The unit vector perpendicular to the plane 4x | vedclass

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(B) The equation of the plane is given by $4x - 3y + 12z = 15$. Comparing this with the general form $ax + by + cz = d$,we get the normal vector $\vec

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$\frac{4\hat{i} - 3\hat{j} + 12\hat{k}}{13}$

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(B) The equation of the plane is given by $4x - 3y + 12z = 15$. Comparing this with the general form $ax + by + cz = d$,we get the normal vector $\vec{n} = 4\hat{i} - 3\hat{j} + 12\hat{k}$. The magnitude of the normal vector is $|\vec{n}| = \sqrt{4^2 + (-3)^2 + 12^2} = \sqrt{16 + 9 + 144} = \sqrt{169} = 13$. The unit vector perpendicular to the plane is given by $\hat{n} = \frac{\vec{n}}{|\vec{n}|} = \frac{4\hat{i} - 3\hat{j} + 12\hat{k}}{13}$.

The unit vector perpendicular to the plane $4x - 3y + 12z = 15$ is

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