Consider the following statements: 1. The magnitude of \(\vec{a}\times \vec{b}\) is same as the area of a triangle with sides \(\vec{a}\) and \(\vec{b}\) 2. If \(\vec{a}\times \vec{b}=\vec{0}\) where \(\vec{a}\ne \vec{0},~\vec{b}\ne \vec{0},\) then \(\vec{a}=\lambda \vec{b}\)
2 only
The question asks us to evaluate two statements related to the vector cross product of two vectors, \(\vec{a}\) and \(\vec{b}\).
Statement 1 says that the magnitude of \(\vec{a}\times \vec{b}\) is the same as the area of a triangle with sides \(\vec{a}\) and \(\vec{b}\).
Let's recall the definition of the magnitude of the cross product:
The magnitude of the cross product of two vectors \(\vec{a}\) and \(\vec{b}\) is given by:
\(|\vec{a}\times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\)
where \(\theta\) is the angle between the vectors \(\vec{a}\) and \(\vec{b}\) (\(0 \le \theta \le \pi\)).
This magnitude is also geometrically interpreted as the area of the parallelogram formed by the vectors \(\vec{a}\) and \(\vec{b}\) as adjacent sides.
Now, consider a triangle with sides \(\vec{a}\) and \(\vec{b}\). If \(\vec{a}\) and \(\vec{b}\) are adjacent sides of a triangle, the area of this triangle is half the area of the parallelogram formed by these vectors.
The area of the triangle with adjacent sides \(\vec{a}\) and \(\vec{b}\) is:
\(\text{Area of triangle} = \frac{1}{2}|\vec{a}||\vec{b}|\sin\theta = \frac{1}{2}|\vec{a}\times \vec{b}|\)
Comparing this with the statement:
Clearly, \(|\vec{a}\times \vec{b}| \ne \frac{1}{2}|\vec{a}\times \vec{b}|\) unless \(|\vec{a}\times \vec{b}|=0\). Therefore, Statement 1 is incorrect. The magnitude of the cross product is the area of the parallelogram, not the triangle.
Statement 2 says that if \(\vec{a}\times \vec{b}=\vec{0}\), where \(\vec{a}\ne \vec{0}\) and \(\vec{b}\ne \vec{0}\), then \(\vec{a}=\lambda \vec{b}\) for some scalar \(\lambda\).
Let's consider the condition \(\vec{a}\times \vec{b}=\vec{0}\).
We know that \(|\vec{a}\times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\).
If \(\vec{a}\times \vec{b}=\vec{0}\), then \(|\vec{a}\times \vec{b}|=0\). This means \(|\vec{a}||\vec{b}|\sin\theta = 0\).
The statement gives the conditions \(\vec{a}\ne \vec{0}\) and \(\vec{b}\ne \vec{0}\). This implies \(|\vec{a}|\ne 0\) and \(|\vec{b}|\ne 0\).
So, for \(|\vec{a}||\vec{b}|\sin\theta = 0\) to be true under the condition that \(|\vec{a}|\ne 0\) and \(|\vec{b}|\ne 0\), we must have \(\sin\theta = 0\).
For \(0 \le \theta \le \pi\), \(\sin\theta = 0\) implies \(\theta = 0\) or \(\theta = \pi\). This means the angle between vectors \(\vec{a}\) and \(\vec{b}\) is either 0 degrees or 180 degrees.
When the angle between two non-zero vectors is 0 or 180 degrees, the vectors are parallel or collinear. Two non-zero vectors are parallel or collinear if and only if one is a scalar multiple of the other.
Thus, if \(\vec{a}\ne \vec{0}\), \(\vec{b}\ne \vec{0}\), and \(\vec{a}\times \vec{b}=\vec{0}\), then \(\vec{a}\) and \(\vec{b}\) are parallel, which means \(\vec{a} = \lambda \vec{b}\) for some scalar \(\lambda\).
Therefore, Statement 2 is correct.
Based on our analysis:
Thus, only Statement 2 is correct.
The correct option is the one stating that only Statement 2 is correct.
| Concept | Formula/Condition | Description |
|---|---|---|
| Magnitude of Cross Product | \(|\vec{a}\times \vec{b}| = |\vec{a}||\vec{b}|\sin\theta\) | Area of the parallelogram formed by \(\vec{a}\) and \(\vec{b}\). |
| Area of Triangle | \(\frac{1}{2}|\vec{a}\times \vec{b}|\) | Area of the triangle with adjacent sides \(\vec{a}\) and \(\vec{b}\). |
| Cross Product is Zero | \(\vec{a}\times \vec{b}=\vec{0}\) | Means vectors \(\vec{a}\) and \(\vec{b}\) are parallel or at least one vector is the zero vector. |
| Parallel Vectors | \(\vec{a} = \lambda \vec{b}\) (\(\vec{a}\ne \vec{0}, \vec{b}\ne \vec{0}\)) | Equivalently, \(\vec{a}\times \vec{b}=\vec{0}\) (\(\vec{a}\ne \vec{0}, \vec{b}\ne \vec{0}\)). |
The vector cross product, also known as the vector product, is a binary operation on two vectors in three-dimensional space. The result is a vector that is perpendicular to both of the input vectors and therefore normal to the plane containing them. The direction of the resulting vector is determined by the right-hand rule. The magnitude, as discussed, relates to the area of a parallelogram.
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