Question

Find angle at which the normal vector to the plane 4x+8y+z=5 is inclined to coordinate axes

Answer 1

Angle with X axis is arc cos 8÷9
Angle with Y axis is arc cos 4÷9
Angle with Z axis is arc cos 1÷9

Answer 2

So, All three angle are $cos^{-1} (\frac{4}{9})$, $cos^{-1} (\frac{8}{9})$ and $cos^{-1} (\frac{1}{9})$

Step-by-step explanation:

Given,

The normal vector to the plane $4x+8y+z=5$  is inclined to coordinate axes.

Normal vector to the plane $ax+by+cz=d$ is given by:

$\overrightarrow n=a \widehat{i}+b\widehat{j}+c\widehat{k}$

For the vector plane $4x+8y+z=5$,

$\overrightarrow n=4 \widehat{i}+8\widehat{j}+\widehat{k}$

⇒$\left | \overrightarrow n \right |=\sqrt{4^{2} +8^{2} +1^{2} }$

⇒$\left | \overrightarrow n \right |=\sqrt{16+64+1 }=9$

For x-axis,

$\overrightarrow x=\widehat{i}$

⇒$\left | \overrightarrow x \right |=1$

For y-axis,

$\overrightarrow y=\widehat{j}$

⇒$\left | \overrightarrow y \right |=1$

For z-axis,

$\overrightarrow z=\widehat{k}$

⇒$\left | \overrightarrow z \right |=1$

Angle $\alpha$ made by $\overrightarrow n$ with x−axis is given by:

$cos\alpha =\frac{(4 \widehat{i}+8\widehat{j}+\widehat{k})(\widehat{i})}{9}=\frac{4}{9}$

∴$\alpha =cos^{-1} (\frac{4}{9})$

Angle $\beta$ made by $\overrightarrow n$ with y−axis is given by:

$cos\alpha =\frac{(4 \widehat{i}+8\widehat{j}+\widehat{k})(\widehat{j})}{9}=\frac{8}{9}$

∴$\beta =cos^{-1} (\frac{8}{9})$

And angle $\gamma$ made by $\overrightarrow n$ with z−axis is given by:

$cos\gamma =\frac{(4 \widehat{i}+8\widehat{j}+\widehat{k})(\widehat{k})}{9}=\frac{1}{9}$

∴$\gamma =cos^{-1} (\frac{1}{9})$

∴ All three angle are $cos^{-1} (\frac{4}{9})$, $cos^{-1} (\frac{8}{9})$ and $cos^{-1} (\frac{1}{9})$