I f → a = ( 3 , − 2 , 1 ) , → b = ( − 1 , 1 , 1 ) then the unit vector parallel to the vector → a + → b is Options: A . &nbsp ( 2 3 , − 1 3 , 2 3 ) B . &nbsp ( 2 5 , − 1 5 , 2 5 ) C . &nbsp ( 2 √ 3 , − 1 √ 3 , 2 √ 3 ) D . &nbsp ( − 2 √ 3 , 1 √ 3 , − 2 √ 3 )

Answer: Option A : A → a + → b = ( 3 , − 2 , 1 ) + ( − 1 , 1 , 1 ) = ( 2 , − 1 , 2 ) ⇒ ∣ ∣ ∣ → a + → b ∣ ∣ ∣ = √ 4 + 1 + 4 = √ 9 = 3 Unit vector parallel to → a + → b is ± a + b ∣ ∣ ∣ → a + → b ∣ ∣ ∣ = ± ( 2 , − 1 , 2 ) 3 = ( 2 3 , − 1 3 , 2 3 )

If →a=(3,−2,1), →b=(−1,1,1) then the uni

Discussion Forum : Vector Algebra

Question -

$I f \to a = (3, - 2, 1), \to b = (- 1, 1, 1)$ then the unit vector parallel to the vector $\to a + \to b$ is

Options:

A .

(\frac{2}{3}, - \frac{1}{3}, \frac{2}{3})

B .

(\frac{2}{5}, - \frac{1}{5}, \frac{2}{5})

C .

(\frac{2}{\sqrt{3}}, - \frac{1}{\sqrt{3}}, \frac{2}{\sqrt{3}})

D .

(- \frac{2}{\sqrt{3}}, \frac{1}{\sqrt{3}}, - \frac{2}{\sqrt{3}})

Answer: Option A
:
A

\to a + \to b = (3, - 2, 1) + (- 1, 1, 1) = (2, - 1, 2)

\Rightarrow ∣ ∣ ∣ \to a + \to b ∣ ∣ ∣ = \sqrt{4 + 1 + 4} = \sqrt{9} = 3

Unit vector parallel to

\to a + \to b

\pm \frac{a + b}{∣ ∣ ∣ \to a + \to b ∣ ∣ ∣} = \pm \frac{(2, - 1, 2)}{3} = (\frac{2}{3}, - \frac{1}{3}, \frac{2}{3})

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Discussion Forum : Vector Algebra

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