Find the angle between the two vectors `vec(a)= 2hat(i) + 3hat(j)- 4hat(k) , vec(b)= 4hat(i)+5hat(j)- 2hat(k)`

Explanation

To find the angle between the two vectors \(\vec{a} = 2\hat{i} + 3\hat{j} - 4\hat{k}\) and \(\vec{b} = 4\hat{i} + 5\hat{j} - 2\hat{k}\), we can use the formula:

\[ \vec{a} \cdot \vec{b} = |\vec{a}| |\vec{b}| \cos \theta \]

where \(\theta\) is the angle between the vectors.

### Step 1: Calculate the dot product \(\vec{a} \cdot \vec{b}\)

The dot product of two vectors \(\vec{a} = a_1\hat{i} + a_2\hat{j} + a_3\hat{k}\) and \(\vec{b} = b_1\hat{i} + b_2\hat{j} + b_3\hat{k}\) is given by:

\[ \vec{a} \cdot \vec{b} = a_1b_1 + a_2b_2 + a_3b_3 \]

Substituting the values from the vectors:

\[ \vec{a} \cdot \vec{b} = (2)(4) + (3)(5) + (-4)(-2) \]

Calculating each term:

\[ = 8 + 15 + 8 = 31 \]

### Step 2: Calculate the magnitudes of \(\vec{a}\) and \(\vec{b}\)

The magnitude of a vector \(\vec{v} = v_1\hat{i} + v_2\hat{j} + v_3\hat{k}\) is given by:

\[ |\vec{v}| = \sqrt{v_1^2 + v_2^2 + v_3^2} \]

Calculating the magnitude of \(\vec{a}\):

\[ |\vec{a}| = \sqrt{2^2 + 3^2 + (-4)^2} = \sqrt{4 + 9 + 16} = \sqrt{29} \]

Calculating the magnitude of \(\vec{b}\):

\[ |\vec{b}| = \sqrt{4^2 + 5^2 + (-2)^2} = \sqrt{16 + 25 + 4} = \sqrt{45} \]

### Step 3: Substitute values into the cosine formula

Now we can substitute the values into the cosine formula:

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

Substituting the values we calculated:

\[ \cos \theta = \frac{31}{\sqrt{29} \cdot \sqrt{45}} \]

Calculating the denominator:

\[ \sqrt{29} \cdot \sqrt{45} = \sqrt{29 \cdot 45} = \sqrt{1305} \]

Thus,

\[ \cos \theta = \frac{31}{\sqrt{1305}} \]

### Step 4: Calculate \(\theta\)

To find \(\theta\), we take the inverse cosine:

\[ \theta = \cos^{-1}\left(\frac{31}{\sqrt{1305}}\right) \]

### Final Result

Using a calculator to find the angle:

\[ \theta \approx \cos^{-1}(0.858) \approx 31.8^\circ \]