Mathematics - 3D-Coordinate Geometry Question with Solution | TestHub

The direction ratios of the line are $1,2,\lambda$.

The direction ratios of the normal to the plane $x+2y+3z=4$ are $1,2,3$.

Let $\theta$ be the angle between the line and the plane. We are given $\cos \theta =\sqrt{\frac{5}{14}}$.

Using the identity ${\sin }^2\theta +{\cos }^2\theta =1$:

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

The angle $\theta$ between a line with direction ratios $a_1,b_1,c_1$ and a plane with normal direction ratios $a_2,b_2,c_2$ is given by:

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

Substituting the values:

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

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

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

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

Squaring both sides:

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

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

$45=25+30\lambda$

$30\lambda =20$

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

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