find the angle formed by vector A = i + 2j - k and vector B = 3i - 4k

To find the angle \(\theta\) between two vectors \(\mathbf{A}\) and \(\mathbf{B}\), we can use the dot product formula:

\[
\cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|}
\]

Where:
- \(\mathbf{A} = i + 2j - k\)
- \(\mathbf{B} = 3i - 4k\)

### Step 1: Calculate the dot product \(\mathbf{A} \cdot \mathbf{B}\)

The dot product of two vectors \(\mathbf{A} = a_1i + a_2j + a_3k\) and \(\mathbf{B} = b_1i + b_2j + b_3k\) is:

\[
\mathbf{A} \cdot \mathbf{B} = a_1b_1 + a_2b_2 + a_3b_3
\]

For \(\mathbf{A} = i + 2j - k\) and \(\mathbf{B} = 3i - 4k\), this becomes:

\[
\mathbf{A} \cdot \mathbf{B} = (1 \cdot 3) + (2 \cdot 0) + (-1 \cdot -4)
\]
\[
\mathbf{A} \cdot \mathbf{B} = 3 + 0 + 4 = 7
\]

### Step 2: Find the magnitudes of \(\mathbf{A}\) and \(\mathbf{B}\)

The magnitude of a vector \(\mathbf{A} = a_1i + a_2j + a_3k\) is given by:

\[
|\mathbf{A}| = \sqrt{a_1^2 + a_2^2 + a_3^2}
\]

For \(\mathbf{A} = i + 2j - k\):

\[
|\mathbf{A}| = \sqrt{1^2 + 2^2 + (-1)^2} = \sqrt{1 + 4 + 1} = \sqrt{6}
\]

For \(\mathbf{B} = 3i - 4k\):

\[
|\mathbf{B}| = \sqrt{3^2 + 0^2 + (-4)^2} = \sqrt{9 + 16} = \sqrt{25} = 5
\]

### Step 3: Use the dot product formula to find \(\cos(\theta)\)

\[
\cos(\theta) = \frac{\mathbf{A} \cdot \mathbf{B}}{|\mathbf{A}| |\mathbf{B}|} = \frac{7}{\sqrt{6} \times 5} = \frac{7}{5\sqrt{6}}
\]

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

To find \(\theta\), we take the inverse cosine (arccos) of \(\frac{7}{5\sqrt{6}}\):

\[
\theta = \cos^{-1}\left(\frac{7}{5\sqrt{6}}\right)
\]

Now, let me calculate the value.The angle \(\theta\) between vectors \(\mathbf{A} = i + 2j - k\) and \(\mathbf{B} = 3i - 4k\) is approximately \(55.14^\circ\).

Would you like more details or explanations on any part of this solution?

Here are some related questions you might find helpful:
1. How do you find the cross product of two vectors?
2. How do you interpret the result of a dot product geometrically?
3. What is the significance of the angle between two vectors being 90 degrees?
4. How can the cosine of an angle between two vectors indicate their relative direction?
5. How can we calculate the projection of one vector onto another?

**Tip:** If the dot product between two vectors is zero, the vectors are orthogonal (perpendicular).

Learn how to calculate the angle formed by vectors A = i + 2j - k and B = 3i - 4k using vector algebra techniques. This problem covers the dot product formula, vector magnitudes, and trigonometric concepts. Suitable for advanced high school or college students.

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