Given two displacement vectors A⃗ =(3.00iˆ−4.00jˆ+4.00kˆ)mA→=(3.00i^− 4.00j^+4.00k^)m and B⃗ =(2.00iˆ+3.00jˆ−7.00kˆ)mB→=(2.00i^+ 3.00j^−7.00k^)m, find the displacements and their magnitudes for (a) C⃗ =A⃗ +B⃗ C→=A→+B→ and (b) D⃗ =2A⃗ −B⃗ D→=2A→−B→.

Displacement C = (5.00iˆ - 1.00jˆ - 3.00kˆ) m

Magnitude = 5.92 m

Displacement D = (4.00iˆ−11.00jˆ+15.00kˆ) m

Magnitude = 19.03 m

Explanation:

Vector A = (3.00iˆ−4.00jˆ+4.00kˆ) m

Vector B = (2.00iˆ+3.00jˆ−7.00kˆ) m

a) Vector C = A + B = (3.00iˆ−4.00jˆ+4.00kˆ) + (2.00iˆ+3.00jˆ−7.00kˆ)

Vector addition is done component by component, that is, do î component, then j component and k component

C = (5.00iˆ - 1.00jˆ - 3.00kˆ) m

Magnitude of C = √[(5²) + (-1)² + (-3)²] = √(35) = 5.92 m

b) Vector D = 2A - B

D = 2(3.00iˆ−4.00jˆ+4.00kˆ) - (2.00iˆ+3.00jˆ−7.00kˆ) = (6.00iˆ−8.00jˆ+8.00kˆ) - (2.00iˆ+3.00jˆ−7.00kˆ) = (4.00iˆ−11.00jˆ+15.00kˆ)

Magnitude of D = √[(4²) + (-11)² + (15)²] = √(362) = 19.03 m

found in the nucleus

has mass of one amu

the answer is a 3.8cm

explanation: