Question 22 of 68beginner🔧 ApplyNumerical2 marks

Find the angle between the vectors î - 2ĵ + 3k̂ and 3î - 2ĵ + k̂.

Correct Answer

Let the given vectors be 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}. To find the angle θ\theta, we use the relation cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}.

Step 1: Calculate the scalar product. a⃗⋅b⃗=(1)(3)+(−2)(−2)+(3)(1)=3+4+3=10\vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10.

Step 2: Calculate the magnitudes. ∣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}.

Step 3: Determine the angle. Substituting these values, cos⁡θ=101414=57\cos \theta = \frac{10}{\sqrt{14} \sqrt{14}} = \frac{5}{7}. Hence, the required angle is θ=cos⁡−1(57)\theta = \cos^{-1}\left(\frac{5}{7}\right).

Exercise: EXERCISE 10.3 | Q: 2 | (Page 24)
For More Understanding

Explanation

This solution follows the standard method shown in Example 14 of the provided context. It calculates the dot product and magnitudes of the vectors to find the cosine of the angle, which is then inverted to find the angle in terms of inverse cosine.

Solution Steps

  1. Step 1: Calculate dot product a⃗⋅b⃗=10\vec{a} \cdot \vec{b} = 10.

  2. Step 2: Calculate magnitudes ∣a⃗∣=14|\vec{a}| = \sqrt{14} and ∣b⃗∣=14|\vec{b}| = \sqrt{14}.

  3. Step 3: Substitute into cos⁡θ=a⃗⋅b⃗∣a⃗∣∣b⃗∣\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} to get 57\frac{5}{7}.

  4. Step 4: State final answer θ=cos⁡−1(5/7)\theta = \cos^{-1}(5/7).

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