Question:
Assertion (A) : The vectors \(\vec{a}\) and \((-2\vec{a})\), where \(\vec{a} \neq \vec{0}\) are collinear vectors.
Reason (R) : \(\vec{a} \cdot (-2\vec{a}) = 0\).
Assertion (A) : The vectors \(\vec{a}\) and \((-2\vec{a})\), where \(\vec{a} \neq \vec{0}\) are collinear vectors.
Reason (R) : \(\vec{a} \cdot (-2\vec{a}) = 0\).
Reason (R) : \(\vec{a} \cdot (-2\vec{a}) = 0\).
Show Hint
Collinear = Proportional (\(\vec{u} = k\vec{v}\)).
Perpendicular = Zero Dot Product (\(\vec{u} \cdot \vec{v} = 0\)).
These are mutually exclusive for non-zero vectors.
Updated On: Sep 10, 2026
- Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).
- Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).
- Assertion (A) is true, but Reason (R) is false.
- Assertion (A) is false, but Reason (R) is true.
Show Solution
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The Correct Option is C
Solution and Explanation
Concept:
• Collinear vectors are vectors that act along the same line or are parallel to each other.
• Mathematically, two vectors \(\vec{u}\) and \(\vec{v}\) are collinear if \(\vec{u} = \lambda \vec{v}\) for some non-zero scalar \(\lambda\).
• The dot product \(\vec{u} \cdot \vec{v} = 0\) implies that the vectors are perpendicular, not collinear (unless one is zero).
Step 1: Evaluate the Assertion
Let \(\vec{b} = -2\vec{a}\).
Since \(\vec{b}\) is a scalar multiple of \(\vec{a}\) (where \(\lambda = -2\)), they are parallel/anti-parallel.
Thus, \(\vec{a}\) and \(-2\vec{a}\) are collinear.
Assertion (A) is true.
Step 2: Evaluate the Reason
Calculate the dot product:
\[ \vec{a} \cdot (-2\vec{a}) = -2(\vec{a} \cdot \vec{a}) = -2|\vec{a}|^2 \] Since it is given that \(\vec{a} \neq \vec{0}\), then \(|\vec{a}| > 0\).
This means \(-2|\vec{a}|^2 < 0\), so it can never be \(0\).
Reason (R) is false.
• Collinear vectors are vectors that act along the same line or are parallel to each other.
• Mathematically, two vectors \(\vec{u}\) and \(\vec{v}\) are collinear if \(\vec{u} = \lambda \vec{v}\) for some non-zero scalar \(\lambda\).
• The dot product \(\vec{u} \cdot \vec{v} = 0\) implies that the vectors are perpendicular, not collinear (unless one is zero).
Step 1: Evaluate the Assertion
Let \(\vec{b} = -2\vec{a}\).
Since \(\vec{b}\) is a scalar multiple of \(\vec{a}\) (where \(\lambda = -2\)), they are parallel/anti-parallel.
Thus, \(\vec{a}\) and \(-2\vec{a}\) are collinear.
Assertion (A) is true.
Step 2: Evaluate the Reason
Calculate the dot product:
\[ \vec{a} \cdot (-2\vec{a}) = -2(\vec{a} \cdot \vec{a}) = -2|\vec{a}|^2 \] Since it is given that \(\vec{a} \neq \vec{0}\), then \(|\vec{a}| > 0\).
This means \(-2|\vec{a}|^2 < 0\), so it can never be \(0\).
Reason (R) is false.
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