Question

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:

• A. \(\left(-\dfrac{1}{3},\dfrac{2}{3},-\dfrac{2}{3}\right)\)
• B. \(\left(\dfrac{2}{3\sqrt{5}},-\dfrac{4}{3\sqrt{5}},\dfrac{5}{3\sqrt{5}}\right)\)
• C. \(\left(\dfrac{-4}{\sqrt{57}},\dfrac{4}{\sqrt{57}},\dfrac{-5}{\sqrt{57}}\right)\)
• D. \(\left(\dfrac{2}{\sqrt{38}},-\dfrac{3}{\sqrt{38}},\dfrac{5}{\sqrt{38}}\right)\)

Hint:

If a plane is given in intercept form \[ \frac{x}{a}+\frac{y}{b}+\frac{z}{c}=1, \] first convert it into standard form \(Ax+By+Cz+D=0\). Then \((A,B,C)\) gives a normal vector to the plane.

Answer 1

The Correct Option is D

Step 1: Write the intercept form of the plane.
A plane with intercepts $a, b, c$ on the axes is $\dfrac{x}{a} + \dfrac{y}{b} + \dfrac{z}{c} = 1$. Here $a = -2$, $b = 4/3$, $c = -4/5$.

Step 2: Substitute the intercepts.
\[ \frac{x}{-2} + \frac{y}{4/3} + \frac{z}{-4/5} = 1 \Rightarrow -\frac{x}{2} + \frac{3y}{4} - \frac{5z}{4} = 1. \]

Step 3: Clear fractions by multiplying by 4.
\[ -2x + 3y - 5z = 4. \] The normal vector is $\vec{n} = (-2, 3, -5)$ or equivalently $(2, -3, 5)$.

Step 4: Compute the magnitude of the normal vector.
\[ |\vec{n}| = \sqrt{4 + 9 + 25} = \sqrt{38}. \]

Step 5: Find the direction cosines.
\[ \ell = \frac{2}{\sqrt{38}},\quad m = \frac{-3}{\sqrt{38}},\quad n = \frac{5}{\sqrt{38}}. \]

Step 6: State the answer.
The direction cosines of the normal are $\left(\dfrac{2}{\sqrt{38}}, \dfrac{-3}{\sqrt{38}}, \dfrac{5}{\sqrt{38}}\right)$.
\[ \boxed{\left(\dfrac{2}{\sqrt{38}},\ -\dfrac{3}{\sqrt{38}},\ \dfrac{5}{\sqrt{38}}\right)} \]