# 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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