Let \(\rm\vec {a}, \vec{b}\) and \( \rm \vec {c}\) be three vectors such that \(\rm\vec {a}, \vec{b}\) and \( \rm \vec {c}\) are co-planar. Which of the following is/are correct? 1. \(\rm(\vec{a}\times \vec{b})\times \vec{c}\) is co-planar with \(\rm\vec {a}\) and \(\rm\vec {b}\) 2. \(\rm(\vec{a}\times \vec{b})\times \vec{c}\) is perpendicular to \(\rm\vec {a}\) and \(\rm\vec {b}\) Select the correct answer using the code given below.
1 only
Understanding Vector Coplanarity and Triple Product
This question deals with properties of vectors, specifically coplanarity and the vector triple product. We are given three vectors \( \vec{a}, \vec{b}, \vec{c} \) that are coplanar. We need to analyze two statements about the vector triple product \( (\vec{a} \times \vec{b}) \times \vec{c} \).
Let's analyze each statement carefully.
Statement 1: \( (\vec{a}\times \vec{b})\times \vec{c} \) is co-planar with \( \vec{a} \) and \( \vec{b} \)
We can use the vector triple product identity to expand \( (\vec{a}\times \vec{b})\times \vec{c} \). The identity is:
\( (\vec{x} \times \vec{y}) \times \vec{z} = (\vec{x} \cdot \vec{z})\vec{y} - (\vec{y} \cdot \vec{z})\vec{x} \)
Applying this identity with \( \vec{x} = \vec{a} \), \( \vec{y} = \vec{b} \), and \( \vec{z} = \vec{c} \), we get:
\( (\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} \)
Let's look at the expression \( (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} \). Here, \( (\vec{a} \cdot \vec{c}) \) and \( (\vec{b} \cdot \vec{c}) \) are scalar values (real numbers). Let \( \lambda = (\vec{a} \cdot \vec{c}) \) and \( \mu = (\vec{b} \cdot \vec{c}) \). Then the expression becomes \( \lambda\vec{b} - \mu\vec{a} \).
A vector that can be written as a linear combination of two vectors \( \vec{a} \) and \( \vec{b} \) in the form \( \alpha\vec{a} + \beta\vec{b} \) lies in the plane spanned by \( \vec{a} \) and \( \vec{b} \). The expression \( \lambda\vec{b} - \mu\vec{a} \) is a linear combination of \( \vec{a} \) and \( \vec{b} \).
Therefore, the vector \( (\vec{a} \times \vec{b}) \times \vec{c} \) lies in the plane formed by \( \vec{a} \) and \( \vec{b} \). This means \( (\vec{a} \times \vec{b}) \times \vec{c} \) is coplanar with \( \vec{a} \) and \( \vec{b} \).
Statement 1 is correct.
Statement 2: \( (\vec{a}\times \vec{b})\times \vec{c} \) is perpendicular to \( \vec{a} \) and \( \vec{b} \)
We know that \( (\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} \). For this vector to be perpendicular to \( \vec{a} \), their dot product must be zero.
Let's calculate the dot product with \( \vec{a} \):
\( ((\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a}) \cdot \vec{a} = (\vec{a} \cdot \vec{c})(\vec{b} \cdot \vec{a}) - (\vec{b} \cdot \vec{c})(\vec{a} \cdot \vec{a}) \)
\( = (\vec{a} \cdot \vec{c})(\vec{a} \cdot \vec{b}) - (\vec{b} \cdot \vec{c})|\vec{a}|^2 \)
For \( (\vec{a} \times \vec{b}) \times \vec{c} \) to be perpendicular to \( \vec{a} \), this expression \( (\vec{a} \cdot \vec{c})(\vec{a} \cdot \vec{b}) - (\vec{b} \cdot \vec{c})|\vec{a}|^2 \) must be equal to zero for any coplanar vectors \( \vec{a}, \vec{b}, \vec{c} \) (where \( \vec{a} \neq \vec{0} \)).
Let's consider a simple example of coplanar vectors. Suppose \( \vec{a} = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \), \( \vec{b} = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \), and \( \vec{c} = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \). These are all in the xy-plane, so they are coplanar.
- \( \vec{a} \cdot \vec{c} = (1)(1) + (0)(1) + (0)(0) = 1 \)
- \( \vec{b} \cdot \vec{c} = (0)(1) + (1)(1) + (0)(0) = 1 \)
- \( \vec{a} \cdot \vec{b} = (1)(0) + (0)(1) + (0)(0) = 0 \)
- \( |\vec{a}|^2 = 1^2 + 0^2 + 0^2 = 1 \)
Now substitute these values into the dot product expression:
\( (1)(0) - (1)(1) = 0 - 1 = -1 \)
Since the dot product is -1 (not 0), \( (\vec{a} \times \vec{b}) \times \vec{c} \) is not perpendicular to \( \vec{a} \) in this case. Therefore, statement 2 is not always correct.
Also, note that \( (\vec{a} \times \vec{b}) \) is a vector perpendicular to the plane containing \( \vec{a} \) and \( \vec{b} \). Since \( \vec{c} \) is coplanar with \( \vec{a} \) and \( \vec{b} \), \( \vec{c} \) lies in this plane. The cross product of a vector perpendicular to a plane \( (\vec{a} \times \vec{b}) \) and a vector in that plane \( \vec{c} \) results in a vector \( (\vec{a} \times \vec{b}) \times \vec{c} \) that also lies in that plane (the plane of \( \vec{a} \) and \( \vec{b} \)). A vector lying in a plane is generally not perpendicular to the vectors spanning that plane, unless specific conditions are met (like orthogonality). Statement 2 is incorrect.
Conclusion on Statements
- Statement 1 is correct.
- Statement 2 is incorrect.
Based on this analysis, only Statement 1 is correct.
| Statement | Analysis | Correctness |
|---|---|---|
| 1. \( (\vec{a}\times \vec{b})\times \vec{c} \) is co-planar with \( \vec{a} \) and \( \vec{b} \) | Vector triple product \( (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} \) is a linear combination of \( \vec{a} \) and \( \vec{b} \). | Correct |
| 2. \( (\vec{a}\times \vec{b})\times \vec{c} \) is perpendicular to \( \vec{a} \) and \( \vec{b} \) | Dot product with \( \vec{a} \) is \( (\vec{a} \cdot \vec{c})(\vec{a} \cdot \vec{b}) - (\vec{b} \cdot \vec{c})|\vec{a}|^2 \), which is not always zero for coplanar vectors. | Incorrect |
Summary of Findings
We evaluated both statements based on the properties of vector operations and identities. The condition that \( \vec{a}, \vec{b}, \vec{c} \) are coplanar was used to understand their geometric relationship but did not change the fundamental vector identities used to evaluate the statements.
Only Statement 1 is supported by vector properties.
Revision Table: Key Vector Concepts
| Concept | Description |
|---|---|
| Coplanar Vectors | Vectors that lie in the same plane. For three vectors \( \vec{a}, \vec{b}, \vec{c} \), they are coplanar if their scalar triple product is zero: \( \vec{a} \cdot (\vec{b} \times \vec{c}) = 0 \). |
| Vector Cross Product \( \vec{a} \times \vec{b} \) | A vector perpendicular to both \( \vec{a} \) and \( \vec{b} \). Its magnitude is \( |\vec{a}||\vec{b}|\sin\theta \), where \( \theta \) is the angle between \( \vec{a} \) and \( \vec{b} \). |
| Vector Dot Product \( \vec{a} \cdot \vec{b} \) | A scalar value \( |\vec{a}||\vec{b}|\cos\theta \). It measures the projection of one vector onto another. If \( \vec{a} \cdot \vec{b} = 0 \) and \( \vec{a}, \vec{b} \) are non-zero, then they are perpendicular. |
| Vector Triple Product \( (\vec{a} \times \vec{b}) \times \vec{c} \) | A vector resulting from the cross product of \( (\vec{a} \times \vec{b}) \) and \( \vec{c} \). Identity: \( (\vec{a} \times \vec{b}) \times \vec{c} = (\vec{a} \cdot \vec{c})\vec{b} - (\vec{b} \cdot \vec{c})\vec{a} \). This vector is always coplanar with \( \vec{a} \) and \( \vec{b} \) (if they are non-collinear). |
Additional Information: Geometric Interpretation
When \( \vec{a}, \vec{b}, \vec{c} \) are coplanar, the vector \( \vec{n} = \vec{a} \times \vec{b} \) is perpendicular to the plane containing \( \vec{a}, \vec{b}, \) and \( \vec{c} \). The vector \( (\vec{a} \times \vec{b}) \times \vec{c} = \vec{n} \times \vec{c} \) is the cross product of a vector \( \vec{n} \) perpendicular to the plane and a vector \( \vec{c} \) lying in the plane. The resulting vector \( \vec{n} \times \vec{c} \) must be perpendicular to both \( \vec{n} \) and \( \vec{c} \). Since \( \vec{c} \) is in the plane, the resulting vector \( \vec{n} \times \vec{c} \) must be in the same plane as \( \vec{a}, \vec{b}, \) and \( \vec{c} \).
This confirms Statement 1 geometrically: if \( \vec{a}, \vec{b}, \vec{c} \) are coplanar, then \( (\vec{a} \times \vec{b}) \times \vec{c} \) lies in the plane formed by \( \vec{a}, \vec{b}, \vec{c} \). Since \( \vec{a} \) and \( \vec{b} \) are also in this plane, \( (\vec{a} \times \vec{b}) \times \vec{c} \) is coplanar with \( \vec{a} \) and \( \vec{b} \).
For Statement 2, \( (\vec{a} \times \vec{b}) \times \vec{c} \) is in the plane of \( \vec{a} \) and \( \vec{b} \). For it to be perpendicular to both \( \vec{a} \) and \( \vec{b} \), it would have to be the zero vector (unless \( \vec{a} \) and \( \vec{b} \) are zero or collinear, which are trivial cases not typically implied unless specified). In general, a vector in a plane is not perpendicular to the vectors forming that plane. For instance, in the xy-plane, the vector \( \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \) is in the plane but not perpendicular to \( \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} \) or \( \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} \).
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