⇒Line and Plane

Correct answer option is (D)

$\begin{array}{rl}& \text{Let }A=(5,-1,4),B=(4,-1,3)\\ \\ & \overline{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{k}}\Rightarrow \mid \overline{\mathrm{AB}}=\sqrt{2}\end{array}$

Projection of $\overline{\mathrm{AB}}$ in the plane $x+y+z=7$ is $|\overline{\mathrm{AB}}|\cos \theta =\left|\overline{{\mathrm{A}}^{\prime }{\mathrm{B}}^{\prime }}\right|\cos \theta$

Direction ratios of normal to the given plane is $1,1,1$.

$\begin{array}{rl}& \cos ({90}^{\circ }-\theta )=\left|\frac{1(-1)+1(0)+1(-1)}{\sqrt{1^2+1^2+1^2}\sqrt{1^2+0^2+1^2}}\right|\\ \\ & \Rightarrow \sin \theta =\frac{2}{\sqrt{6}}\Rightarrow \cos \theta =\sqrt{1-\frac{4}{6}}=\sqrt{\frac{1}{3}}\\ \\ & \text{ Required projection }=|\overline{\mathrm{AB}}|\cos \theta \\ \\ & =\sqrt{2}\times \frac{1}{\sqrt{3}}=\sqrt{\frac{2}{3}}\end{array}$

Correct answer option is (C)

The acute angle $\theta$ between line $\overline{a}+\lambda \overline{b}$ and the plane $\bar{\mathrm{r}}\cdot \bar{\mathrm{n}}=\mathrm{p}$ is given by

$\sin \theta =\left|\frac{\bar{\mathrm{b}}\cdot \bar{\mathrm{n}}}{|\bar{\mathrm{b}}|\cdot |\bar{\mathrm{n}}|}\right|$ ..... (i)

Here, $\overline{b}=2\hat{i}+\hat{j}-2\hat{k}$ and $\overline{n}=\hat{i}-2\hat{j}-\lambda \hat{k}$

Also, $\theta ={\cos }^{-1}\left(\frac{2\sqrt{2}}{3}\right)={\sin }^{-1}\left(\frac{1}{3}\right)$

$\begin{array}{rl}& \therefore \text{ (i) }\Rightarrow \sin ({\sin }^{-1}\left(\frac{1}{3}\right))=\left|\frac{2-2+2\lambda }{\sqrt{4+1+4}\sqrt{1+4+{\lambda }^2}}\right|\\ & \Rightarrow \frac{1}{3}=\left|\frac{2\lambda }{3\sqrt{5+{\lambda }^2}}\right|\\ & \Rightarrow 5+{\lambda }^2=4{\lambda }^2\\ & \Rightarrow {\lambda }^2=\frac{5}{3}\\ & \Rightarrow \lambda =\sqrt{\frac{5}{3}}\end{array}$