Question 45 of 68intermediate💡 UnderstandShort Answer2 marks

If either a⃗ = 0⃗ or b⃗ = 0⃗, then a⃗ × b⃗ = 0⃗. Is the converse true? Justify your answer with an example.

Correct Answer

No, the converse is not true. The given statement says that if a vector is zero, the cross product is zero. However, the converse would imply that a⃗×b⃗=0⃗\vec{a} \times \vec{b} = \vec{0} only if one of the vectors is zero. According to the provided context, for two nonzero vectors, a⃗×b⃗=0⃗\vec{a} \times \vec{b} = \vec{0} if and only if a⃗\vec{a} and b⃗\vec{b} are parallel (or collinear) to each other.

Therefore, a⃗×b⃗=0⃗\vec{a} \times \vec{b} = \vec{0} does not necessarily mean that either a⃗=0⃗\vec{a} = \vec{0} or b⃗=0⃗\vec{b} = \vec{0}; they could be non-zero parallel vectors. For example, let a⃗=i^\vec{a} = \hat{i} and b⃗=2i^\vec{b} = 2\hat{i}. Here, neither vector is zero, but since they are parallel, a⃗×b⃗=0⃗\vec{a} \times \vec{b} = \vec{0}.

Exercise: EXERCISE 10.4 | Q: 8 | (Page 31)
For More Understanding

Explanation

The question tests the understanding of conditions for the vector product to be zero. The provided context explicitly states in Observation 2 that for nonzero vectors, the cross product is zero if and only if the vectors are parallel. This observation directly contradicts the converse statement, which would incorrectly assert that the cross product is zero only if one of the vectors is the zero vector. The example using unit vectors illustrates the parallel case.

Solution Steps

  1. Step 1: Identify the converse statement: If a⃗×b⃗=0⃗\vec{a} \times \vec{b} = \vec{0}, then either a⃗=0⃗\vec{a} = \vec{0} or b⃗=0⃗\vec{b} = \vec{0}.

  2. Step 2: Use Observation 2 from the context: a⃗×b⃗=0⃗⇔a⃗∥b⃗\vec{a} \times \vec{b} = \vec{0} \Leftrightarrow \vec{a} \parallel \vec{b} for nonzero vectors.

  3. Step 3: Conclude that the cross product is zero for parallel non-zero vectors as well, so the converse is false.

  4. Step 4: Justify with an example of parallel non-zero vectors, such as a⃗=i^\vec{a} = \hat{i} and b⃗=2i^\vec{b} = 2\hat{i}.

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