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# How do you find the unit vector perpendicular to the vector (4i - 3j)?

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10
Jul 10, 2016

Therefore, the reqd vectors are $\left(3 \frac{i}{5} + 4 \frac{j}{5}\right) , \mathmr{and} , \left(- \frac{3}{5} i - \frac{4}{5} j\right)$

#### Explanation:

Let us name the given vctr. $\left(4 i - 3 j\right) = \left(4 , - 3\right) = \vec{u} ,$ and, suppose that, $\vec{v} = \left(x , y\right)$ is the reqd. unit vector perp. to $\vec{u}$.

Since, $\vec{v}$ is unit vctr., $| | \vec{v} | | = 1 \Rightarrow \sqrt{{x}^{2} + {y}^{2}} = 1$

$\therefore {x}^{2} + {y}^{2} = 1. \ldots \ldots \ldots \ldots . . \left(1\right)$

Next, $\vec{v}$ is perp. to $\vec{u}$, $\vec{v} . \vec{u} = 0$

$\therefore \left(x , y\right) \cdot \left(4 , - 3\right) = 0 \Rightarrow 4 x - 3 y = 0.$ or, $x = \frac{3}{4} y \ldots \ldots \ldots \ldots \left(2\right)$

We solve (1) & (2) : (3/4y)^2+y^2=1 rArr y=+-4/5, giving, $x = \pm \frac{3}{5}$

Therefore, the reqd vectors are $\left(3 \frac{i}{5} + 4 \frac{j}{5}\right) , \mathmr{and} , \left(- 3 \frac{i}{5} - 4 \frac{j}{5}\right)$

Enjoy Maths.!

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5
Jul 10, 2016

$\pm \frac{1}{5} \left(3 i + 4 j\right)$

#### Explanation:

Any unit vector is of the form (cos a, sin a ) = i cos a + j sin a..

This has to be perpendicular to the given vector.

So, the dot product of the two, (4, -3).( cos a, sin a)

$4 \cos a - 3 \sin a = 0$..

And so,, tan a = 4/3 > 0.

Both sin and cos have the same sign.

Thus, $\left(\cos a , \sin a\right) = \pm \left(\frac{3}{5} , \frac{4}{5}\right)$.

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