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$ \vec A and \vec B $ are in opposite direction so they are
When two vectors \( \vec{A} \) and \( \vec{B} \) are in opposite directions, they are referred to as antiparallel vectors. Antiparallel vectors have the same magnitude but opposite directions. This is different from parallel vectors, which have the same direction.
$ \vec C = \vec A + \vec B $ and A = B = C . Find the angle between $ \vec A and \vec B $
$$ \vec C = \vec A + \vec B $$ $$ C^2 = A^2 + B^2 + 2AB cos \theta $$
The resultant of two forces of magnitude 2N and 3N can never be.
The resultant of two forces can be found using the triangle law of vector addition. The magnitude of the resultant force \( \vec{R} \) of two forces of magnitudes 2N and 3N can range from the absolute difference of the two forces to the sum of the two forces. Hence, \( |2N - 3N| \leq R \leq 2N + 3N \), which simplifies to \( 1N \leq R \leq 5N \). Therefore, a resultant force of 0.5N is not possible.
The sum of $ \vec P and \vec Q $ is at right agnles to their difference then
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 $A = B$.
Out of the following pairs of forces, the resultant of which can not be 18N
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:
Thus, the pair 11N and 5N cannot have a resultant of 18N.
$ \vec A = 2 \hat i + 2 \hat j - \hat k $ $ \vec B = 2 \hat i - \hat j - 2 \hat k $ Find $ 3 \vec A - 2 \vec B $
Linear momentajm of a particle is $ (3 \hat i + 2 \hat j - \hat k ) kgms^{-1} $. Find its magnitude
$ \vec A \times \vec B = \vec C $ Then $ \vec C $ is perpendicular to
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.
Which statement is true ?
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}.
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The bank mixes NEET previous year questions (PYQs) with practice questions, each tagged with its exam appearances where applicable.