MHT CET · Maths · Three Dimensional Geometry

The distance of the point \((3,4,5)\) from the point of intersection of the line
\(\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}\) and plane \(x+y+z=2\) is

A 6 units
B 13 units
C 10 units
D 7 units

Verified Solution

Answer & Solution

Correct Answer

(A) 6 units

Step-by-step Solution

Detailed explanation

Given \(\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}=\lambda \quad \ldots\) (say) ...(1)
Hence coordinates of any point on this line are \(\therefore x=\lambda+3, y=2 \lambda+4, z=2 \lambda+5\)
Since this point lies on the plane, we write
\(\begin{array}{l}
(\lambda+3)+(2 \lambda+4)+(2 \lambda+5)=2 \\
5 \lambda+12=2 \Rightarrow \lambda=-2
\end{array}\)
Hence coordinates of point of inter section are \(\equiv(-2+3,-4+4,-4+5)\) i.e. \((1,0,1)\) \(\therefore\) Distance between \((3,4,5)\) and \((1,0,1)\) is
\(=\sqrt{(3-1)^{2}+(4-0)^{2}+(5-1)^{2}}=6\)

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

Same subject

Explore more questions on app

From MHT CET

Explore more questions on app

The distance of the point \((3,4,5)\) from the point of intersection of the line\(\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}\) and plane \(x+y+z=2\) is

Step-by-step Solution

The distance of the point \((3,4,5)\) from the point of intersection of the line
\(\frac{x-3}{1}=\frac{y-4}{2}=\frac{z-5}{2}\) and plane \(x+y+z=2\) is