If the plane 2x – 3y – 6z = 13 makes an angle with the x-axis, then find the value of λ.
If the plane 2x – 3y – 6z = 13 makes an angle with the x-axis, then find the value of λ.

Direction ratios of the x-axis are 1,0,0 and direction ratios of normal to the plane are 2,−3,−6.

The angle between the line and the plane,

$\mathrm{sin}\varphi =\frac{|}{\stackrel{\to }{a}·\stackrel{\to }{n}\mid }\stackrel{\to }{a}||\stackrel{\to }{n}\mid$

\sin \phi=\frac|{\vec{a} \cdot \vec{n} \mid}{\vec{a}|| \vec{n} \mid}

$\mathrm{sin}\varphi =\frac{|}{{a}_{1},{a}_{2}+{b}_{1}·{b}_{2}+{c}_{1}·{c}_{2}\mid }\sqrt{{a}_{1}^{2}+{b}_{1}^{2}–{c}_{1}^{2}·\sqrt{{a}_{2}^{2}–{b}_{2}^{2}+{c}_{2}^{2}}}$

\sin \phi=\frac|{a_{1}, a_{2}+b_{1} \cdot b_{2}+c_{1} \cdot c_{2} \mid}{\sqrt{a_{1}^{2}+b_{1}^{2}-c_{1}^{2} \cdot \sqrt{a_{2}^{2}-b_{2}^{2}+c_{2}^{2}}}}

Let be the angle between the -axis and the given plane.

$\begin{array}{l}\mathrm{sin}\varphi =\frac{|1×2+0×\left(–3\right)+0×\left(6\right)|}{\left\{\sqrt{{1}^{2}+{0}^{2}+{0}^{2}}}\left\{\sqrt{{2}^{2}+\left(–3{\right)}^{2}+\left(–6{\right)}^{2}}\right\}\right\}=\frac{2}{7}\\ ⇒\varphi ={\mathrm{sin}}^{1}\left(\frac{2}{7}\right)\end{array}$

\begin{array}{l}

\sin \phi=\frac{|1 \times 2+0 \times(-3)+0 \times(6)|}{\left\{\sqrt{\left.1^{2}+0^{2}+0^{2}\right\}}\left\{\sqrt{2^{2}+(-3)^{2}+(-6)^{2}}\right\}\right.}=\frac{2}{7} \\

\Rightarrow \phi=\sin ^{1}\left(\frac{2}{7}\right)

\end{array}

Hence, .