JEE PHYSICS MCQ: MOTION IN PLANE

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Analyzing +4 | -1

\overrightarrow P \,\,.\,\,\overrightarrow Q = 0\,and\,\overrightarrow P \,\, \times \,\,\overrightarrow Q = 0

, then the angle between

\overrightarrow Q \,and\,\overrightarrow R

View answer and explanation

Correct option: B

{90^0}

Option B (

{90^0}

) is correct because if the dot product is 0, the vectors are perpendicular, and if the cross product is 0, they are parallel or anti-parallel. Since

\overrightarrow Q

is perpendicular to

\overrightarrow P

and

\overrightarrow R

is parallel to

\overrightarrow P

\overrightarrow Q

and

\overrightarrow R

are perpendicular.

More Questions from MOTION IN PLANE

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\mathrm{\vec P} = \mathrm{\vec A} + \mathrm{\vec B}\,\,\mathrm{and}\,P = A + B.

\mathrm{\vec A}\,and\,\mathrm{\vec B}

0^{\circ}

90^{\circ}

\mathrm{If}\,(\mathbf{a}+3 \mathbf{b}) \cdot(7 \mathbf{a}-5 \mathbf{b})=0

(\mathbf{a}-4 \mathbf{b}) \cdot(7 \mathbf{a}-2 \mathbf{b})=0

\mathbf{a}

\mathbf{b}

\theta

\left(90^{\circ}-\theta\right)

39.2 \mathrm{~ms}^{-1}

50 \mathrm{~m}

{g}=9.8 \mathrm{~ms}^{-2}