Let vec a =2 hat i -2 hat j + hat k vec b = hat j - hat k and vec c =- hat i +3 hat j +2 hat k be given vectors. Find





(v) angle between vec a and vec b

(vi) sine of the angle between vec a and vec c

Let's solve for the following:

**(v) Angle between \(\vec{a}\) and \(\vec{b}\)**

To find the angle \(\theta\) between two vectors \(\vec{a}\) and \(\vec{b}\), we use the formula:

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

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

\[
\vec{a} \cdot \vec{b} = (2\hat{i} - 2\hat{j} + \hat{k}) \cdot (\hat{j} - \hat{k}) = (2 \times 0) + (-2 \times 1) + (1 \times -1) = -2 - 1 = -3
\]

**Step 2: Calculate the magnitudes \(\|\vec{a}\|\) and \(\|\vec{b}\|\)**

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

\[
\|\vec{b}\| = \sqrt{(0)^2 + (1)^2 + (-1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2}
\]

**Step 3: Calculate \(\cos \theta\)**

\[
\cos \theta = \frac{-3}{3 \times \sqrt{2}} = \frac{-3}{3\sqrt{2}} = -\frac{1}{\sqrt{2}}
\]

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

\[
\theta = \cos^{-1} \left(-\frac{1}{\sqrt{2}}\right) = \cos^{-1} \left(-\frac{\sqrt{2}}{2}\right)
\]

This implies \(\theta = 135^\circ\) or \(\frac{3\pi}{4}\) radians.

---

**(vi) Sine of the angle between \(\vec{a}\) and \(\vec{c}\)**

To find \(\sin \phi\), where \(\phi\) is the angle between \(\vec{a}\) and \(\vec{c}\), we use the formula:

\[
\sin \phi = \sqrt{1 - \cos^2 \phi}
\]

**Step 1: Calculate the dot product \(\vec{a} \cdot \vec{c}\)**

\[
\vec{a} \cdot \vec{c} = (2\hat{i} - 2\hat{j} + \hat{k}) \cdot (-\hat{i} + 3\hat{j} + 2\hat{k}) = (2 \times -1) + (-2 \times 3) + (1 \times 2) = -2 - 6 + 2 = -6
\]

**Step 2: Calculate the magnitudes \(\|\vec{a}\|\) and \(\|\vec{c}\|\)**

We already know \(\|\vec{a}\| = 3\).

\[
\|\vec{c}\| = \sqrt{(-1)^2 + (3)^2 + (2)^2} = \sqrt{1 + 9 + 4} = \sqrt{14}
\]

**Step 3: Calculate \(\cos \phi\)**

\[
\cos \phi = \frac{\vec{a} \cdot \vec{c}}{\|\vec{a}\| \|\vec{c}\|} = \frac{-6}{3 \times \sqrt{14}} = \frac{-6}{3\sqrt{14}} = -\frac{2}{\sqrt{14}} = -\frac{2}{\sqrt{14}} \cdot \frac{\sqrt{14}}{\sqrt{14}} = -\frac{2\sqrt{14}}{14} = -\frac{\sqrt{14}}{7}
\]

**Step 4: Calculate \(\sin \phi\)**

\[
\sin \phi = \sqrt{1 - \left(-\frac{\sqrt{14}}{7}\right)^2} = \sqrt{1 - \frac{14}{49}} = \sqrt{1 - \frac{2}{7}} = \sqrt{\frac{5}{7}}
\]

Thus, the sine of the angle between \(\vec{a}\) and \(\vec{c}\) is \(\boxed{\sqrt{\frac{5}{7}}}\).

Would you like more details or have any questions?

Here are some related questions:
1. How do you find the cosine of the angle between two vectors?
2. What is the geometric interpretation of the dot product?
3. How do vectors relate to projections in space?
4. How can the cross product help find angles between vectors?
5. What are some applications of vectors in physics and engineering?

**Tip:** The dot product can determine not only the magnitude of the angle between two vectors but also the direction (acute or obtuse) based on the sign of the result.

Learn how to calculate the angle between vectors \(\vec{a}\) and \(\vec{b}\), and the sine of the angle between vectors \(\vec{a}\) and \(\vec{c}\). Detailed step-by-step solutions using dot product, magnitude, and trigonometric functions. Suitable for undergraduate students studying vectors and trigonometry.

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