A Vector Perpendicular To The Vector ( I^+2 J^ )And Having Magnitude 3 Sqrt 5 Units Is _____.

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

Therefore, dot product of r and A must be zero. This is because cos 90°= 0

= (xi + yj) * (i + 2j) = x + 2y = 0 —[1]

As given in the question:

The magnitude of r is \(3\sqrt{5} \) \(\left | r \right | = \sqrt{(x^{2} + y^{2})} = 3\sqrt{5} \)

Now, by taking square on both sides, we get:

\(x^{2} + y^{2} = 9 * 5 = 45 \) \(x^{2} + y^{2} = 45 \)

From equation [1] :

\((-2y)^{2} + y^{2} = 45 \) \(\Rightarrow 4y^{2} + y^{2} = 45 \) \(\Rightarrow 5y^{2} = 45 \) \(\Rightarrow y^{2} = 9 \) \(\Rightarrow y = 3 \)

Therefore, x = -2y = 6

\(\Rightarrow r = (6i – 3j) \)

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