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Q2:
Find the angle between the vectors $\hat{i} - 2\hat{j} + 3\hat{k}$ and $3\hat{i} - 2\hat{j} + \hat{k}$

Solution :

Step 1: Identification of Vectors

Let the given vectors be represented as $\vec{a}$ and $\vec{b}$:

  • $\vec{a} = 1\hat{i} - 2\hat{j} + 3\hat{k}$
  • $\vec{b} = 3\hat{i} - 2\hat{j} + 1\hat{k}$

Step 2: Theoretical Foundation for Angle Calculation

The angle $\theta$ between two non-zero vectors $\vec{a}$ and $\vec{b}$ is determined by the dot product formula [Per the definition of the Scalar Product]:

$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|}$

Where $\theta = \cos^{-1} \left( \frac{\vec{a} \cdot \vec{b}}{|\vec{a}| |\vec{b}|} \right)$

Step 3: Calculating the Dot Product ($\vec{a} \cdot \vec{b}$)

Using the component-wise multiplication rule $(a_1\hat{i} + a_2\hat{j} + a_3\hat{k}) \cdot (b_1\hat{i} + b_2\hat{j} + b_3\hat{k}) = a_1b_1 + a_2b_2 + a_3b_3$:

$\vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1)$

$\vec{a} \cdot \vec{b} = 3 + 4 + 3 = 10$

Step 4: Calculating Magnitudes ($|\vec{a}|$ and $|\vec{b}|$)

The magnitude of a vector $\vec{v} = x\hat{i} + y\hat{j} + z\hat{k}$ is given by $\sqrt{x^2 + y^2 + z^2}$ [By the Euclidean Norm definition]:

$|\vec{a}| = \sqrt{(1)^2 + (-2)^2 + (3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$

$|\vec{b}| = \sqrt{(3)^2 + (-2)^2 + (1)^2} = \sqrt{9 + 4 + 1} = \sqrt{14}$

Step 5: Computing the Angle $\theta$

Substitute the derived values into the cosine formula:

$\cos \theta = \frac{10}{\sqrt{14} \cdot \sqrt{14}}$

$\cos \theta = \frac{10}{14} = \frac{5}{7}$

$\theta = \cos^{-1} \left( \frac{5}{7} \right)$

θ a b

Final Solution: The angle between the vectors is $\theta = \cos^{-1} \left( \frac{5}{7} \right)$ radians.


More Questions from Class 12 Mathematics Vector Algebra EXERCISE 10.3


CBSE Solutions for Class 12 Mathematics Vector Algebra


Chapters in CBSE - Class 12 Mathematics


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