If -2, frac{4}{3}, frac{-4}{5} are the interc | vedclass

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(D) The equation of a plane with intercepts $a, b, c$ on the $X, Y, Z$-axes is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Given intercept

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$\left(\frac{2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}, \frac{5}{\sqrt{38}}\right)$

Vedclass · Answer

(D) The equation of a plane with intercepts $a, b, c$ on the $X, Y, Z$-axes is given by $\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1$. Given intercepts are $a = -2, b = \frac{4}{3}, c = \frac{-4}{5}$. Substituting these values,the equation of the plane is $\frac{x}{-2} + \frac{y}{4/3} + \frac{z}{-4/5} = 1$. This simplifies to $-\frac{x}{2} + \frac{3y}{4} - \frac{5z}{4} = 1$. Multiplying by $4$,we get $-2x + 3y - 5z = 4$,or $2x - 3y + 5z + 4 = 0$. The direction ratios of the normal to the plane are $(2, -3, 5)$. The direction cosines $(l, m, n)$ are given by $\frac{a}{\sqrt{a^2+b^2+c^2}}, \frac{b}{\sqrt{a^2+b^2+c^2}}, \frac{c}{\sqrt{a^2+b^2+c^2}}$. Here,$\sqrt{2^2 + (-3)^2 + 5^2} = \sqrt{4 + 9 + 25} = \sqrt{38}$. Thus,the direction cosines are $\left(\frac{2}{\sqrt{38}}, \frac{-3}{\sqrt{38}}, \frac{5}{\sqrt{38}}\right)$.

If $-2, \frac{4}{3}, \frac{-4}{5}$ are the intercepts made by a plane on $X, Y, Z$-axes respectively,then the direction cosines of a normal to this plane are

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