Mathematics - 3D Coordinate Geometry Question with Solution | TestHub

Let the given points be $A(5,-1,4)$ and $B(4,-1,3)$.

The vector joining $A$ and $B$ is $\overrightarrow{AB}=(4-5)\hat{i}+(-1-(-1))\hat{j}+(3-4)\hat{k}=-\hat{i}-\hat{k}$.

The length of the line segment $AB$ is $|\overrightarrow{AB}|=\sqrt{(-1{)}^2+0^2+(-1{)}^2}=\sqrt{2}$.

The normal vector to the given plane $x+y+z=7$ is $\vec{n}=\hat{i}+\hat{j}+\hat{k}$.

The magnitude of the normal vector is $|\vec{n}|=\sqrt{1^2+1^2+1^2}=\sqrt{3}$.

The length of the projection of $\overrightarrow{AB}$ along the normal $\vec{n}$ is given by $d=\frac{|\overrightarrow{AB}\cdot \vec{n}|}{|\vec{n}|}$.

$d=\frac{|(-1)(1)+(0)(1)+(-1)(1)|}{\sqrt{3}}=\frac{2}{\sqrt{3}}$.

The length of the projection of the line segment on the plane is $l=\sqrt{|\overrightarrow{AB}{|}^2-d^2}$.

$l=\sqrt{(\sqrt{2}{)}^2-{\left(\frac{2}{\sqrt{3}}\right)}^2}=\sqrt{2-\frac{4}{3}}=\sqrt{(}\frac{2}{3})$.

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