What are the direction cosines of the normal | vedclass

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Question

(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 normal vector $\vec{n}

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 normal vector $\vec{n} = (1, 2, -3)$. The magnitude of the normal vector is $|\vec{n}| = \sqrt{1^2 + 2^2 + (-3)^2} = \sqrt{1 + 4 + 9} = \sqrt{14}$. The direction cosines $(l, m, n)$ are given by $\frac{A}{|\vec{n}|}, \frac{B}{|\vec{n}|}, \frac{C}{|\vec{n}|}$. Thus,the direction cosines are $\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{-3}{\sqrt{14}}$. Alternatively,multiplying the equation by $-1$,we get $-x - 2y + 3z - 4 = 0$,which gives the direction cosines as $-\frac{1}{\sqrt{14}}, -\frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}}$. Comparing this with the given options,option $A$ is correct.

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

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