Question

The length of the projection of the line segment joining points $(5,-1,4)$ and $(4,-1,8)$ on the plane $x+y+z=7$.

• A. $\frac{2}{\sqrt{3}}$
• B. $\frac{2}{3}$
• C. $\sqrt{5}$
• D. $\sqrt{\frac{2}{3}}$

Answer 1

The correct option is D $\sqrt{\frac{2}{3}}$

Let $A(5,-1,4)$ and $B(4,-1,3)$ be a line

then $\overrightarrow{AB}=(4-5)\hat{i}+(-1+1)\hat{j}+(3-4)\hat{k}$

$=-\hat{i}-\hat{j}$

$\left|AB\right|=\sqrt{1^2+1^2}=\sqrt{2}$ units

Given:Equation of the plane is $x+y+z=7$

Normal vector$\overrightarrow{n}=\hat{i}+\hat{j}+\hat{k}$

Equation of plane is $\overrightarrow{r}.\hat{n}=d$

$(x\hat{i}+y\hat{j}+z\hat{k}).(\hat{i}+\hat{j}+\hat{k})=7$

$\left|n\right|=\sqrt{1+1+1}=\sqrt{3}$

Length of Projection $AB$ along the plane is $P=\left|AB\right|\cos (90-\theta )=\left|AB\right|\sin \theta$

Here $\cos \theta =\frac{\overrightarrow{AB}.\overrightarrow{n}}{\left|AB\right|.\left|\overrightarrow{n}\right|}$

$=\frac{(-\hat{i}-\hat{j}).(\hat{i}+\hat{j}+\hat{k})}{\left(\sqrt{2}\right).\left(\sqrt{3}\right)}$

$=\frac{-1-1+0}{\sqrt{6}}=-\frac{2}{\sqrt{6}}\times \frac{\sqrt{2}}{\sqrt{2}}=\frac{2\sqrt{2}}{2\sqrt{3}}=\frac{\sqrt{2}}{\sqrt{3}}$

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

Now,$P=\left|AB\right|\sin \theta =\sqrt{2}\times \frac{1}{\sqrt{3}}=\sqrt{\frac{2}{3}}$

$\therefore$Length of projection of line segment on the plane is $\sqrt{\frac{2}{3}}$