JEE PHYSICS MCQ: MOTION IN PLANE

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Analyzing +4 | -1

Let A, B and

\mathbf{C}

be unit vectors. Suppose

\mathbf{A} \cdot \mathbf{B}=\mathbf{A} \cdot \mathbf{C}=0

and the angle between

\mathbf{B}

and

\mathbf{C}

\frac{\pi}{4}

. Then,

View answer and explanation

Correct option: B

\mathbf{A}=\pm \sqrt{2}(\mathbf{B} \times \mathbf{C})

The correct answer is B (

\mathbf{A}=\pm \sqrt{2}(\mathbf{B} \times \mathbf{C})

) because it is derived from the fact that A is perpendicular to both B and C, and A is a unit vector.

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}