If \(\vec a + 2\vec b + 3\vec c = \vec 0\) and \(\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a = \lambda \left( {\vec b \times \overrightarrow {c\;} } \right),\) then what is the value of λ?
6
Understanding the Vector Problem
This problem involves solving a system of vector equations. We are given one equation relating three vectors \(\vec a\), \(\vec b\), and \(\vec c\) by vector addition, and another equation involving the cross products of these vectors. Our goal is to find the value of the scalar \(\lambda\) that satisfies the second equation.
Given Information
We are provided with two vector equations:
- Equation 1: \(\vec a + 2\vec b + 3\vec c = \vec 0\)
- Equation 2: \(\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a = \lambda \left( {\vec b \times \overrightarrow {c\;} } \right)\)
Solving the Vector Equation for \(\lambda\)
The strategy is to use the first equation to express one vector in terms of the others and substitute this into the second equation. This substitution will allow us to simplify the cross product expression and compare it with the right side of the second equation to find \(\lambda\).
Step-by-Step Calculation
From Equation 1, we can express \(\vec a\) in terms of \(\vec b\) and \(\vec c\):
\( \vec a = -2\vec b - 3\vec c \)Now, substitute this expression for \(\vec a\) into the left side of Equation 2:
\( \vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a \) \( = (-2\vec b - 3\vec c) \times \vec b + \vec b \times \vec c + \vec c \times (-2\vec b - 3\vec c) \)Using the distributive property of the cross product (\(\vec u \times (\vec v + \vec w) = \vec u \times \vec v + \vec u \times \vec w\) and \((\vec u + \vec v) \times \vec w = \vec u \times \vec w + \vec v \times \vec w\)), we expand the expression:
\( = (-2\vec b \times \vec b) + (-3\vec c \times \vec b) + (\vec b \times \vec c) + (\vec c \times -2\vec b) + (\vec c \times -3\vec c) \)Now, we use the properties of the cross product:
- The cross product of a vector with itself is the zero vector: \(\vec v \times \vec v = \vec 0\).
- The cross product is anti-commutative: \(\vec u \times \vec v = -\vec v \times \vec u\).
- Scalars can be factored out: \((k\vec u) \times \vec v = k(\vec u \times \vec v)\) and \(\vec u \times (k\vec v) = k(\vec u \times \vec v)\).
Applying these properties:
- \(-2\vec b \times \vec b = -2(\vec b \times \vec b) = -2(\vec 0) = \vec 0\)
- \(-3\vec c \times \vec b = -3(-\vec b \times \vec c) = 3(\vec b \times \vec c)\)
- \(\vec b \times \vec c\) remains as is.
- \(\vec c \times (-2\vec b) = -2(\vec c \times \vec b) = -2(-\vec b \times \vec c) = 2(\vec b \times \vec c)\)
- \(\vec c \times (-3\vec c) = -3(\vec c \times \vec c) = -3(\vec 0) = \vec 0\)
Substitute these back into the expanded expression:
\( = \vec 0 + 3(\vec b \times \vec c) + (\vec b \times \vec c) + 2(\vec b \times \vec c) + \vec 0 \)Combine the terms involving \((\vec b \times \vec c)\):
\( = (3 + 1 + 2)(\vec b \times \vec c) \) \( = 6(\vec b \times \vec c) \)So, the left side of Equation 2 simplifies to \(6(\vec b \times \vec c)\). Now, we equate this to the right side of Equation 2:
\( 6(\vec b \times \vec c) = \lambda (\vec b \times \vec c) \)Assuming that \((\vec b \times \vec c) \neq \vec 0\) (which implies that vectors \(\vec b\) and \(\vec c\) are not parallel and are not zero vectors), we can compare the scalar coefficients on both sides.
\( \lambda = 6 \)This value of \(\lambda\) satisfies the given vector equation involving cross products.
Verification
If \(\lambda = 6\), the second equation becomes \(\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a = 6 \left( {\vec b \times \overrightarrow {c\;} } \right)\). Our calculation showed that substituting \(\vec a = -2\vec b - 3\vec c\) into the left side indeed resulted in \(6 \left( {\vec b \times \overrightarrow {c\;} } \right)\), confirming our value for \(\lambda\).
| Given Equation 1 | \(\vec a + 2\vec b + 3\vec c = \vec 0\) |
|---|---|
| Derived \(\vec a\) | \(\vec a = -2\vec b - 3\vec c\) |
| Given Equation 2 (LHS) | \(\vec a \times \vec b + \vec b \times \vec c + \vec c \times \vec a\) |
| LHS after substitution | \((-2\vec b - 3\vec c) \times \vec b + \vec b \times \vec c + \vec c \times (-2\vec b - 3\vec c)\) |
| LHS after simplification | \(6 (\vec b \times \vec c)\) |
| Given Equation 2 (RHS) | \(\lambda (\vec b \times \vec c)\) |
| Equating LHS and RHS | \(6 (\vec b \times \vec c) = \lambda (\vec b \times \vec c)\) |
| Value of \(\lambda\) | \(6\) |
Conclusion
Based on the step-by-step simplification of the vector cross product equation using the relationship provided by the first vector equation, the value of \(\lambda\) is found to be 6.
Revision Table - Vector Algebra Review
| Concept | Description | Property Used |
|---|---|---|
| Vector Addition | Combining vectors head-to-tail. Commutative and Associative. | \(\vec a + 2\vec b + 3\vec c = \vec 0\) allows isolating one vector. |
| Vector Cross Product | Operation yielding a vector perpendicular to the plane of the two vectors. Magnitude is related to the area of the parallelogram. | Used in the second equation. |
| Anti-commutativity | \(\vec u \times \vec v = -\vec v \times \vec u\) | Crucial for simplifying terms like \(\vec c \times \vec b\). |
| Cross Product with Self | \(\vec v \times \vec v = \vec 0\) | Used to eliminate terms like \(\vec b \times \vec b\) and \(\vec c \times \vec c\). |
| Distributivity | \(\vec u \times (\vec v + \vec w) = \vec u \times \vec v + \vec u \times \vec w\) | Used to expand terms like \((\vec a) \times \vec b\) after substituting \(\vec a\). |
Additional Information - Vector Operations
Vector algebra is a fundamental tool in physics and engineering. Understanding vector operations is key to solving problems involving forces, velocities, areas, and volumes in three dimensions.
- Vector Addition: Geometrically, adding vectors involves placing the tail of the second vector at the head of the first. Algebraically, you add corresponding components.
- Scalar Multiplication: Multiplying a vector by a scalar changes its magnitude but not its direction (unless the scalar is negative).
- Dot Product (Scalar Product): Results in a scalar. \(\vec u \cdot \vec v = |\vec u| |\vec v| \cos \theta\). Useful for finding the angle between vectors or the projection of one vector onto another.
- Cross Product (Vector Product): Results in a vector perpendicular to both original vectors. \(|\vec u \times \vec v| = |\vec u| |\vec v| \sin \theta\). Direction is given by the right-hand rule. Useful for finding torques, angular momentum, and areas of parallelograms.
In this problem, the cross product properties, specifically anti-commutativity and the cross product of a vector with itself being zero, were essential for simplification.
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