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Comparing Vectors

If the unit v...

Question

If the unit vectors $\to a$ and $\to b$ are inclined at an angular are $| \to a - \to b | < 1$ , then $0 \leq θ \leq π, θ$ lies in the

A

[0, \frac{π}{3}]

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B

[\frac{2 π}{3}, π]

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C

[\frac{π}{3}, \frac{π}{2}]

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D

[\frac{π}{2}, \frac{2 π}{3}]

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Solution

The correct option is A $[0, \frac{π}{3}]$
$\begin{matrix} ∣ ∣ \to a - \to b ∣ ∣ < 1 {∣ ∣ \to a - \to b ∣ ∣}^{2} < 1 (\to a - \to b) \cdot (\to a - \to b) < 1 2 - 2 \to a \cdot \to b < 1 1 < 2 \to a \cdot \to b \frac{1}{2} < \to a \cdot \to b \frac{1}{2} < cos θ cos θ > \frac{1}{2} θ \in [0, \frac{π}{3}) H e n c e, t h e c o r r e c t o p t i o n i s (A) \end{matrix}$

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Similar questions

Q. If the unit vectors

\to a

\to b

are inclined at an angle 2

θ

| \to a - \to b |

\leq θ \leq π

θ

lies in the interval

Let and be two unit vectors andθ is the angle between them. Then is a unit vector if

(A) (B) (C) (D)

\to a, \to b

are two unit vectors inclined at an angle

\frac{π}{3}

{\to a \times (\to b + \to a \times \to b)} . \to b =

| \to a + \to b | = | \to a - \to b |

(\to a, \to b) = \frac{π}{2}

\to a, \to b, \to a + \to b

are unit vectors, then

(\to a, \to b) = \frac{2 π}{3}

.

\to a, \to b, \to c

are unit vectors and the angles between

\to a, \to b

\to b, \to c

\to c, \to a

\frac{π}{6}, \frac{π}{3}

\frac{π}{4}

respectively, then

| \to a + \to b + \to c |

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