Motion in a Lane MCQs for NEET — Physics Questions with Answers

Given that the sum of vectors $\vec{P}$ and $\vec{Q}$ is at right angles to their difference, we have:

$\vec{P} + \vec{Q} \perp \vec{P} - \vec{Q}$.

This implies that their dot product is zero:

$(\vec{P} + \vec{Q}) \cdot (\vec{P} - \vec{Q}) = 0$.

Expanding this, we get:

$\vec{P} \cdot \vec{P} - \vec{P} \cdot \vec{Q} + \vec{Q} \cdot \vec{P} - \vec{Q} \cdot \vec{Q} = 0$.

Simplifying, we obtain:

$P^2 - Q^2 = 0$, or $P^2 = Q^2$.

Hence, $P = Q$, which corresponds to $A = B$.

To determine which pair of forces cannot have a resultant of 18N, we use the triangle inequality theorem for vectors. For two vectors with magnitudes $a$ and $b$:

The resultant $R$ satisfies $|a - b| \leq R \leq a + b$.

For the given options:

1. $|11N - 7N| \leq 18N \leq 11N + 7N$ ⟹ $4N \leq 18N \leq 18N$ (Possible)
2. $|11N - 8N| \leq 18N \leq 11N + 8N$ ⟹ $3N \leq 18N \leq 19N$ (Possible)
3. $|11N - 29N| \leq 18N \leq 11N + 29N$ ⟹ $18N \leq 18N \leq 40N$ (Possible)
4. $|11N - 5N| \leq 18N \leq 11N + 5N$ ⟹ $6N \leq 18N \leq 16N$ (Not Possible)

Thus, the pair 11N and 5N cannot have a resultant of 18N.

The cross product $ \\vec{A} \\times \\vec{B} = \\vec{C} $ results in a vector \vec{C} that is perpendicular to both \vec{A} and \vec{B} regardless of the angle between them. This is a fundamental property of the cross product. Hence, \vec{C} is perpendicular to \vec{A} and \vec{B} whatever the angle between them.

The cross product $ \\vec{A} \\times \\vec{B} $ is anticommutative, which means that $ \\vec{A} \\times \\vec{B} = - \\vec{B} \\times \\vec{A} $. This is a fundamental property of the cross product in vector algebra. Therefore, the correct statement is \vec{A} \times \vec{B} = - \vec{B} \times \vec{A}.

For two vectors \( \vec{A} \) and \( \vec{B} \), if \( |\vec{A} + \vec{B}| = |\vec{A} - \vec{B}| \), then the vectors are perpendicular to each other. This is because the magnitudes of the sum and difference of the vectors are equal only when the angle between them is 90°. Therefore, the angle between \( \vec{A} \) and \( \vec{B} \) is 90°.

To find the unit vector in the direction of \( \hat{i} + 2 \hat{j} - 3 \hat{k} \), we first find the magnitude of the vector: \( \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14} \). The unit vector is then given by dividing each component by the magnitude: \( \frac{1}{\sqrt{14}} ( \hat{i} + 2 \hat{j} - 3 \hat{k} ) \). Therefore, the correct answer is \( \frac{1}{\sqrt{14}} ( \hat{i} + 2 \hat{j} - 3 \hat{k} ) \).

To find a unit vector that is perpendicular to both vectors \( \vec{A} \) and \( \vec{B} \), we use the cross product \( \vec{A} \times \vec{B} \). The magnitude of the cross product is given by \( AB \sin \theta \). Therefore, the unit vector perpendicular to both \( \vec{A} \) and \( \vec{B} \) is \( \frac{ \vec{A} \times \vec{B} }{ AB \sin \theta } \).