Question

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
 $$\begin{array}{l}A.\sqrt{{\displaystyle \frac{2}{3}}}\\ B.{\displaystyle \frac{2}{\sqrt{3}}}\\ C.{\displaystyle \frac{2}{3}}\\ D.{\displaystyle \frac{1}{3}}\end{array}$$

Answer 1

Hint: First of all, we will find the angle between the plane and the line joining the given two points, using the formula given as follows:
 $$\cos \theta ={\displaystyle \frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}\times \sqrt{{a_2}^2+{b_2}^2+{c_2}^2}}}$$
Where, $$\theta$$ is the angle between the plane and line, $$\left(a_1,b_1,c_1\right)$$ are direction ratios of a line and $$\left(a_2,b_2,c_2\right)$$ are direction ratios of plane. A clear figure will be helpful. Then, we can make use of the formula for projection as $$AB\sin \theta .$$

Complete step-by-step answer:
We have been asked, to find the projection of the line segment joining the point (5, -1, 4) and (4, -1, 3) on the plane $$x+y+z=7.$$

We know that, direction ratio of a is equal to the difference of their respective coordinates.
Direction ratio of line joining points A and B is given by,
 $$\begin{array}{l}5-4=1\\ -1-\left(-1\right)=-1+1=0\\ 4-3=1\end{array}$$
 $$\Rightarrow 1,0,1$$ direction ratio of the line.
Now, the equation of the plane is $$x+y+z=7$$ therefore, the direction ratio of the plane is (1, 1, 1)
Using the formula, $$\cos \theta ={\displaystyle \frac{a_1{a}_2+b_1{b}_2+c_1{c}_2}{\sqrt{{a_1}^2+{b_1}^2+{c_1}^2}\times \sqrt{{a_2}^2+{b_2}^2+{c_2}^2}}}$$
Here, we have, $$a_1=1,b_1=0,c_1=1\mathrm{a}\mathrm{n}\mathrm{d}{a}_2=1,b_2=1,c_2=1$$ .Therefore, substituting these in the above formula, we get,
 $$\begin{array}{l}\Rightarrow \cos \theta ={\displaystyle \frac{1\times 1+0\times 1+1\times 1}{\sqrt{1^2+0^2+1^2}\times \sqrt{1^2+1^2+1^2}}}\\ \Rightarrow {\displaystyle \frac{2}{\sqrt{2}\times \sqrt{3}}}\end{array}$$
Multiplying the numerator and denominator with 2, we get,
 $$\begin{array}{l}\Rightarrow 2\times {\displaystyle \frac{\sqrt{2}}{\sqrt{2}}}\times \sqrt{3}\times \sqrt{2}\\ \Rightarrow 2\times {\displaystyle \frac{\sqrt{2}}{2}}\times \sqrt{3}\end{array}$$
Cancelling similar terms, we get,
 $$\begin{array}{l}\Rightarrow {\displaystyle \frac{\sqrt{2}}{\sqrt{3}}}\\ \Rightarrow \sqrt{{\displaystyle \frac{2}{3}}}\end{array}$$
Using the distance formula between the two points i.e. $$d=\sqrt{{\left(x_2-x_1\right)}^2+{\left(y_2-y_1\right)}^2+{\left(z_2-z_1\right)}^2}$$
 $$\begin{array}{l}AB=\sqrt{{\left(5-4\right)}^2+{\left(-1-\left(-1\right)\right)}^2+{\left(4-3\right)}^2}\\ AB=\sqrt{1^2+0^2+1^2}\\ AB=\sqrt{2}\end{array}$$
We can clearly see that, the projection of the line segment AB on the plane is equal to $$AB\sin \theta$$
 $$\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{j}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}=AB\sin \theta$$
Using, $$\sin \theta =\sqrt{1-{\cos }^2\theta }$$ we get,
 $$\begin{array}{l}\mathrm{P}\mathrm{r}\mathrm{o}\mathrm{j}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}=\sqrt{2}\times \sqrt{1-{\left(\sqrt{{\displaystyle \frac{2}{3}}}\right)}^2}\\ \Rightarrow \sqrt{2}\times \sqrt{1-{\displaystyle \frac{2}{3}}}\\ \Rightarrow \sqrt{2}\times \sqrt{{\displaystyle \frac{3-2}{3}}}\\ \Rightarrow \sqrt{2}\times \sqrt{{\displaystyle \frac{1}{3}}}\\ \Rightarrow \sqrt{{\displaystyle \frac{2}{3}}}\end{array}$$
Therefore, the correct option is A.

Note: The diagram is very important in this question, as it will help you to think how to proceed further, to get the required result.
Be careful while finding the projection because sometimes we just miss the square root over 2 and choose the answer as $${\displaystyle \frac{2}{\sqrt{3}}}$$ which is wrong.