Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

The direction ratios of the given line are $a_1=1$, $b_1=2$, $c_1=\lambda$.

The direction ratios of the normal to the given plane $x+2y+3z=6$ are $a_2=1$, $b_2=2$, $c_2=3$.

The angle $\theta$ between a line and a plane is given by:

$\sin \theta =\frac{|a_1{a}_2+b_1{b}_2+c_1{c}_2|}{\sqrt{a_1^2+b_1^2+c_1^2}\sqrt{a_2^2+b_2^2+c_2^2}}$

Given $\theta ={\cos }^{-1}\sqrt{\frac{5}{14}}$, we have $\cos \theta =\sqrt{\frac{5}{14}}$.

$\sin \theta =\sqrt{1-{\cos }^2\theta }=\sqrt{1-\frac{5}{14}}=\sqrt{\frac{9}{14}}=\frac{3}{\sqrt{14}}$

Substituting the values into the formula:

$\frac{3}{\sqrt{14}}=\frac{|(1)(1)+(2)(2)+(\lambda )(3)|}{\sqrt{1^2+2^2+{\lambda }^2}\sqrt{1^2+2^2+3^2}}$

$\frac{3}{\sqrt{14}}=\frac{|1+4+3\lambda |}{\sqrt{5+{\lambda }^2}\sqrt{14}}$

$\frac{3}{\sqrt{14}}=\frac{|5+3\lambda |}{\sqrt{14}\sqrt{5+{\lambda }^2}}$

Multiplying both sides by $\sqrt{14}$:

$3=\frac{|5+3\lambda |}{\sqrt{5+{\lambda }^2}}$

Squaring both sides:

$9=\frac{(5+3\lambda {)}^2}{5+{\lambda }^2}$

$9(5+{\lambda }^2)=25+30\lambda +9{\lambda }^2$

$45+9{\lambda }^2=25+30\lambda +9{\lambda }^2$

$45=25+30\lambda$

$30\lambda =20$

$\lambda =\frac{2}{3}$

Answer: $\frac{2}{3}$