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JEE PHYSICS MCQ: MOTION IN PLANE

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Analyzing +4 | -1
Statement-I: a⃗=i^+pj^+2k^\vec{a}=\hat{i}+p \hat{j}+2 \hat{k} and b⃗=2i^+3j^+qk^\vec{b}=2 \hat{i}+3 \hat{j}+q \hat{k} are parallel vectors if p=32,q=4p=\frac{3}{2}, q=4.
Statement-II: If a⃗=a1i^+a2j^+a3k^\vec{a}=a_1 \hat{i}+a_2 \hat{j}+a_3 \hat{k} and b⃗=b1i^+b2j^+b3k^\vec{b}=b_1 \hat{i}+b_2 \hat{j}+b_3 \hat{k} are parallel, then a1b1=a2b2−a3b3\frac{a_1}{b_1}=\frac{a_2}{b_2}-\frac{a_3}{b_3}
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Correct option: A
Statement-I is true and statement-II is true
Option D (Statement-I is true and statement-II is true) is correct because both the given statements are correct about parallel vectors.

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Given, P⃗=A⃗+B⃗  and P=A+B.\mathrm{\vec P} = \mathrm{\vec A} + \mathrm{\vec B}\,\,\mathrm{and}\,P = A + B. The angle between A⃗ and B⃗\mathrm{\vec A}\,and\,\mathrm{\vec B} is Assertion: The necessary condition for motion in two dimensions, velocity and acceleration vectors should have any angle between 0∘0^{\circ} and 90∘90^{\circ} between them.
Reason: For two-dimensional or three-dimensional motion, velocity and acceleration of an object works in opposite direction.
If (a+3b)⋅(7a−5b)=0\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0 and
(a−4b)⋅(7a−2b)=0(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0. Then, the angle between a\mathbf{a} and b\mathbf{b} is
At the top of the trajectory of a projectile, the direction of its velocity and acceleration are A ball is thrown at an angle θ\theta and another ball is thrown at an angle (90∘−θ)\left(90^{\circ}-\theta\right) with the horizontal direction from the same point with velocity 39.2 ms−139.2 \mathrm{~ms}^{-1}. The second ball reaches 50 m50 \mathrm{~m} higher than the first ball. Find their individual heights. Take g=9.8 ms−2{g}=9.8 \mathrm{~ms}^{-2}. Assertion: The drover in a vehicle moving with a constant speed on a straight road is in a non-inertial frame of reference.

Reason: A reference frame in which Newton's laws of motion are applicable is non-inertial.
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