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Young's Modulus of Elasticity

lf the magnit...

Question

lf the magnitudes of $\to A, \to B, \to C$ are $12, 5, 13$ units respectively and $\to A + \to B = \to C$ , then the angle between $\to A, \to B$ is:

A

0

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B

π

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C

\frac{π}{2}

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D

\frac{π}{4}

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Solution

The correct option is A $\frac{π}{2}$
$\to A + \to B = \to C$ suggest that they make a triangle and $5, 12, 13$ make a right angled one.
Hence $\to A, \to B$ make 90 degrees as they are the smaller sides.

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\to A = \to B + \to C

and the magnitudes of

\to A, \to B, \to C

5

4

3

units respectively. Then the angle between

\to A

\to C

\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 |

\to a

\to b

\to c

are unit vectors such that

\to a + \to b + \to c = 0,

\to a \cdot \to b + \to b \cdot \to c + \to c \cdot \to a

The magnitudes of vectors $\to A, \to B a n d \to C$ are 3, 4 and 5 units respectively. If $\to A + \to B = \to C$ , the angle between $\to A$ and $\to B$ is

The magnitude of vector $\to A, \to B$ and $\to C$ are respectively 12, 5 and 13 units and $\to A + \to B = \to C$ then the angle between $\to A$ and $\to B$ is

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