The sum of two unit vectors is a unit vector. The magnitude of their difference is

Question

The sum of two unit vectors is a unit vector. The magnitude of their difference is (A) 2 (B) √3 (C) √2 (D) 1

Tardigrade Admin · Accepted Answer

Let vecc= veca+ vecb where veca, vecb, vecc are all unit vectors ∴ vecc2= veca2+ vecb2+2 veca⋅ vecb or 1=1+1+2 veca . vecb ⇒ 2 veca . vecb=-1 Again | veca- vecb|2= veca2+ vecb2-2 veca . vecb =1+1-(-1)=3 ∴ =| veca- vecb|=√3

Q. The sum of two unit vectors is a unit vector. The magnitude of their difference is

2

$3$

$2$

1

Let $c = a + b$ where $a, b, c$ are all unit vectors
$∴ c^{2} = a^{2} + b^{2} + 2 a \cdot b$
or $1 = 1 + 1 + 2 a . b$
$\Rightarrow 2 a . b = - 1$
Again $∣ ∣ a - b ∣ ∣^{2} = a^{2} + b^{2} - 2 a . b$
$= 1 + 1 - (- 1) = 3$
$∴ = ∣ ∣ a - b ∣ ∣ = 3$

Q. The sum of two unit vectors is a unit vector. The magnitude of their difference is

2

3​

2​

1

Let c=a+b where a,b,c are all unit vectors ∴c2=a2+b2+2a⋅b or 1=1+1+2a.b ⇒2a.b=−1 Again ∣∣​a−b∣∣​2=a2+b2−2a.b =1+1−(−1)=3 ∴=∣∣​a−b∣∣​=3​

$3$

$2$

Let $c = a + b$ where $a, b, c$ are all unit vectors
$∴ c^{2} = a^{2} + b^{2} + 2 a \cdot b$
or $1 = 1 + 1 + 2 a . b$
$\Rightarrow 2 a . b = - 1$
Again $∣ ∣ a - b ∣ ∣^{2} = a^{2} + b^{2} - 2 a . b$
$= 1 + 1 - (- 1) = 3$
$∴ = ∣ ∣ a - b ∣ ∣ = 3$