Question

The angle between the planes $3x + 4y + 5z = 3$ and $4 x-3 y + 5z = 9$ is equal to

• A. $\frac{\pi}{2}$
• B. $\frac{\pi}{4}$
• C. $\frac{\pi}{6}$
• D. $\frac{\pi}{3}$

Answer 1

Answer: (D)

$\frac{\pi}{3}$

Answer 2

Given planes are $3 x+4 y+5 z=3$ and
$4 x-3 y+5 z=9$. Since, DR's of normal to the planes are
$\left(a_{1}, b_{1}, c_{1}\right)=(3,4,5)$
and $\left(a_{2}, b_{2}, c_{2}\right)=(4,-3,5)$
$\therefore$ Angle between the planes $=$ Angle between their normal
$\therefore \cos \theta =\frac{a_{1} a_{2}+b_{1} b_{2}+c_{1} c_{2}}{\sqrt{a_{1}^{2}+b_{1}^{2}+c_{1}^{2}} \sqrt{a_{2}^{2}+b_{2}^{2}+c_{2}^{2}}}$
$=\frac{3 \times 4+4 \times(-3)+5 \times 5}{\sqrt{3^{2}+4^{2}+5^{2}} \sqrt{4^{2}+(-3)^{2}+5^{2}}}$
$=\frac{12-12+25}{\sqrt{9+16+25} \sqrt{16+9+25}}=\frac{25}{50}$
$\Rightarrow \cos \theta=\frac{1}{2} \Rightarrow \theta=\frac{\pi}{3}$