The direction cosines of the normal to the pl | vedclass

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(C) The equation of the plane is given by $3x + 4y + 12z = 52$.Comparing this with the general form $Ax + By + Cz = D$,the direction ratios of the nor

Vedclass · Accepted Answer

$\frac{3}{13}, \frac{4}{13}, \frac{12}{13}$

Vedclass · Answer

(C) The equation of the plane is given by $3x + 4y + 12z = 52$.Comparing this with the general form $Ax + By + Cz = D$,the direction ratios of the normal to the plane are $(A, B, C) = (3, 4, 12)$.To find the direction cosines $(l, m, n)$,we divide the direction ratios by the magnitude of the normal vector $\sqrt{A^2 + B^2 + C^2}$.Magnitude $= \sqrt{3^2 + 4^2 + 12^2} = \sqrt{9 + 16 + 144} = \sqrt{169} = 13$.Thus,the direction cosines are $\left( \frac{3}{13}, \frac{4}{13}, \frac{12}{13} \right)$.

The direction cosines of the normal to the plane $3x + 4y + 12z = 52$ will be

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