Question

find a vector perpendicular to i+2j and having a magnitude of 3 root 5.

Answer 1

Let a vector , r = x i + y j is perpendicular to A = (i + 2j)
So, dot product of r and A must be zero. Because cos90° = 0
e.g., r.A = (x i + y j).(i + 2j) = x + 2y = 0 -----(1)

A/C to question,
Magnitude of r is 3√5
So, |r| = √(x² + y²) = 3√5
Taking square both sides,
x² + y² = 9 ×5 = 45
x² + y² = 45
From equation (1),
(-2y)² + y² = 45
⇒4y² + y² = 45
⇒ 5y² = 45
⇒y² = 9⇒y = ± 3
∴ x = -2y = $\mp$6

Hence, r = ±(6i - 3j)

Answer 2

Answer:

Let a vector , r = x i + y j is perpendicular to A = (i + 2j)

So, dot product of r and A must be zero. Because cos90° = 0

e.g., r.A = (x i + y j).(i + 2j) = x + 2y = 0 -----(1)

A/C to question,

Magnitude of r is 3√5

So, |r| = √(x² + y²) = 3√5

Taking square both sides,

x² + y² = 9 ×5 = 45

x² + y² = 45

From equation (1),

(-2y)² + y² = 45

⇒4y² + y² = 45

⇒ 5y² = 45

⇒y² = 9⇒y = ± 3

∴ x = -2y = 6

Hence, r = ±(6i - 3j)