Question

Plz solve this question

Answer 1

Answer:

Explanation:

F = -2√3 i - 4j - 6k

Magnitude = √((-2√3)² + (-4)² + (-6)²)  = √(12 + 16 + 36) = √64 = 8

m = 2 kg

a = F/m  => a = 8/2 = 4

Magnitude of acceleration = 4

A = i - 2j + 3k

B = 3i + 5j + k

A + B =  4i + 3j + 4k

A - B = -2i - 7j + 2k

B - A = 2i + 7j - 2k

P = 7i - 2j + 3k

q = 3i  + j + 5k

p - 2q =  i - 4j - 7k

Answer 2

4. force , F = -2√3 i - 4j - 6k is applied on a 2kg mass.

first find magnitude of force, |F| = √{(-2√3)² + (-4)² + (-6)²}

= √{12 + 16 + 36}

= √64 = 8N

now using formula, F = ma,

where F = 8N and m = 2kg

so, a = 8/2 = 4m/s²

hence, magnitude of acceleration = 4m/s²

5.(A) A + B = (i - 2j + 3k) + (3i + 5j + k)

= (1 + 3)i + (-2 + 5)j + (3 + 1)k

= 4i + 3j + 4k

A - B = (i - 2j + 3k) - (3i + 5j + k)

= (1 - 3)i + (-2 - 5)j + (3 - 1)k

= -2i - 7j + 2k

B - A = (3i + 5j + k) - (i - 2j + 3k)

= (3 - 1)i + (5 + 2)j + (1 - 3)k

= 2i + 7j - 2k

(B) P - 2q = (7i - 2j + 3k) - 2(3i + j + 5k)

= (7i - 2j + 3k) - (6i + 2j + 10k)

= (7 - 6)i + (-2 - 2)j + (3 - 10)k

= i - 4j - 7k