Question

The vector parallel to 4î - 3j+5k and whose
length is the arithmetic mean of lengths of two
vectors 2î - 4j+ 4k and i + Joj +3k is
1) 4i-37 +5h 2) (4 – 3ſ + 5k)/13
3) (4ì –39 +5£)/V2 4) (4ỉ –3j+5k)/15​

Answer 1

Given:

The vector parallel to 4i - 3j + 5k and lengths is the arithmetic mean of lengths of two vectors 2i - 4j + 4k and i+ 6j + 3k.

To find:

The parallel vector.

Solution:

From given, we have,

The vector parallel to 4i - 3j + 5k

⇒ The parallel vector is given by,

= (4i - 3j + 5k) × 1/√(4² + 3² + 5²)

= (4i - 3j + 5k)/√50

The lengths is the arithmetic mean of lengths of two vectors 2i - 4j + 4k and i+ 6j + 3k.

The magnitude of 2i - 4j + 4k is

= √(2² + 4² + 4²)

= 6

The magnitude of i +6j + 3k is

= √(1² + 6² + 3²)

= √46

⇒ The arithmetic mean of lengths

(6 + √46)/2

So, the required vector is,

= (4i - 3j + 5k)/√50 × (6 + √46)/2

= (3√2 + √23)/10 (4i - 3j + 5k)