Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

$\begin{array}{rl}& \text{ Let }A=(5,-1,4),B=(4,-1,3)\\ & \overline{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{k}}\Rightarrow |\overline{\mathrm{AB}}|=\sqrt{2}\end{array}$
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 $1,1,1$. $$\begin{array}{rl}& \cos \left(90^{\circ }-\theta \right)=\left|\frac{1(-1)+1(0)+1(-1)}{\sqrt{1^2+1^2+1^2}\sqrt{1^2+0^2+1^2}}\right|\\ & \begin{array}{r}\sin \theta =\frac{2}{\sqrt{6}}\Rightarrow \cos \theta =\sqrt{1-\frac{4}{6}}=\sqrt{\frac{1}{3}}\\ \text{ Required projection }=|\overline{\mathrm{AB}}|\cos \theta \\ =\sqrt{2}\times \frac{1}{\sqrt{3}}=\sqrt{\frac{2}{3}}\end{array}\end{array}$$