NEET Practice Questions (MCQs) with Answers & Solutions

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 \( \frac{ \vec{A} \times \vec{B} }{ AB \sin \theta } \).

$ \vec A + \vec B + \vec C = \hat j $

$ x = 2t^2 = 8 $ $ \therefore t = 2s $ $ y = { 3 \over 2 } t^2 = {3 \over2 } (2)^2 = 6m $

$$ x = A coswt $$ $$ y = A ( i - sin wt ) $$ $$ V \alpha = { dx \over dv } = - A \omega sin wt $$ $$Vy = - A \omega cos wt $$ $$ V = \sqrt { Vx^2 + Vy^2 } $$

The time of flight for a projectile is given by \( \frac{2V_0 \sin \theta}{g} \). Since the angle of projection \( \theta \) will repeat at half the time of flight, the time when the angle will be \( \theta \) again is \( \frac{2V_0 \sin \theta}{g} \).

$ \vec V_0 = V_0cosθ \hat i + V_0 sinθ \hat j$ $ \vec V = V_0 cosθ \hat i + (V_0sm \theta – gt) \hat j$ $ \vec V_0 . \vec V = 0 $ $ \therefore t = { V_0 \over gsin \theta } $ Now find x

At time t $ V_ x = V_0 = V_0 cos \theta $ $ V_y = Visinθ – gt$
$ tan \alpha = { Vy \over Vx } = {V_0sin \theta – gt \over V_0cos \theta } $