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

Concept: Dot Product Angle — the cosine of the angle between two vectors is given by their dot product divided by the product of their magnitudes.

Let $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$.

Step 1: Dot product

$\vec{a} \cdot \vec{b} = (1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10$.

Step 2: Magnitudes

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

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

Step 3: Cosine of angle

$\cos\theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}|\,|\vec{b}|} = \frac{10}{\sqrt{14} \cdot \sqrt{14}} = \frac{10}{14} = \frac{5}{7}$.

The dot product gives us a direct link between two vectors and the angle between them. When you take the dot product of two vectors, you're essentially multiplying their magnitudes and the cosine of the angle between them. This means if we can compute the dot product and the individual magnitudes, we can solve for the angle.

Let's call the first vector $\vec{a} = \hat{i} - 2\hat{j} + 3\hat{k}$ and the second $\vec{b} = 3\hat{i} - 2\hat{j} + \hat{k}$.

1. Compute the dot product $\vec{a} \cdot \vec{b}$

   Multiply corresponding components and add:

   $(1)(3) + (-2)(-2) + (3)(1) = 3 + 4 + 3 = 10$

2. Find the magnitude of each vector

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

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

   Notice both magnitudes are equal — that's a nice symmetry here.

3. Apply the dot product formula

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

   $10 = (\sqrt{14})(\sqrt{14})\cos\theta = 14\cos\theta$

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

4. Write the angle

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

Method: Angle Between Two Vectors Given in Component Form

Use this when both vectors are given by components and you need the angle between them.

Steps

Step 1: Compute the dot product from components.

Multiply matching components and add — track every sign.

Step 2: Compute each magnitude.

Step 3: Assemble the cosine.

Step 4: Take the inverse cosine.

If the result is not a standard angle, leaving it as $\cos^{-1}(\cdot)$ is the correct exact answer. (Shortcut: when $|\vec a|=|\vec b|$, the denominator is $|\vec a|^2$.)

Common Mistakes

Mistake 1: Sign errors in the dot product.

Why it's wrong: $(-2)(-2)=+4$, so $\vec{a}\cdot\vec{b}=3+4+3=10$; treating it as $-4$ gives the wrong cosine. Correct approach: a negative times a negative is positive — multiply signed components carefully.

Mistake 2: Reporting $\tfrac57$ as the angle.

Why it's wrong: $\tfrac57$ is $\cos\theta$, not $\theta$. Correct approach: the angle is $\cos^{-1}\!\left(\tfrac57\right)$.

Mistake 3: Forgetting the square root in a magnitude.

Why it's wrong: $|\vec a|=\sqrt{1+4+9}=\sqrt{14}$, not $14$; dropping the root changes the denominator. Correct approach: take the square root of the sum of squares for each vector.