The length of the projection of the line segment joining the points \((5,-1,4)\) and \((4,-1,3)\) on the plane \(x+y+z=7\) is

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

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