MHT CET · Maths · Three Dimensional Geometry
The length (in units) 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
- A \(\frac{2}{\sqrt{3}}\)
- B \(\frac{2}{3}\)
- C \(\frac{\sqrt{2}}{3}\)
- D \(\sqrt{\frac{2}{3}}\)
Answer & Solution
Correct Answer
(D) \(\sqrt{\frac{2}{3}}\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \text { Let } A=(5,-1,4), B=(4,-1,3) \\ & \overline{\mathrm{AB}}=-\hat{\mathrm{i}}-\hat{\mathrm{k}} \Rightarrow|\overline{\mathrm{AB}}|=\sqrt{2}\end{aligned}\)

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

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