Question

Find the equation of a plane which is at a distance of $3$ units from the origin and which is normal to the vector $3\hat{i}+4\hat{j}+12\hat{k}$. Also find the angle between the line $\frac{x}{1}=\frac{y}{-4}=\frac{z}{-8}$

Answer 1

Given plane normal to vector

$\overrightarrow{n}=3\hat{i}+4\hat{j}+12\hat{k}$

Let eq of plane

$3x+4y+12z+d=0$

distance of plane from origin is 3

$distance=\left|\frac{a{x}_1+b{y}_1+c{z}_1+d}{\sqrt{a^2+b^2+c^2}}\right|$

$3=\frac{3\left(0\right)+4\left(0\right)+12\left(0\right)+d}{\sqrt{3^2+4^2+{12}^2}}$

$3=\frac{0+d}{\sqrt{9+16+144}}$

$3=\frac{d}{\sqrt{169}}$

$3=\frac{d}{13}$

$39=d$

so eq of plane

$3x+4y+12z+39=0$

angle beween line

$\frac{x}{1}=\frac{y}{-4}=\frac{z}{-8}$

and given plane

$\sin \theta =\frac{a{n}_1+b{n}_2+c{n}_3}{\sqrt{a^2+b^2+c^2}\sqrt{n_1^2+n_2^2+n_3^2}}$

$\sin \theta =\frac{3\left(1\right)+4(-4)+12(-8)}{\sqrt{3^2+4^2+{12}^2}\sqrt{1^2+(-4{)}^2+(-8{)}^2}}$

$\sin \theta =\frac{3-16-96}{\sqrt{169}\sqrt{1+16+64}}$

$\sin \theta =\frac{-109}{13\sqrt{81}}$

$\sin \theta =\frac{-109}{117}$

$\theta ={\sin }^{-1}\left(\frac{-109}{117}\right)$