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

Exercise 10.3 · Q2: Angle between i^−2j^+3k^\hat i-2\hat j+3\hat k and 3i^−2j^+k^3\hat i-2\hat j+\hat k

Exercise 10.3Scalar product in components, angle between vectors · Section 10.6.1
The question

Find the angle between the vectors i^−2j^+3k^\hat i - 2\hat j + 3\hat k and 3i^−2j^+k^3\hat i - 2\hat j + \hat k⁠.

Exam answer

Let a⃗=i^−2j^+3k^\vec a = \hat i - 2\hat j + 3\hat k and b⃗=3i^−2j^+k^\vec b = 3\hat i - 2\hat j + \hat k⁠, and let θ\theta be the angle between them.

a⃗⋅b⃗=(1)(3)+(−2)(−2)+(3)(1)=3+4+3=10\begin{aligned} \vec a\cdot\vec b &= (1)(3) + (-2)(-2) + (3)(1) \\ &= 3 + 4 + 3 = 10 \end{aligned} ∣a⃗∣=12+(−2)2+32=14|\vec a| = \sqrt{1^2 + (-2)^2 + 3^2} = \sqrt{14} ∣b⃗∣=32+(−2)2+12=14|\vec b| = \sqrt{3^2 + (-2)^2 + 1^2} = \sqrt{14} cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=1014×14=1014=57\begin{aligned} \cos\theta &= \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} \\ &= \frac{10}{\sqrt{14} \times \sqrt{14}} \\ &= \frac{10}{14} = \frac57 \end{aligned}

Since 0≤θ≤π0 \le \theta \le \pi⁠,

∴ θ=cos⁡−1(57)\therefore\ \theta = \cos^{-1}\Big(\dfrac57\Big)⁠.

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

The one idea this problem is built on

When the vectors are given in components, the scalar product is the sum of the products of matching components:

a⃗⋅b⃗=a1b1+a2b2+a3b3.\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3.

Dividing it by ∣a⃗∣∣b⃗∣|\vec a||\vec b| gives cos⁡θ\cos\theta⁠, and the angle is the one value in [0,π][0, \pi] with that cosine. A positive scalar product means an acute angle.

Given, to find, and the plan
Given
a⃗=i^−2j^+3k^\vec a = \hat i - 2\hat j + 3\hat k⁠, b⃗=3i^−2j^+k^\vec b = 3\hat i - 2\hat j + \hat k⁠.
To find
The angle θ\theta between a⃗\vec a and b⃗\vec b⁠.
Plan
(1) Compute a⃗⋅b⃗\vec a\cdot\vec b from the components. (2) Compute ∣a⃗∣|\vec a| and ∣b⃗∣|\vec b|⁠. (3) Form cos⁡θ\cos\theta and take cos⁡−1\cos^{-1}⁠. (4) Read off whether the angle is acute or obtuse.

Basics you need

1Scalar product in components

For a⃗=a1i^+a2j^+a3k^\vec a = a_1\hat i + a_2\hat j + a_3\hat k and b⃗=b1i^+b2j^+b3k^\vec b = b_1\hat i + b_2\hat j + b_3\hat k⁠, a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3 (Section 10.6.1, from i^⋅i^=1\hat i\cdot\hat i = 1⁠, i^⋅j^=0\hat i\cdot\hat j = 0⁠, …). Lesson: the dot product

2The angle formula

cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}⁠, with 0≤θ≤π0 \le \theta \le \pi (Section 10.6.1, Observation 6). Lesson: Examples 13–21

3Magnitude from components

∣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

Solution

  1. The scalar product

    a⃗⋅b⃗=(1)(3)+(−2)(−2)+(3)(1)=3+4+3=10.\begin{aligned} \vec a\cdot\vec b &= (1)(3) + (-2)(-2) + (3)(1) \\ &= 3 + 4 + 3 = 10. \end{aligned}
    Multiply the i^\hat i-components, the j^\hat j-components and the k^\hat k-components, and add. The middle product is (−2)(−2)=+4(-2)(-2) = +4⁠: two negative components give a positive product, and this is where most sign slips happen.
    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 two lengths

    ∣a⃗∣=12+(−2)2+32=1+4+9=14,∣b⃗∣=32+(−2)2+12=9+4+1=14.\begin{aligned} |\vec a| &= \sqrt{1^2 + (-2)^2 + 3^2} \\ &= \sqrt{1 + 4 + 9} = \sqrt{14}, \\ |\vec b| &= \sqrt{3^2 + (-2)^2 + 1^2} \\ &= \sqrt{9 + 4 + 1} = \sqrt{14}. \end{aligned}
    Each length is the square root of the sum of the squares of the components. The two vectors use the same three numbers 1, 2, 3 (with signs), so their lengths are equal; squaring removes the signs.
    ∣r⃗∣=x2+y2+z2|\vec r| = \sqrt{x^2 + y^2 + z^2} (Section 10.5.1). Basics: use Pythagoras twice

  3. The cosine of the angle

    cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣=1014×14=1014=57.\begin{aligned} \cos\theta &= \frac{\vec a\cdot\vec b}{|\vec a||\vec b|} = \frac{10}{\sqrt{14} \times \sqrt{14}} \\ &= \frac{10}{14} = \frac57. \end{aligned}
    The product of the lengths is 14×14=14\sqrt{14} \times \sqrt{14} = 14⁠, a whole number, so the cosine is the fraction 1014\frac{10}{14}⁠, which cancels by 2 to 57\frac57⁠. It lies in [−1,1][-1, 1]⁠, as every cosine must.
    Section 10.6.1, Observation 6. Lesson: Examples 13–21

  4. The angle

    θ=cos⁡−1(57).\theta = \cos^{-1}\Big(\frac57\Big).

    So θ=cos⁡−157\theta = \cos^{-1}\frac57, about 44.42∘44.42^\circ⁠. Since cos⁡θ=57>0\cos\theta = \frac57 > 0⁠, the angle is acute.

    θ ≈ 44.42° O a, length √14 b, length √14 10/√14 cos θ = 5/7 > 0 so θ is acute
    The two vectors drawn in their own plane, true to scale: both have length 14\sqrt{14}⁠, and the angle between them is cos⁡−157≈44.42∘\cos^{-1}\frac57 \approx 44.42^\circ⁠. The perpendicular from the tip of b⃗\vec b meets the line of a⃗\vec a at distance ∣b⃗∣cos⁡θ=1014|\vec b|\cos\theta = \frac{10}{\sqrt{14}} from O, in front of O because the angle is acute.
    The angle between two vectors is taken in [0,π][0, \pi]⁠, the principal range of cos⁡−1\cos^{-1}⁠, where cosine takes each value exactly once, so cos⁡−157\cos^{-1}\frac57 is the one answer. 57\frac57 is not the cosine of a standard angle, so the exact answer is left as cos⁡−157\cos^{-1}\frac57⁠, as in the book's answer.
    0≤θ≤π0 \le \theta \le \pi (Section 10.6.1, Definition 2). Basics: cos θ is a shadow
Result

The angle between the vectors is θ=cos⁡−157\theta = \cos^{-1}\frac57 (about 44.42∘44.42^\circ⁠).

(Textbook answer: cos⁡−1(57)\cos^{-1}\left(\frac57\right)⁠.)

Check it with numbers

∣a⃗∣=∣b⃗∣=14≈3.741657|\vec a| = |\vec b| = \sqrt{14} \approx 3.741657⁠, and 57≈0.714286\frac57 \approx 0.714286⁠. Then cos⁡−1(0.714286)≈0.775193\cos^{-1}(0.714286) \approx 0.775193 radians ≈44.415∘\approx 44.415^\circ⁠, between 0∘0^\circ and 90∘90^\circ⁠: acute, as the positive scalar product says.

Back in the definition: ∣a⃗∣∣b⃗∣cos⁡θ=14×57=10=a⃗⋅b⃗|\vec a||\vec b|\cos\theta = 14 \times \frac57 = 10 = \vec a\cdot\vec b⁠. The shadow of b⃗\vec b on a⃗\vec a is 1014≈2.672612\frac{10}{\sqrt{14}} \approx 2.672612⁠, shorter than ∣b⃗∣≈3.741657|\vec b| \approx 3.741657⁠, as it must be.

Common mistakes
  1. A sign slip in the scalar product. Taking (−2)(−2)=−4(-2)(-2) = -4 gives a⃗⋅b⃗=2\vec a\cdot\vec b = 2 and cos⁡θ=17\cos\theta = \frac17⁠, a wrong angle.
  2. Dividing by 14\sqrt{14} only once. 1014\frac{10}{\sqrt{14}} is the projection of one vector on the other, not cos⁡θ\cos\theta⁠; it is bigger than 1.
  3. Adding the vectors instead of multiplying components. The scalar product multiplies matching components; it is a number, not a vector.
What to take away
  1. In components, a⃗⋅b⃗=a1b1+a2b2+a3b3\vec a\cdot\vec b = a_1b_1 + a_2b_2 + a_3b_3⁠, and cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}⁠.
  2. When the cosine is not a standard value, the exact answer is θ=cos⁡−1(…)\theta = \cos^{-1}(\ldots)⁠.
  3. The sign of a⃗⋅b⃗\vec a\cdot\vec b tells acute (positive), right (zero) or obtuse (negative).

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
cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos\theta = \frac{\vec a\cdot\vec b}{|\vec a||\vec b|}Step 3Section 10.6.1, Observation 6
0≤θ≤π0 \le \theta \le \pi⁠; principal range of cos⁡−1\cos^{-1}Step 4Section 10.6.1, Definition 2 (Basics: cos θ is a shadow)
The answerResultNCERT answer key, Exercise 10.3 Q2 (Lesson: the dot product)
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 →