Consider the following for the next two (02) items that follow : Let \(\vec{a}=3 \hat{i}+3 \hat{j}+3 \hat{k} \) and \( \vec{c}=\hat{j}-\hat{k} \text {. Let } \vec{b}\) be such that \(\vec{a} \cdot \vec{b}=27 \) and \( \vec{a} \times \vec{b}=\overrightarrow{9 c}\)
What is the angle between \((\vec{a}+\vec{b})\) and \(\vec{c}\) ?
Understanding the Problem: Finding the Angle Between Vectors
The problem asks us to determine the angle between the vector sum \((\vec{a}+\vec{b})\) and the vector \(\vec{c}\), given specific properties relating vectors \(\vec{a}\) and \(\vec{b}\), and the explicit form of vectors \(\vec{a}\) and \(\vec{c}\). We are given:
- Vector \(\vec{a} = 3 \hat{i} + 3 \hat{j} + 3 \hat{k}\)
- Vector \(\vec{c} = \hat{j} - \hat{k}\)
- The dot product: \(\vec{a} \cdot \vec{b} = 27\)
- The cross product: \(\vec{a} \times \vec{b} = 9 \vec{c}\)
To find the angle between two vectors, say \(\vec{V}\) and \(\vec{W}\), we can use the dot product formula:
\[ \vec{V} \cdot \vec{W} = |\vec{V}| |\vec{W}| \cos \theta \] where \(\theta\) is the angle between \(\vec{V}\) and \(\vec{W}\). So, \(\cos \theta = \frac{\vec{V} \cdot \vec{W}}{|\vec{V}| |\vec{W}|}\).
In our case, we need the angle between \((\vec{a}+\vec{b})\) and \(\vec{c}\). Let \(\vec{V} = \vec{a}+\vec{b}\). We need to calculate \(\vec{V} \cdot \vec{c}\) and the magnitudes \(|\vec{V}|\) and \(|\vec{c}|\). First, we need to find vector \(\vec{b}\).
Determining Vector \( \vec{b} \) using Vector Properties
We have two equations involving \(\vec{b}\): a scalar equation (\(\vec{a} \cdot \vec{b} = 27\)) and a vector equation (\(\vec{a} \times \vec{b} = 9\vec{c}\)). We can use a standard vector identity involving the vector triple product to find \(\vec{b}\):
\[ (\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a} \]
Let's calculate the necessary components:
- Calculate \(\vec{a} \cdot \vec{a}\):
\(\vec{a} = 3\hat{i} + 3\hat{j} + 3\hat{k} = 3(\hat{i} + \hat{j} + \hat{k})\)
\( \vec{a} \cdot \vec{a} = |\vec{a}|^2 = (3)^2 + (3)^2 + (3)^2 = 9 + 9 + 9 = 27 \) - We are given \(\vec{a} \cdot \vec{b}\):
\( \vec{a} \cdot \vec{b} = 27 \) - We are given \(\vec{a} \times \vec{b}\):
\( \vec{a} \times \vec{b} = 9 \vec{c} = 9 (\hat{j} - \hat{k}) = 9\hat{j} - 9\hat{k} \) - Calculate \( (\vec{a} \times \vec{b}) \times \vec{a} \):
\( (\vec{a} \times \vec{b}) \times \vec{a} = (9\hat{j} - 9\hat{k}) \times (3\hat{i} + 3\hat{j} + 3\hat{k}) \)
\( = 9(\hat{j} - \hat{k}) \times 3(\hat{i} + \hat{j} + \hat{k}) \)
\( = 27 (\hat{j} - \hat{k}) \times (\hat{i} + \hat{j} + \hat{k}) \)
Using the cross product rules (\(\hat{j} \times \hat{i} = -\hat{k}\), \(\hat{j} \times \hat{j} = \vec{0}\), \(\hat{j} \times \hat{k} = \hat{i}\), \(\hat{k} \times \hat{i} = \hat{j}\), \(\hat{k} \times \hat{j} = -\hat{i}\), \(\hat{k} \times \hat{k} = \vec{0}\)):
\( = 27 [(\hat{j} \times \hat{i}) + (\hat{j} \times \hat{j}) + (\hat{j} \times \hat{k}) - (\hat{k} \times \hat{i}) - (\hat{k} \times \hat{j}) - (\hat{k} \times \hat{k})] \)
\( = 27 [-\hat{k} + \vec{0} + \hat{i} - \hat{j} - (-\hat{i}) - \vec{0}] \)
\( = 27 [-\hat{k} + \hat{i} - \hat{j} + \hat{i}] \)
\( = 27 [2\hat{i} - \hat{j} - \hat{k}] \)
Now substitute these results back into the vector triple product identity:
\[ 27 (2\hat{i} - \hat{j} - \hat{k}) = (27)\vec{b} - (27)\vec{a} \]
Divide both sides by 27 (since \(27 \neq 0\)):
\[ 2\hat{i} - \hat{j} - \hat{k} = \vec{b} - \vec{a} \]
Solve for \(\vec{b}\):
\[ \vec{b} = \vec{a} + 2\hat{i} - \hat{j} - \hat{k} \]
Substitute the value of \(\vec{a}\):
\[ \vec{b} = (3\hat{i} + 3\hat{j} + 3\hat{k}) + (2\hat{i} - \hat{j} - \hat{k}) \]
\[ \vec{b} = (3+2)\hat{i} + (3-1)\hat{j} + (3-1)\hat{k} \]
\[ \vec{b} = 5\hat{i} + 2\hat{j} + 2\hat{k} \]
We have successfully found vector \(\vec{b}\).
Calculating \( (\vec{a}+\vec{b}) \) and its Dot Product with \( \vec{c} \)
Now, let's find the vector sum \( (\vec{a}+\vec{b}) \):
\[ \vec{a}+\vec{b} = (3\hat{i} + 3\hat{j} + 3\hat{k}) + (5\hat{i} + 2\hat{j} + 2\hat{k}) \]
\[ \vec{a}+\vec{b} = (3+5)\hat{i} + (3+2)\hat{j} + (3+2)\hat{k} \]
\[ \vec{a}+\vec{b} = 8\hat{i} + 5\hat{j} + 5\hat{k} \]
Now, calculate the dot product of \( (\vec{a}+\vec{b}) \) and \( \vec{c} \):
\[ (\vec{a}+\vec{b}) \cdot \vec{c} = (8\hat{i} + 5\hat{j} + 5\hat{k}) \cdot (\hat{j} - \hat{k}) \]
\[ (\vec{a}+\vec{b}) \cdot \vec{c} = (8)(0) + (5)(1) + (5)(-1) \]
\[ (\vec{a}+\vec{b}) \cdot \vec{c} = 0 + 5 - 5 = 0 \]
Finding the Angle Between \( (\vec{a}+\vec{b}) \) and \( \vec{c} \)
Let \(\vec{V} = \vec{a}+\vec{b}\) and \(\vec{W} = \vec{c}\). The angle \(\theta\) between \(\vec{V}\) and \(\vec{W}\) is given by:
\[ \cos \theta = \frac{\vec{V} \cdot \vec{W}}{|\vec{V}| |\vec{W}|} \]
We found that \(\vec{V} \cdot \vec{W} = (\vec{a}+\vec{b}) \cdot \vec{c} = 0\).
For the angle to be defined, the magnitudes of the vectors must be non-zero. Let's check:
- \( |\vec{a}+\vec{b}| = |8\hat{i} + 5\hat{j} + 5\hat{k}| = \sqrt{8^2 + 5^2 + 5^2} = \sqrt{64 + 25 + 25} = \sqrt{114} \). This is non-zero.
- \( |\vec{c}| = |\hat{j} - \hat{k}| = \sqrt{0^2 + 1^2 + (-1)^2} = \sqrt{0 + 1 + 1} = \sqrt{2} \). This is non-zero.
Since the dot product \( (\vec{a}+\vec{b}) \cdot \vec{c} = 0 \) and both vectors \( (\vec{a}+\vec{b}) \) and \( \vec{c} \) are non-zero vectors, the cosine of the angle between them is 0.
\[ \cos \theta = 0 \]
The angle \(\theta\) whose cosine is 0 is \( \frac{\pi}{2} \) radians (or \(90^\circ\)).
Therefore, the angle between \( (\vec{a}+\vec{b}) \) and \( \vec{c} \) is \( \frac{\pi}{2} \).
Summary of Steps
Here is a quick recap of the steps taken to find the angle between \( (\vec{a}+\vec{b}) \) and \( \vec{c} \):
- Used the vector triple product identity \( (\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a} \) to find an expression for \(\vec{b}\).
- Calculated the required dot products and cross products using the given information about \(\vec{a}\), \(\vec{c}\), \(\vec{a} \cdot \vec{b}\), and \(\vec{a} \times \vec{b}\).
- Solved the vector equation obtained in step 1 to find the components of vector \(\vec{b}\).
- Calculated the vector sum \( (\vec{a}+\vec{b}) \).
- Calculated the dot product of \( (\vec{a}+\vec{b}) \) and \(\vec{c}\).
- Used the dot product formula for the angle, \(\cos \theta = \frac{(\vec{a}+\vec{b}) \cdot \vec{c}}{|(\vec{a}+\vec{b})| |\vec{c}|}\).
- Determined the angle \(\theta\) based on the value of \(\cos \theta\).
| Vector Quantity | Calculated/Given Value |
|---|---|
| \( \vec{a} \) | \( 3\hat{i} + 3\hat{j} + 3\hat{k} \) |
| \( \vec{c} \) | \( \hat{j} - \hat{k} \) |
| \( \vec{a} \cdot \vec{b} \) | \( 27 \) |
| \( \vec{a} \times \vec{b} \) | \( 9\vec{c} = 9\hat{j} - 9\hat{k} \) |
| \( \vec{a} \cdot \vec{a} \) | \( 27 \) |
| \( (\vec{a} \times \vec{b}) \times \vec{a} \) | \( 27(2\hat{i} - \hat{j} - \hat{k}) \) |
| Vector \( \vec{b} \) | \( 5\hat{i} + 2\hat{j} + 2\hat{k} \) |
| Vector \( \vec{a}+\vec{b} \) | \( 8\hat{i} + 5\hat{j} + 5\hat{k} \) |
| \( (\vec{a}+\vec{b}) \cdot \vec{c} \) | \( 0 \) |
| \( \cos \theta \) | \( \frac{0}{|(\vec{a}+\vec{b})| |\vec{c}|} = 0 \) |
| Angle \( \theta \) | \( \frac{\pi}{2} \) |
Revision Table: Key Vector Concepts
| Concept | Description | Formula Example |
|---|---|---|
| Dot Product | Scalar product, measures the projection of one vector onto another. If dot product is 0, vectors are perpendicular. | \( \vec{u} \cdot \vec{v} = u_x v_x + u_y v_y + u_z v_z \) |
| Cross Product | Vector product, results in a vector perpendicular to the plane of the two vectors. Magnitude relates to the area of a parallelogram. | \( \vec{u} \times \vec{v} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ u_x & u_y & u_z \\ v_x & v_y & v_z \end{vmatrix} \) |
| Vector Triple Product Identity | Relates the cross product and dot product of three vectors. Useful for manipulating vector equations. | \( \vec{A} \times (\vec{B} \times \vec{C}) = (\vec{A} \cdot \vec{C})\vec{B} - (\vec{A} \cdot \vec{B})\vec{C} \) |
| Angle Between Vectors | Determined using the dot product formula. | \( \cos \theta = \frac{\vec{u} \cdot \vec{v}}{|\vec{u}| |\vec{v}|} \) |
Additional Information: Importance of Vector Identities
Vector identities, like the vector triple product identity used in this problem, are powerful tools in vector algebra. They allow us to manipulate complex vector expressions and solve for unknown vectors or relationships between them. In this case, the identity \( (\vec{a} \times \vec{b}) \times \vec{a} = (\vec{a} \cdot \vec{a})\vec{b} - (\vec{a} \cdot \vec{b})\vec{a} \) was crucial because it provided a way to isolate and solve for vector \(\vec{b}\) using the given dot and cross product conditions. Without such identities, solving for \(\vec{b}\) from just \(\vec{a} \cdot \vec{b}\) and \(\vec{a} \times \vec{b}\) would be significantly more challenging. Understanding and being able to apply these identities is key to solving many problems in physics and engineering that involve vector quantities.
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