The direction cosines of the normal to the pl | vedclass

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(A) The equation of the plane is $x + 2y - 3z + 4 = 0$. Comparing this with the general form $Ax + By + Cz + D = 0$,we get the direction ratios of the

Vedclass · Accepted Answer

$-\frac{1}{\sqrt{14}}, -\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$

Vedclass · Answer

(A) The equation of the plane is $x + 2y - 3z + 4 = 0$. Comparing this with the general form $Ax + By + Cz + D = 0$,we get the direction ratios of the normal as $(A, B, C) = (1, 2, -3)$. The magnitude of the normal vector is $\sqrt{A^2 + B^2 + C^2} = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$. The direction cosines $(l, m, n)$ are given by $\pm \frac{A}{\sqrt{A^2 + B^2 + C^2}}, \pm \frac{B}{\sqrt{A^2 + B^2 + C^2}}, \pm \frac{C}{\sqrt{A^2 + B^2 + C^2}}$. Taking the negative sign,we get $l = -\frac{1}{\sqrt{14}}$,$m = -\frac{2}{\sqrt{14}}$,and $n = -\frac{-3}{\sqrt{14}} = \frac{3}{\sqrt{14}}$. Thus,the direction cosines are $(-\frac{1}{\sqrt{14}}, -\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}})$,which matches option $A$.

Point	Plane
$(-6, 0, 0)$	$2x - 3y + 6z - 2 = 0$

The direction cosines of the normal to the plane $x + 2y - 3z + 4 = 0$ are

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