt3}.

    A number whose absolute value is 13\frac{1}{\sqrt3} is either 13\frac{1}{\sqrt3} or −13-\frac{1}{\sqrt3}⁠. Each step is an "if and only if", so these two values are all the solutions and both are genuine.
    Unit vector: magnitude 1 (Section 10.3). Lesson: six kinds of vectors

  4. Check both values

    For x=±13x = \pm\frac{1}{\sqrt3}⁠, each component squared is 13\frac13⁠, and 13+13+13=1\sqrt{\frac13 + \frac13 + \frac13} = 1⁠. The two vectors are

    ±13(i^+j^+k^),\pm\tfrac{1}{\sqrt3}(\hat i + \hat j + \hat k),

    the unit vector along i^+j^+k^\hat i + \hat j + \hat k and its negative, pointing the opposite way.

    The positive value gives the unit vector in the direction of i^+j^+k^\hat i + \hat j + \hat k⁠; the negative value reverses the direction but keeps the length 1. Both make equal angles with the three axes, as in Exercise 10.2 Q14.
    Unit vector in the direction of a⃗\vec a⁠: a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec a (Section 10.5); negative of a vector (Section 10.3). Lesson: put it all together
Result

x=±13x = \pm\frac{1}{\sqrt3} (equivalently ±33\pm\frac{\sqrt3}{3}⁠).

(Textbook answer: ±13\pm\frac{1}{\sqrt3}⁠.)

Check it with numbers

13≈0.577350\frac{1}{\sqrt3} \approx 0.577350⁠. With x=0.577350x = 0.577350 the length is 3×0.577350\sqrt3 \times 0.577350⁠; since 3≈1.732051\sqrt3 \approx 1.732051⁠, this is about 1.

Exactly: 3×(13)2=3×13=13 \times \left(\frac{1}{\sqrt3}\right)^2 = 3 \times \frac13 = 1⁠.

Common mistakes
  1. Losing the negative value. 3x2=3 ∣x∣\sqrt{3x^2} = \sqrt3\,|x|⁠, not 3 x\sqrt3\,x⁠; writing 3 x=1\sqrt3\,x = 1 gives only x=13x = \frac{1}{\sqrt3}⁠.
  2. Forgetting the square root. 3x2=13x^2 = 1 comes from squaring ∣r⃗∣=1|\vec r| = 1⁠; stopping at 3x=13x = 1 gives the wrong x=13x = \frac13⁠.
  3. Using xx as the length. The length of x(i^+j^+k^)x(\hat i + \hat j + \hat k) is 3 ∣x∣\sqrt3\,|x|⁠, not ∣x∣|x|⁠.
What to take away
  1. ∣λa⃗∣=∣λ∣ ∣a⃗∣|\lambda\vec a| = |\lambda|\,|\vec a|⁠.
  2. λa⃗\lambda\vec a is a unit vector exactly when λ=±1∣a⃗∣\lambda = \pm\frac{1}{|\vec a|}⁠.
  3. x2=∣x∣\sqrt{x^2} = |x|⁠: keep both signs.

Every fact used, and where it comes from

FactUsed inSource
Unit vector: magnitude 1; negative of a vectorSteps 3, 4Section 10.3 (Lesson: six kinds of vectors)
Scalar multiple by componentsStep 1Section 10.5.1, (iv)
∣ai^+bj^+ck^∣=a2+b2+c2|a\hat i + b\hat j + c\hat k| = \sqrt{a^2 + b^2 + c^2}Step 2Section 10.5.1 (Basics: use Pythagoras twice)
∣λa⃗∣=∣λ∣∣a⃗∣|\lambda\vec a| = |\lambda||\vec a|⁠; a^=1∣a⃗∣a⃗\hat a = \frac{1}{|\vec a|}\vec aSteps 2, 4Section 10.5
The answerResultNCERT answer key, Miscellaneous Exercise on Chapter 10, Q5 (Lesson: put it all together)
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 →