Question

The position vectors of two particles ejected simultaneously from the same source are r1→=3i^+4j^+5k^ and r2→=2i^+6j^+8k^. The cosine of the angle between r1→ and r2→ is

Answer 1

sin → cos ; sec → cosec ; cosec → sec; tan → cot; cos ... A vector x is added to two vectors A = 3i -5j+7k and B = 2i + 4j -3k. 11. ... If particle moves from r1 position to r2 position then.

Answer 2

The cosine of the angle between r1→ and r2→ is  7 / 2√13.

Given: The position vectors given as, r1 = 3i^ + 4j^ + 5k^ and r2 = 2i^ + 6j^ + 8k^

To Find: The cosine of the angle between r1→ and r2→

Solution:

• We know that the value of the cosine of the angle between two position vectors can be found using the formula,

        cos Ф = ( r1 . r2 ) / ( | r1 | × | r2 | )                                  ...(1)

where Ф = angle between the vectors, ( r1 . r2 ) = dot product of the 2 vectors.

Coming to the numerical, we are given;

r1 = 3i^ + 4j^ + 5k^ and r2 = 2i^ + 6j^ + 8k^

So, putting respective values in (1), we get;

     cos Ф = ( r1 . r2 ) / ( | r1 | × | r2 | )      

 ⇒ cos Ф = ( 3i^ + 4j^ + 5k^ ) . ( 2i^ + 6j^ + 8k^ ) ] / ( | r1 | × | r2 | )

 ⇒ cos Ф = ( 6 + 24 + 40 ) / ( | r1 | × | r2 | )

 ⇒ cos Ф = 70 / [√( 3² + 4² + 5²) × √( 2² + 6² + 8² ) ]

 ⇒ cos Ф = 70 / [ √50 × √104 ]

 ⇒ cos Ф = 7 / 2√13

Hence, the cosine of the angle between r1→ and r2→ is  7 / 2√13.

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